problem solving distance formula

The Distance Formula

Distance Formula and Pythagorean Theorem

The distance formula is derived from the Pythagorean theorem. To find the distance between two points ($$x_1, y_1$$) and ($$x_2, y_2$$), all that you need to do is use the coordinates of these ordered pairs and apply the formula pictured below.

The distance formula is $ \text{ Distance } = \sqrt{(x_2 -x_1)^2 + (y_2- y_1)^2} $

Below is a diagram of the distance formula applied to a picture of a line segment

Video Tutorial on the Distance Formula

Practice Problems

What is the distance between the the points $$(0,0)$$ and $$(6,8)$$ plotted on the graph?

Connect the two points and draw a right triangle. Set up the Pythagorean theorem

$ a^2 + b^2 = \red c^2 \\ 5^2 + 24^2 = \red c^2 $

Now, let's find the distance. .

You might be wondering does it matter which $$ \blue x $$ value is $$ \blue{ x_1} $$. For instance, up above we chose $$ \blue {6} $$, from the $$ \boxed {(\blue 6, \red 8) } $$ as $$ \blue {x_1}$$

What if we chose $$ \blue 0 $$ from $$ \boxed { (\blue 0, \red 0) }$$ as $$ \blue {x_1}$$?

$ \\ \text{Distance } = \sqrt{(\blue {x_2} -\blue{x_1})^2 + (\red{ y_2} - \red{ y_1})^2} \\ \sqrt{(\blue 6 - \blue 0 )^2 + (\red 8 -\red{ 0 } )^2} \\ \sqrt{(6 )^2 + ( 8 )^2} \\ \sqrt{36 + 64} = \sqrt{100} \\ \fbox{10} $

As you can see it does not matter which $$\blue x$$ value you use first . This is because after you take difference of the $$ \blue x $$ values, you then square them. And $$ (\red -8)^2 $$ has the same value as $$ (8)^2 $$

What is the distance between the points $$ (2, 4) $$ and $$ (26, 9)$$ ? Round your answer to the nearest tenth

Graph the two points

Plug in the side lengths into the Pythagoren Theorem

Note, it does not matter which 'way' you draw the right triangle . It can be either up above or down below .

$ a^2 + b^2 = c^2 \\ 5^2 + 24^2 = \red c^2 $

Solve for the Hypotenuse

Notice the line colored green that shows the same exact mathematical equation both up above, using the pythagorean theorem, and down below using the formula.

The Distance between the points $$(\blue 2, \red 4) \text{ and } ( \blue{ 26} , \red 9)$$ $ \\ \text{d} = \sqrt{(\blue {x_2} -\blue{x_1})^2 + (\red{ y_2} - \red{ y_1})^2} \\ \text{d} = \sqrt{(\blue 2 -\blue { 26} )^2 + (\red 4 - \red{9} )^2} \\ \text{d} = \sqrt{(\blue {-24 })^2 + (\red{- 5} )^2} \\ \color{green}{ \text{d} = \sqrt{576 + 25 }} \\ \boxed{ \text{d} =\text{distance} = \sqrt{601}= 24.5 } $

What is the distance between the points $$ (4, 6) $$ and $$ (28, 13) $$ ? Round your answer to the nearest tenth

Now, create a Right Triangle

$ a^2 + b^2 = c^2 \\ 24^2 + 7^2 = \red c^2 $

Solve for the Hyptenuse

The Distance between the points $$(\blue 4, \red 6) \text{ and } ( \blue{ 28} , \red {13} )$$ $ \\ \text{d} = \sqrt{(\blue {x_2} -\blue{x_1})^2 + (\red{ y_2} - \red{ y_1})^2} \\ \text{d} =\sqrt{(\blue 4 -\blue { 28} )^2 + (\red 6 - \red{ 13 } )^2} \\ \text{d} = \sqrt{(\blue {-24 })^2 + (\red{-7} )^2} \\ \text{d} = \color{green}{ \sqrt{576 + 49 }} \\ \boxed{ \text{d} = \sqrt{625}= 25 } $

The point $$ (4, 8) $$ lies on a circle centered at $$ (12, 14)$$. What is the radius of this circle ? Round your answer to the nearest tenth

Stuck? Click here for a big hint

The Distance between the points $$(\blue 4, \red 8 ) \text{ and } ( \blue{ 12} , \red {14} )$$ $ \\ \text{d} = \sqrt{(\blue {x_2} -\blue{x_1})^2 + (\red{ y_2} - \red{ y_1})^2} \\ \text{d} = \sqrt{(\blue{12} -\blue { 4} )^2 + (\red 14 - \red{8} )^2} \\ \text{d} = \sqrt{(\blue {8} )^2 + (\red{6} )^2} \\ \color{green}{ \text{d} = \sqrt{64 + 36 }} \\ \text{d} = \sqrt{100 } \\ \boxed{ \text{d} =\text{d} =10 } $

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Distance Formula Questions

Distance formula questions with solutions are provided here for students to practice and understand how to find the distance between the two points in a Cartesian plane .

In coordinate geometry, the distance between two points A(x 1 , y 1 ) and B(x 2 , y 2 ) is given by

\(\begin{array}{l}AB = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\end{array} \)

This formula is used to find the distance between two points in a two-dimensional plane.

The distance between two points A(x 1 , y 2 , z 1 ) and B(x 2 , y 2 , z 2 ) in three dimensional plane is given by –

\(\begin{array}{l}AB = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}\end{array} \)

To see the derivation of the distance formula, click here .

Distance Formula Questions with Solutions

Let us apply the distance formula by solving some questions.

Question 1:

Find the distance between the following points:

(I) (-1, 2) and (2, 3)

(II) (0, 1) and (6, –1)

(III) (1, 0, –1) and (2, 0, 7)

(I) Let the distance between the points (-1, 2) and (2, 3) be d, then

d = √[(2 – ( –1)) 2 + (3 – 2) 2 ] = √[9 + 1] = √10 units.

(II) Let the distance between the points (0, 1) and (6, –1) be d, then

d = √[(6 – 0) 2 + ( –1 – 1) 2 ] = √[36 + 4] = √40 = 2√10 units.

(III) Let the distance between the points (1, 0, –1) and (2, 0, 7) be d, then

d = √[(2 – 1) 2 + (0 – 0) 2 + (7 – ( –1)) 2 ] = √[1 + 0 + 64] = √65 units.

Question 2:

Find a point on the x-axis which is equidistant from the points ( –5, 2) and (9, –2).

Let the points be P( –5, 2) and Q(9, –2) and let R(x , 0) be the point on x-axis which is equidistant from P and Q.

Now, PR = QR ⇒ PR 2 = QR 2

⇒ [(x – ( –5)) 2 + (0 – 2) 2 ] = [(x – 9) 2 + (0 – ( –2)) 2 ]

⇒ (x + 5) 2 + 4 = (x – 9) 2 + 4

⇒ x 2 + 10x + 25 = x 2 – 18x + 81

⇒ 10x + 18x = 81 – 25

⇒ x = 56/28 = 2

Therefore, the point is R(2, 0).

Question 3:

Show that the points P( –1, 6, 6), Q(0, 7, 10) and R( –4, 9, 6) form an isosceles right triangle.

Now, PQ = √[(0 +1) 2 + (7 – 6) 2 + (10 – 6) 2 ]

= √[1 + 1 + 16] = √18 = 3√2 units.

QR = √[( –4 – 0) 2 + (9 – 7) 2 + (6 – 10) 2 ]

= √[16 + 4 + 16] = √36 = 6 units.

PR = √[( –4 – ( –1)) 2 + (9 – 6) 2 + (6 – 6) 2 ]

= √[9 + 9 + 0] = √18 = 3√2 units.

Also, PQ 2 + PR 2 = QR 2 . Hence, PQR is a right isosceles triangle.

Question 4:

Show that the points A(2, 3), B(3, 4) and C(4, 5) are collinear.

Now, AB = √[(3 – 2) 2 + (4 – 3) 2 ]

= √[1 + 1] =√2 units

BC = √[(4 – 3) 2 + (5 – 4) 2 ]

= √[1 + 1] = √2 units

AC = √[(4 –2) 2 + (5 – 3) 2 ]

= √[2 2. + 2 2 ] = √8 = 2√2 units.

Clearly, AB + BC = AC

∴ the points are collinear.

Question 5:

The vertices of a triangle ABC are A(1, 2), B(3, –4), and C(5, –6). Find its circumcentre and circum-radius.

The circumcentre of a triangle is an interior point of a triangle equidistant from the vertices of the triangle, and the circum-radius is that equal distance.

Let O(x, y) be the circumcentre of the triangle ABC. the

AO = BO = CO ⇒ AO 2 = BO 2 = CO 2

AO 2 = BO 2

⇒ (x – 1) 2 + (y – 2) 2 = (x – 3) 2 + (y + 4) 2

⇒ x 2 – 2x + 1 + y 2 – 4y + 4 = x 2 – 6x + 9 + y 2 + 8y + 16

⇒ 4x – 12y = 25 – 5

⇒ x – 3y = 5 ….(i)

Again taking BO 2 = CO 2.

⇒ (x – 3) 2 + (y + 4) 2 = (x – 5) 2 + (y + 6) 2

⇒ –6x + 8y + 10x – 12y = 25 + 36 – 9 – 16

⇒ 4x – 4y = 36

⇒ x – y = 9 ….(ii)

Subtracting (ii) from (i) we get, y = 2 and x = 11

∴ the circumcentre is O(11, 2)

And the circum-radius is AO = √[(11 – 1) 2 + (2 – 2) 2 ] = 10 units.

• Centroid of a Triangle
• Orthocentre of a Triangle
• Section Formula

Question 6:

Find the area of the square whose one pair of opposite vertices are (3, 4) and (5, 6).

Let ABCD be the square and let A(3, 4) and C(5, 6) be the pair of opposite vertices of the square.

Now, AC makes the diagonal of the square and the area of a square is also given by (diagonal) 2 /2/

AC =√[(5 – 3) 2 + (6 – 4) 2 ] = √[2 2 + 2 2 ] = 2√2 units

Area of the square ABCD = AC 2 /2 = 8/2 = 4 square units.

Question 7:

Find the orthocentre of the triangle ABC such that A(2, 2), B(6, 3) and C(4, 11).

Let us find the length of each side of the triangle

AB = √[(6 – 2) 2 + (3 – 2) 2 ] = √[16 + 1] = √17 units.

BC = √[(4 – 6) 2 + (11 – 3) 2 ] = √4 + 64 = √68 units

AC = √[(4 – 2) 2 + (11 – 2) 2 ] = √[4 + 81] = √85 units

We see that AB 2 + BC 2 = AC 2

∴ triangle ABC is a right-angled triangle at B

Thus, the ortho-centre of the triangle ABC is at B(6, 3).

Question 8:

Find the center of the circle passing through points A(0, 5), B(3, 8), and C(6, 5).

Let O(x, y) be the center of the circle.

Then AO 2 = BO 2 = CO 2

Taking AO 2 = BO 2

⇒ (x – 0) 2 + (y – 5) 2 = (x – 3) 2 + (y – 8) 2

⇒ –10y + 6x + 16y = 9 + 64 – 25

⇒ 6x + 6y = 48

⇒ x + y = 8 ….(i)

Again taking BO 2 = CO 2

⇒ (x – 3) 2 + (y – 8) 2 = (x – 6) + (y – 5) 2

⇒ –6x –16y + 12x + 10y = 36 + 25 – 9 – 64

⇒ 6x – 6y = – 12

⇒ x – y = –2 ….(ii)

Adding (i) and (ii), we get

2x = 6 ⇒ x = 3 and y = 5

∴ O(3, 5) is the centre of the circle passing through points A(0, 5), B(3, 8) and C(6, 5).

Question 9:

If the point P(2, 1) lies on the line segment joining the points A(4, 2) and B(8, 4), then prove that AP = ½ AB.

Let us find the length of AP and AB,

AP = √[(2 – 4) 2 + (1 – 2) 2 ] = √[4 + 1] = √5 units.

AB = √[(8 – 4) 2 + (4 – 2) 2 ] = √[16 + 4] = 2√5 units.

Clearly, 2AP = AB

⇒ AP = ½ AB.

Question 10:

Find the coordinates of a point that is equidistant from the three vertices of a triangle (0, 0), (0, 2y), and (2x, 0).

Let the triangle be ABC such that A(0, 0), B(2x, 0), and C(0, 2y), and let P(h, k) be the point equidistant from all the points.

Then, PA 2 = PB 2 = PC 2

Taking, PA 2 = PB 2

⇒ (h – 0) 2 + (k – 0) 2 = (h – 2x) 2 + (k – 0) 2.

⇒ h 2 + k 2 = h 2 + 4x 2 – 4hx + k 2

⇒ 4hx = 4x 2.

Again taking, PB 2 = PC 2

⇒ (h – 2x) 2 + (k – 0) 2 = (h – 0) 2 + (k – 2y) 2

⇒ h 2 + 4x 2 – 4hx + k 2 = h 2 + k 2 – 4ky + 4y 2

⇒ 4x 2 – 4x 2 = –4ky + 4y 2 { ∵ h = x}

∴ the coordinates of the point are P(x, y).

Check your answers with the distance formula calculator .

<h2>Related Video Lesson</h2>

<div class=”youtube-video” data-video-thumbnail=””><img class=”tumbnailYT” src=” https://cdn1.byjus.com/wp-content/uploads/2020/01/section-formula-in-coordinate-geometry.jpg ” data-video=” https://www.youtube.com/embed/KowWdT5IcaQ ” /></div>

Practice Questions on Distance Formula

1. Find the distance between the following points:

(I) (2, 5) and (3, 6)

(II) ( –9, 3) and (3, 2)

(III) (4, 5, –2) and (0, –2, 3)

2. Find a point on the y-axis which is equidistant from the points ( –5, 2) and (9, –2).

3. Find the area of the square whose one pair of opposite vertices are (2, 6) and (6, 2).

4. Find the orthocentre of the triangle PQR such that P( –1, 6, 6), Q(0, 7, 10) and R( –4, 9, 6).

5. Find the coordinates of the fourth vertex of the parallelogram whose rest of the three vertices are ( –2, 3), (6, 7) and (8, 3).

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Distance formula

Here you will learn about the distance formula, including how to find the distance between two coordinates.

Students first learn about the distance formula in 8th grade as a part of geometry, and again in high school geometry as a part of expressing geometric properties with equations.

What is the distance formula?

The distance formula (also known as the Euclidean distance formula ) is an application of the Pythagorean theorem a^2+b^2=c^2 in coordinate geometry.

It will calculate the distance between two cartesian coordinates on a two-dimensional plane , or coordinate plane .

To do this, find the differences between the x- coordinates and the difference between the y- coordinates , square them, then find the square root of the answer.

This can be written as the distance formula,

where d is the distance between the points \left(x_1, y_1\right) and \left(x_2, y_2\right).

For example,

The line segment between the first point and the second point forms the hypotenuse of a right angled triangle.

The length of the hypotenuse of the right triangle is the distance between the two end points of the line segment.

Common Core State Standards

How does this relate to 8 th grade math?

• Grade 8 – Geometry (8.G.B.8) Apply the Pythagorean Theorem to find the distance between two points in a coordinate system.

[FREE] Algebra Worksheet (Grade 6 to 8)

Use this quiz to check your grade 6 – grade 8 students’ understanding of algebra. 10+ questions with answers covering a range of 6th to 8th grade algebra topics to identify areas of strength and support!

How to use the distance formula

In order to use the distance formula, you need to:

Identify the two points and label them \bf{\left(\textbf{x}_{1}, \textbf{y}_{1}\right)} and \bf{\left(\textbf{x}_{2}, \textbf{y}_{2}\right)}.

Substitute the values into the formula, \bf{\textbf{d}=\sqrt{\left(\textbf{x}_2-\textbf{x}_1\right)^2+\left(\textbf{y}_2-\textbf{y}_1\right)^2}}.

Solve the equation.

Distance formula examples

Example 1: distance between two points on a coordinate axes in the first quadrant.

Find the distance between the points A and B.

A=(3,1) and B=(6,5).

Let \left(x_{1}, y_{1}\right)=(3,1) and \left(x_{2}, y_{2}\right)=(6,5).

2 Substitute the values into the formula, \bf{\textbf{d}=\sqrt{\left(\textbf{x}_2-\textbf{x}_1\right)^2+\left(\textbf{y}_2-\textbf{y}_1\right)^2}}.

3 Solve the equation.

Example 2: find the distance between two points on a coordinate axes

Give your answer to 1 decimal place.

A=(1,-5) and B=(6,2).

Let \left(x_{1}, y_{1}\right)=(1,-5) and \left(x_{2}, y_{2}\right)=(6,2).

Example 3: find the distance between two given points with positive coordinates

Find the distance between the points (1,4) and (7,12).

Let \left(x_{1}, y_{1}\right)=(1,4) and \left(x_{2}, y_{2}\right)=(7,12).

Example 4: find the distance between any two given points

Find the distance between the points (-2,5) and (6,-7).

Let \left(x_{1}, y_{1}\right)=(-2,5) and \left(x_{2}, y_{2}\right)=(6,-7).

Example 5: finding a missing value given the distance

The distance between the points (1,5) and (16,k) is 17.

Find the value of k, where k is negative.

Let \left(x_{1}, y_{1}\right)=(1,5) and \left(x_{2}, y_{2}\right)=(16,k).

It is also stated that d=17.

The question states that k is negative so k=-3.

Example 6: finding a missing value given the distance

The distance between the points (2,9) and (f,10) is 15.

Find the value of f, where f is positive.

Let \left(x_{1}, y_{1}\right)=(2,9) and \left(x_{2}, y_{2}\right)=(f, 10).

It is also stated that d=15.

The question states that f is positive so f=17.0 (1dp).

Teaching tips for distance formula

• Use visual aids such as coordinate planes that highlight the x- axis and y- axis, graphs, or geometric shapes to visually represent the distance formula.
• Introduce real-world scenarios where distance calculations are essential. For example, discuss scenarios involving mapping or measuring distances between points in various contexts to allow students to see the relevance of the concept.
• Allow students to explore the distance formula through hands-on activities such as measuring distances on a coordinate plane or calculating distances between objects in the classroom.

Easy mistakes to make

• Confusing the distance formula with the midpoint formula An easy mistake to make is to find the midpoint instead of the distance. The midpoint formula is \left(\cfrac{x_{1}+x_{2}}{2}, \cfrac{y_{1}+y_{2}}{2}\right).
• Squaring negative numbers to give a negative When using the distance formula, it is common to get negative values after the subtraction step. These values will be squared, so it is important to remember that the square of a negative value is positive. For example, (-3)^2=9.
• Subtracting in the wrong order When subtracting, a common misconception is to switch the order of subtraction when plugging in the coordinates ( for example, using \left(x_{2}-x_{1}\right) and \left(y_{1}-y_{2}\right). Ensure that the subtraction is done in the same order for both coordinate values: \left(x_{2}-x_{1}\right)  and \left(y_{2}-y_{1}\right).
• Forgetting to simplify When solving, neglecting to simplify the expression inside the square root is a common mistake. After squaring each term, simplify the expression inside the square root before taking the square root.

Related graphing linear equation lessons

• Graphing linear equations
• Slope intercept form
• How to find midpoint
• How to find the y intercept
• How to find the slope of a line
• Linear interpolation

Practice distance formula questions

1. Find the distance between the point (6,8) and the origin.

The origin is (0,0) so let (x_{1},y_{1})=(0,0) and (x_{2},y_{2})=(6,8).

2. Find the distance between the points (0,10) and (24,0).

Let \left(x_{1},y_{1}\right)=(0,10) and \left(x_{2},y_{2}\right)=(24,0).

3. Find the distance between the points (5,3) and (14,10).

Let \left(x_{1},y_{1}\right)=(5,3) and \left(x_{2},y_{2}\right)=(14,10).

4. Find the distance between the points (-2,4) and (-8,-9).

Let \left(x_{1},y_{1}\right)=(-2,4) and \left(x_{2},y_{2}\right)=(-8,-9).

5. The distance between the points (8,-3) and (15,a) is 25.

Find the value of a, where a is positive.

As a is positive, a=21.

6. The distance between the points (b,4) and (6,-8) is 15.

Find the value of b, where b is negative.

As b is negative, b=-3.

Distance formula FAQs

The distance formula calculates the distance between two points by treating the vertical and horizontal distances as sides of a right triangle, and then finding the length of the line (hypotenuse of a right triangle) using the Pythagorean Theorem.

The theorem is named after the ancient Greek mathematician, Pythagoras, and describes the relationships between the sides of a right triangle. It states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two other sides.

Yes, the distance formula can be extended to three dimensions. To find the distance between the two points \left(x_{1},y_{1},z_{1}\right) and \left(x_{2},y_{2},z_{2}\right) and using the following formula: d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}

The next lessons are

• Angles in parallel lines
• Angles in polygons
• Rate of change
• Systems of equations
• Number patterns

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Distance Problems with Solutions

The concept of the distance between two points is an important one in mathematics. A set of distance problems with detailed solutions (at the bottom of this page) are presented.

11.1 Distance and Midpoint Formulas; Circles

Learning objectives.

By the end of this section, you will be able to:

Use the Distance Formula

Use the Midpoint Formula

• Write the equation of a circle in standard form
• Graph a circle

Be Prepared 11.1

Before you get started, take this readiness quiz.

Find the length of the hypotenuse of a right triangle whose legs are 12 and 16 inches. If you missed this problem, review Example 2.34 .

Be Prepared 11.2

Factor: x 2 − 18 x + 81 . x 2 − 18 x + 81 . If you missed this problem, review Example 6.24 .

Be Prepared 11.3

Solve by completing the square: x 2 − 12 x − 12 = 0 . x 2 − 12 x − 12 = 0 . If you missed this problem, review Example 9.22 .

In this chapter we will be looking at the conic sections, usually called the conics, and their properties. The conics are curves that result from a plane intersecting a double cone—two cones placed point-to-point. Each half of a double cone is called a nappe.

There are four conics—the circle , parabola , ellipse , and hyperbola . The next figure shows how the plane intersecting the double cone results in each curve.

Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.

We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.

Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions . We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.

Example 11.1

Use the rectangular coordinate system to find the distance between the points ( 6 , 4 ) ( 6 , 4 ) and ( 2 , 1 ) . ( 2 , 1 ) .

Try It 11.1

Use the rectangular coordinate system to find the distance between the points ( 6 , 1 ) ( 6 , 1 ) and ( 2 , −2 ) . ( 2 , −2 ) .

Try It 11.2

Use the rectangular coordinate system to find the distance between the points ( 5 , 3 ) ( 5 , 3 ) and ( −3 , −3 ) . ( −3 , −3 ) .

The method we used in the last example leads us to the formula to find the distance between the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( x 2 , y 2 ) .

When we found the length of the horizontal leg we subtracted 6 − 2 6 − 2 which is x 2 − x 1 . x 2 − x 1 .

When we found the length of the vertical leg we subtracted 4 − 1 4 − 1 which is y 2 − y 1 . y 2 − y 1 .

If the triangle had been in a different position, we may have subtracted x 1 − x 2 x 1 − x 2 or y 1 − y 2 . y 1 − y 2 . The expressions x 2 − x 1 x 2 − x 1 and x 1 − x 2 x 1 − x 2 vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say | x 2 − x 1 | | x 2 − x 1 | and | y 2 − y 1 | . | y 2 − y 1 | .

In the Pythagorean Theorem, we substitute the general expressions | x 2 − x 1 | | x 2 − x 1 | and | y 2 − y 1 | | y 2 − y 1 | rather than the numbers.

This is the Distance Formula we use to find the distance d between the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( x 2 , y 2 ) .

Distance Formula

The distance d between the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) ( x 2 , y 2 ) is

Example 11.2

Use the Distance Formula to find the distance between the points ( −5 , −3 ) ( −5 , −3 ) and ( 7 , 2 ) . ( 7 , 2 ) .

Try It 11.3

Use the Distance Formula to find the distance between the points ( −4 , −5 ) ( −4 , −5 ) and ( 5 , 7 ) . ( 5 , 7 ) .

Try It 11.4

Use the Distance Formula to find the distance between the points ( −2 , −5 ) ( −2 , −5 ) and ( −14 , −10 ) . ( −14 , −10 ) .

Example 11.3

Use the Distance Formula to find the distance between the points ( 10 , −4 ) ( 10 , −4 ) and ( −1 , 5 ) . ( −1 , 5 ) . Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Try It 11.5

Use the Distance Formula to find the distance between the points ( −4 , −5 ) ( −4 , −5 ) and ( 3 , 4 ) . ( 3 , 4 ) . Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Try It 11.6

Use the Distance Formula to find the distance between the points ( −2 , −5 ) ( −2 , −5 ) and ( −3 , −4 ) . ( −3 , −4 ) . Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the x -coordinates and the average of the y -coordinates of the endpoints.

Midpoint Formula

The midpoint of the line segment whose endpoints are the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) ( x 2 , y 2 ) is

To find the midpoint of a line segment, we find the average of the x -coordinates and the average of the y -coordinates of the endpoints.

Example 11.4

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are ( −5 , −4 ) ( −5 , −4 ) and ( 7 , 2 ) . ( 7 , 2 ) . Plot the endpoints and the midpoint on a rectangular coordinate system.

Try It 11.7

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are ( −3 , −5 ) ( −3 , −5 ) and ( 5 , 7 ) . ( 5 , 7 ) . Plot the endpoints and the midpoint on a rectangular coordinate system.

Try It 11.8

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are ( −2 , −5 ) ( −2 , −5 ) and ( 6 , −1 ) . ( 6 , −1 ) . Plot the endpoints and the midpoint on a rectangular coordinate system.

Both the Distance Formula and the Midpoint Formula depend on two points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( x 2 , y 2 ) . It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, it may be easier to remember the formulas.

Write the Equation of a Circle in Standard Form

As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.

We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, ( h , k ) , ( h , k ) , and the fixed distance is called the radius , r , of the circle.

A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center , ( h , k ) , ( h , k ) , and the fixed distance is called the radius , r , of the circle.

This is the standard form of the equation of a circle with center, ( h , k ) , ( h , k ) , and radius, r .

Standard Form of the Equation a Circle

The standard form of the equation of a circle with center, ( h , k ) , ( h , k ) , and radius, r , is

Example 11.5

Write the standard form of the equation of the circle with radius 3 and center ( 0 , 0 ) . ( 0 , 0 ) .

Try It 11.9

Write the standard form of the equation of the circle with a radius of 6 and center ( 0 , 0 ) . ( 0 , 0 ) .

Try It 11.10

Write the standard form of the equation of the circle with a radius of 8 and center ( 0 , 0 ) . ( 0 , 0 ) .

In the last example, the center was ( 0 , 0 ) . ( 0 , 0 ) . Notice what happened to the equation. Whenever the center is ( 0 , 0 ) , ( 0 , 0 ) , the standard form becomes x 2 + y 2 = r 2 . x 2 + y 2 = r 2 .

Example 11.6

Write the standard form of the equation of the circle with radius 2 and center ( −1 , 3 ) . ( −1 , 3 ) .

Try It 11.11

Write the standard form of the equation of the circle with a radius of 7 and center ( 2 , −4 ) . ( 2 , −4 ) .

Try It 11.12

Write the standard form of the equation of the circle with a radius of 9 and center ( −3 , −5 ) . ( −3 , −5 ) .

In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.

Example 11.7

Write the standard form of the equation of the circle with center ( 2 , 4 ) ( 2 , 4 ) that also contains the point ( −2 , 1 ) . ( −2 , 1 ) .

The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it. We will use the center ( 2 , 4 ) ( 2 , 4 ) and point ( −2 , 1 ) ( −2 , 1 )

Now that we know the radius, r = 5 , r = 5 , and the center, ( 2 , 4 ) , ( 2 , 4 ) , we can use the standard form of the equation of a circle to find the equation.

Try It 11.13

Write the standard form of the equation of the circle with center ( 2 , 1 ) ( 2 , 1 ) that also contains the point ( −2 , −2 ) . ( −2 , −2 ) .

Try It 11.14

Write the standard form of the equation of the circle with center ( 7 , 1 ) ( 7 , 1 ) that also contains the point ( −1 , −5 ) . ( −1 , −5 ) .

Graph a Circle

Any equation of the form ( x − h ) 2 + ( y − k ) 2 = r 2 ( x − h ) 2 + ( y − k ) 2 = r 2 is the standard form of the equation of a circle with center, ( h , k ) , ( h , k ) , and radius, r. We can then graph the circle on a rectangular coordinate system.

Note that the standard form calls for subtraction from x and y . In the next example, the equation has x + 2 , x + 2 , so we need to rewrite the addition as subtraction of a negative.

Example 11.8

Find the center and radius, then graph the circle: ( x + 2 ) 2 + ( y − 1 ) 2 = 9 . ( x + 2 ) 2 + ( y − 1 ) 2 = 9 .

Try It 11.15

ⓐ Find the center and radius, then ⓑ graph the circle: ( x − 3 ) 2 + ( y + 4 ) 2 = 4 . ( x − 3 ) 2 + ( y + 4 ) 2 = 4 .

Try It 11.16

ⓐ Find the center and radius, then ⓑ graph the circle: ( x − 3 ) 2 + ( y − 1 ) 2 = 16 . ( x − 3 ) 2 + ( y − 1 ) 2 = 16 .

To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of x 2 , y 2 x 2 , y 2 to be one.

Example 11.9

Find the center and radius and then graph the circle, 4 x 2 + 4 y 2 = 64 . 4 x 2 + 4 y 2 = 64 .

Try It 11.17

ⓐ Find the center and radius, then ⓑ graph the circle: 3 x 2 + 3 y 2 = 27 3 x 2 + 3 y 2 = 27

Try It 11.18

ⓐ Find the center and radius, then ⓑ graph the circle: 5 x 2 + 5 y 2 = 125 5 x 2 + 5 y 2 = 125

If we expand the equation from Example 11.8 , ( x + 2 ) 2 + ( y − 1 ) 2 = 9 , ( x + 2 ) 2 + ( y − 1 ) 2 = 9 , the equation of the circle looks very different.

This form of the equation is called the general form of the equation of the circle .

General Form of the Equation of a Circle

The general form of the equation of a circle is

If we are given an equation in general form, we can change it to standard form by completing the squares in both x and y . Then we can graph the circle using its center and radius.

Example 11.10

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 4 x − 6 y + 4 = 0 . x 2 + y 2 − 4 x − 6 y + 4 = 0 .

We need to rewrite this general form into standard form in order to find the center and radius.

Try It 11.19

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 6 x − 8 y + 9 = 0 . x 2 + y 2 − 6 x − 8 y + 9 = 0 .

Try It 11.20

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 + 6 x − 2 y + 1 = 0 . x 2 + y 2 + 6 x − 2 y + 1 = 0 .

In the next example, there is a y -term and a y 2 y 2 -term. But notice that there is no x -term, only an x 2 x 2 -term. We have seen this before and know that it means h is 0. We will need to complete the square for the y terms, but not for the x terms.

Example 11.11

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 + 8 y = 0 . x 2 + y 2 + 8 y = 0 .

Try It 11.21

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 2 x − 3 = 0 . x 2 + y 2 − 2 x − 3 = 0 .

Try It 11.22

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 12 y + 11 = 0 . x 2 + y 2 − 12 y + 11 = 0 .

Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.

• Distance-Midpoint Formulas and Circles
• Finding the Distance and Midpoint Between Two Points
• Completing the Square to Write Equation in Standard Form of a Circle

Section 11.1 Exercises

Practice makes perfect.

In the following exercises, find the distance between the points. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

( 2 , 0 ) ( 2 , 0 ) and ( 5 , 4 ) ( 5 , 4 )

( −4 , −3 ) ( −4 , −3 ) and ( 2 , 5 ) ( 2 , 5 )

( −4 , −3 ) ( −4 , −3 ) and ( 8 , 2 ) ( 8 , 2 )

( −7 , −3 ) ( −7 , −3 ) and ( 8 , 5 ) ( 8 , 5 )

( −1 , 4 ) ( −1 , 4 ) and ( 2 , 0 ) ( 2 , 0 )

( −1 , 3 ) ( −1 , 3 ) and ( 5 , −5 ) ( 5 , −5 )

( 1 , −4 ) ( 1 , −4 ) and ( 6 , 8 ) ( 6 , 8 )

( −8 , −2 ) ( −8 , −2 ) and ( 7 , 6 ) ( 7 , 6 )

( −3 , −5 ) ( −3 , −5 ) and ( 0 , 1 ) ( 0 , 1 )

( −1 , −2 ) ( −1 , −2 ) and ( −3 , 4 ) ( −3 , 4 )

( 3 , −1 ) ( 3 , −1 ) and ( 1 , 7 ) ( 1 , 7 )

( −4 , −5 ) ( −4 , −5 ) and ( 7 , 4 ) ( 7 , 4 )

In the following exercises, ⓐ find the midpoint of the line segments whose endpoints are given and ⓑ plot the endpoints and the midpoint on a rectangular coordinate system.

( 0 , −5 ) ( 0 , −5 ) and ( 4 , −3 ) ( 4 , −3 )

( −2 , −6 ) ( −2 , −6 ) and ( 6 , −2 ) ( 6 , −2 )

( 3 , −1 ) ( 3 , −1 ) and ( 4 , −2 ) ( 4 , −2 )

( −3 , −3 ) ( −3 , −3 ) and ( 6 , −1 ) ( 6 , −1 )

In the following exercises, write the standard form of the equation of the circle with the given radius and center ( 0 , 0 ) . ( 0 , 0 ) .

Radius: 2 2

Radius: 5 5

In the following exercises, write the standard form of the equation of the circle with the given radius and center

Radius: 1, center: ( 3 , 5 ) ( 3 , 5 )

Radius: 10, center: ( −2 , 6 ) ( −2 , 6 )

Radius: 2.5 , 2.5 , center: ( 1.5 , −3.5 ) ( 1.5 , −3.5 )

Radius: 1.5 , 1.5 , center: ( −5.5 , −6.5 ) ( −5.5 , −6.5 )

For the following exercises, write the standard form of the equation of the circle with the given center with point on the circle.

Center ( 3 , −2 ) ( 3 , −2 ) with point ( 3 , 6 ) ( 3 , 6 )

Center ( 6 , −6 ) ( 6 , −6 ) with point ( 2 , −3 ) ( 2 , −3 )

Center ( 4 , 4 ) ( 4 , 4 ) with point ( 2 , 2 ) ( 2 , 2 )

Center ( −5 , 6 ) ( −5 , 6 ) with point ( −2 , 3 ) ( −2 , 3 )

In the following exercises, ⓐ find the center and radius, then ⓑ graph each circle.

( x + 5 ) 2 + ( y + 3 ) 2 = 1 ( x + 5 ) 2 + ( y + 3 ) 2 = 1

( x − 2 ) 2 + ( y − 3 ) 2 = 9 ( x − 2 ) 2 + ( y − 3 ) 2 = 9

( x − 4 ) 2 + ( y + 2 ) 2 = 16 ( x − 4 ) 2 + ( y + 2 ) 2 = 16

( x + 2 ) 2 + ( y − 5 ) 2 = 4 ( x + 2 ) 2 + ( y − 5 ) 2 = 4

x 2 + ( y + 2 ) 2 = 25 x 2 + ( y + 2 ) 2 = 25

( x − 1 ) 2 + y 2 = 36 ( x − 1 ) 2 + y 2 = 36

( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25 ( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25

( x − 1 ) 2 + ( y − 3 ) 2 = 9 4 ( x − 1 ) 2 + ( y − 3 ) 2 = 9 4

x 2 + y 2 = 64 x 2 + y 2 = 64

x 2 + y 2 = 49 x 2 + y 2 = 49

2 x 2 + 2 y 2 = 8 2 x 2 + 2 y 2 = 8

6 x 2 + 6 y 2 = 216 6 x 2 + 6 y 2 = 216

In the following exercises, ⓐ identify the center and radius and ⓑ graph.

x 2 + y 2 + 2 x + 6 y + 9 = 0 x 2 + y 2 + 2 x + 6 y + 9 = 0

x 2 + y 2 − 6 x − 8 y = 0 x 2 + y 2 − 6 x − 8 y = 0

x 2 + y 2 − 4 x + 10 y − 7 = 0 x 2 + y 2 − 4 x + 10 y − 7 = 0

x 2 + y 2 + 12 x − 14 y + 21 = 0 x 2 + y 2 + 12 x − 14 y + 21 = 0

x 2 + y 2 + 6 y + 5 = 0 x 2 + y 2 + 6 y + 5 = 0

x 2 + y 2 − 10 y = 0 x 2 + y 2 − 10 y = 0

x 2 + y 2 + 4 x = 0 x 2 + y 2 + 4 x = 0

x 2 + y 2 − 14 x + 13 = 0 x 2 + y 2 − 14 x + 13 = 0

Writing Exercises

Explain the relationship between the distance formula and the equation of a circle.

Is a circle a function? Explain why or why not.

In your own words, state the definition of a circle.

In your own words, explain the steps you would take to change the general form of the equation of a circle to the standard form.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction

• Authors: Lynn Marecek, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Intermediate Algebra 2e
• Publication date: May 6, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/11-1-distance-and-midpoint-formulas-circles

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Distance Formula

Learn about the distance formula., distance formula lesson, what is the distance formula.

The distance formula is a way of finding the distance between two points. It does this by creating a virtual right triangle and using the Pythagorean theorem. The distance formula has a 2D (two-dimensional) variation and a 3D (three-dimensional) variation.

The 2D distance formula is given as:

$$\begin{align}& d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \end{align}$$

The 3D distance formula is given as:

$$\begin{align}& d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}} \end{align}$$

Where d is the distance between the points, ( x 1 , y 1 , z 1 ) is point 1, and ( x 2 , y 2 , z 2 ) is point 2.

How the Distance Formula Works

The 2D distance formula is the Pythagorean formula applied to two points in the x-y coordinate plane. The first component inside of the square root is (x 2 - x 1 ) 2 . This is the horizontal leg of the right triangle. The second component inside the square root is (y 2 - y 1 ) 2 . This is the vertical leg of the right triangle.

We can visualize the 3D distance formula as a right triangle that happens to reside in the x-y-z 3D coordinate system. Because the two points we are measuring between do not sit flat on a 2D plane, we add in the third term with the variable z . The third term inside of the square root is (z 2 - z 1 ) 2 . It allows the distance between the points to be accurately calculated when they are in 3D space.

2D Distance Formula Example Problem

Find the distance between the points (2, 5) and (7, 3).

$$\begin{align}& \text{1.) The points lie in a 2D system/plane. So, we will use the 2D formula.} \\ \\ & \text{2.) Let's substitute the points into the equation and then simplify.} \\ \\ & \text{3.) } d = \sqrt{(7 - 2)^{2} + (3 - 5)^{2}} \\ \\ & \text{4.) } d = \sqrt{(5)^{2} + (-2)^{2}} \\ \\ & \text{5.) } d = \sqrt{25 + 4} \\ \\ & \text{6.) The distance between the points is: } \; \boxed{d = \sqrt{29}} \end{align}$$

3D Distance Formula Example Problem

Find the distance between the points (1, 4, 11) and (2, 6, 18).

$$\begin{align}& \text{1.) The points are in 3D space, so we will use the 3D distance formula..} \\ \\ & \text{2.) Let's substitute the points into the equation and then simplify.} \\ \\ & \text{3.) } d = \sqrt{(2 - 1)^{2} + (6 - 4)^{2} + (18 - 11)^{2}} \\ \\ & \text{4.) } d = \sqrt{(1)^{2} + (2)^{2} + (7)^{2}} \\ \\ & \text{5.) } d = \sqrt{1 + 4 + 49} \\ \\ & \text{6.) The distance between the points is: } \; \boxed{d = \sqrt{54}} \end{align}$$

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11.2: Distance and Midpoint Formulas and Circles

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Learning Objectives

By the end of this section, you will be able to:

Use the Distance Formula

Use the midpoint formula.

• Write the equation of a circle in standard form
• Graph a circle

Before you get started, take this readiness quiz.

• Find the length of the hypotenuse of a right triangle whose legs are \(12\) and \(16\) inches. If you missed this problem, review Example 2.34.
• Factor: \(x^{2}-18 x+81\). If you missed this problem, review Example 6.24.
• Solve by completing the square: \(x^{2}-12 x-12=0\). If you missed this problem, review Example 9.22.

In this chapter we will be looking at the conic sections, usually called the conics, and their properties. The conics are curves that result from a plane intersecting a double cone—two cones placed point-to-point. Each half of a double cone is called a nappe.

There are four conics—the circle , parabola , ellipse , and hyperbola . The next figure shows how the plane intersecting the double cone results in each curve.

Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.

We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.

Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions. We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.

Example \(\PageIndex{1}\)

Use the rectangular coordinate system to find the distance between the points \((6,4)\) and \((2,1)\).

Exercise \(\PageIndex{1}\)

Use the rectangular coordinate system to find the distance between the points \((6,1)\) and \((2,-2)\).

Exercise \(\PageIndex{2}\)

Use the rectangular coordinate system to find the distance between the points \((5,3)\) and \((-3,-3)\).

The method we used in the last example leads us to the formula to find the distance between the two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\).

When we found the length of the horizontal leg we subtracted \(6−2\) which is \(x_{2}-x_{1}\).

When we found the length of the vertical leg we subtracted \(4−1\) which is \(y_{2}-y_{1}\).

If the triangle had been in a different position, we may have subtracted \(x_{1}-x_{2}\) or \(y_{1}-y_{2}\). The expressions \(x_{2}-x_{1}\) and \(x_{1}-x_{2}\) vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say \(\left|x_{2}-x_{1}\right|\) and \(\left|y_{2}-y_{1}\right|\).

In the Pythagorean Theorem, we substitute the general expressions \(\left|x_{2}-x_{1}\right|\) and \(\left|y_{2}-y_{1}\right|\) rather than the numbers.

\(\begin{array}{l c}{} & {a^{2}+b^{2}=c^{2}} \\ {\text {Substitute in the values. }}&{(|x_{2}-x_{1}|)^{2}+(|y_{2}-y_{1}|)^{2}=d^{2}} \\ {\text{Squaring the expressions makes}}&{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}=d^{2}} \\ \text{them positive, so we eliminate} \\\text{the absolute value bars.}\\ {\text{Use the Square Root Property.}}&{d=\pm\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}\\ {\text{Distance is positive, so eliminate}}&{d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}\\\text{the negative value.}\end{array}\)

This is the Distance Formula we use to find the distance \(d\) between the two points \((x_{1},y_{1})\) and \((x_{2}, y_{2})\).

Definition \(\PageIndex{1}\)

Distance Formula

The distance \(d\) between the two points \((x_{1},y_{1})\) and \((x_{2}, y_{2})\) is

\(d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

Example \(\PageIndex{2}\)

Use the Distance Formula to find the distance between the points \((-5,-3)\) and \((7,2)\).

Write the Distance Formula.

Label the points, \(\left( \begin{array}{c}{x_{1}, y_{1}} \\ {-5,-3}\end{array}\right)\), \(\left( \begin{array}{l}{x_{2}, y_{2}} \\ {7,2}\end{array}\right)\) and substitute.

\(d=\sqrt{(7-(-5))^{2}+(2-(-3))^{2}}\)

\(d=\sqrt{12^{2}+5^{2}}\) \(d=\sqrt{144+25}\) \(d=\sqrt{169}\) \(d=13\)

Exercise \(\PageIndex{3}\)

Use the Distance Formula to find the distance between the points \((-4,-5)\) and \((5,7)\).

Exercise \(\PageIndex{4}\)

Use the Distance Formula to find the distance between the points \((-2,-5)\) and \((-14,-10)\).

Example \(\PageIndex{3}\)

Use the Distance Formula to find the distance between the points \((10,−4)\) and \((−1,5)\). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Label the points, \(\left( \begin{array}{c}{x_{1}, y_{1}} \\ {10,-4}\end{array}\right)\), \(\left( \begin{array}{c}{x_{2}, y_{2}} \\ {-1,5}\end{array}\right)\) and substitute.

\(d=\sqrt{(-1-10)^{2}+(5-(-4))^{2}}\)

\(d=\sqrt{(-11)^{2}+9^{2}}\) \(d=\sqrt{121+81}\) \(d=\sqrt{202}\)

Since \(202\) is not a perfect square, we can leave the answer in exact form or find a decimal approximation.

\(d=\sqrt{202}\) or \(d \approx 14.2\)

Exercise \(\PageIndex{5}\)

Use the Distance Formula to find the distance between the points \((−4,−5)\) and \((3,4)\). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

\(d=\sqrt{130}, d \approx 11.4\)

Exercise \(\PageIndex{6}\)

Use the Distance Formula to find the distance between the points \((−2,−5)\) and \((−3,−4)\). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

\(d=\sqrt{2}, d \approx 1.4\)

It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the \(x\)-coordinates and the average of the \(y\)-coordinates of the endpoints.

Definition \(\PageIndex{2}\)

Midpoint Formula

The midpoint of the line segment whose endpoints are the two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is

\(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

To find the midpoint of a line segment, we find the average of the \(x\)-coordinates and the average of the \(y\)-coordinates of the endpoints.

Example \(\PageIndex{4}\)

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are \((−5,−4)\) and \((7,2)\). Plot the endpoints and the midpoint on a rectangular coordinate system.

Exercise \(\PageIndex{7}\)

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are \((−3,−5)\) and \((5,7)\). Plot the endpoints and the midpoint on a rectangular coordinate system.

Exercise \(\PageIndex{8}\)

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are \((−2,−5)\) and \((6,−1)\). Plot the endpoints and the midpoint on a rectangular coordinate system.

Both the Distance Formula and the Midpoint Formula depend on two points, \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\). It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, is may be easier to remember the formulas.

Write the Equation of a Circle in Standard Form

As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.

We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, \((h,k)\), and the fixed distance is called the radius , \(r\), of the circle.

Definition \(\PageIndex{3}\)

A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center , \((h,k)\), and the fixed distance is called the radius , \(r\), of the circle.

This is the standard form of the equation of a circle with center, \((h,k)\), and radius, \(r\).

Definition \(\PageIndex{4}\)

The standard form of the equation of a circle with center, \((h,k)\), and radius, \(r\), is

Example \(\PageIndex{5}\)

Write the standard form of the equation of the circle with radius \(3\) and center \((0,0)\).

Exercise \(\PageIndex{9}\)

Write the standard form of the equation of the circle with a radius of \(6\) and center \((0,0)\).

\(x^{2}+y^{2}=36\)

Exercise \(\PageIndex{10}\)

Write the standard form of the equation of the circle with a radius of \(8\) and center \((0,0)\).

\(x^{2}+y^{2}=64\)

In the last example, the center was \((0,0)\). Notice what happened to the equation. Whenever the center is \((0,0)\), the standard form becomes \(x^{2}+y^{2}=r^{2}\).

Example \(\PageIndex{6}\)

Write the standard form of the equation of the circle with radius \(2\) and center \((−1,3)\).

Exercise \(\PageIndex{11}\)

Write the standard form of the equation of the circle with a radius of \(7\) and center \((2,−4)\).

\((x-2)^{2}+(y+4)^{2}=49\)

Exercise \(\PageIndex{12}\)

Write the standard form of the equation of the circle with a radius of \(9\) and center \((−3,−5)\).

\((x+3)^{2}+(y+5)^{2}=81\)

In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.

Example \(\PageIndex{7}\)

Write the standard form of the equation of the circle with center \((2,4)\) that also contains the point \((−2,1)\).

The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it. We will use the center \((2,4)\) and point \((−2,1)\)

Use the Distance Formula to find the radius.

\(r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

Substitute the values. \(\left( \begin{array}{l}{x_{1}, y_{1}} \\ {2,4}\end{array}\right), \left( \begin{array}{c}{x_{2}, y_{2}} \\ {-2,1}\end{array}\right)\)

\(r=\sqrt{(-2-2)^{2}+(1-4)^{2}}\)

\(r=\sqrt{(-4)^{2}+(-3)^{2}}\) \(r=\sqrt{16+9}\) \(r=\sqrt{25}\) \(r=5\)

Now that we know the radius, \(r=5\), and the center, \((2,4)\), we can use the standard form of the equation of a circle to find the equation.

Use the standard form of the equation of a circle.

\((x-h)^{2}+(y-k)^{2}=r^{2}\)

Substitute in the values.

\((x-2)^{2}+(y-4)^{2}=5^{2}\)

\((x-2)^{2}+(y-4)^{2}=25\)

Exercise \(\PageIndex{13}\)

Write the standard form of the equation of the circle with center \((2,1)\) that also contains the point \((−2,−2)\).

\((x-2)^{2}+(y-1)^{2}=25\)

Exercise \(\PageIndex{14}\)

Write the standard form of the equation of the circle with center \((7,1)\) that also contains the point \((−1,−5)\).

\((x-7)^{2}+(y-1)^{2}=100\)

Graph a Circle

Any equation of the form \((x-h)^{2}+(y-k)^{2}=r^{2}\) is the standard form of the equation of a circle with center, \((h,k)\), and radius, \(r\) . We can then graph the circle on a rectangular coordinate system.

Note that the standard form calls for subtraction from \(x\) and \(y\). In the next example, the equation has \(x+2\), so we need to rewrite the addition as subtraction of a negative.

Example \(\PageIndex{8}\)

Find the center and radius, then graph the circle: \((x+2)^{2}+(y-1)^{2}=9\).

Exercise \(\PageIndex{15}\)

• Find the center and radius, then
• Graph the circle: \((x-3)^{2}+(y+4)^{2}=4\).
• The circle is centered at \((3,-4)\) with a radius of \(2\).

Exercise \(\PageIndex{16}\)

• Graph the circle: \((x-3)^{2}+(y-1)^{2}=16\).
• The circle is centered at \((3,1)\) with a radius of \(4\).

To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of \(x^{2}, y^{2}\) to be one.

Example \(\PageIndex{9}\)

Find the center and radius and then graph the circle, \(4 x^{2}+4 y^{2}=64\).

Exercise \(\PageIndex{17}\)

• Graph the circle: \(3 x^{2}+3 y^{2}=27\)
• The circle is centered at \((0,0)\) with a radius of \(3\).

Exercise \(\PageIndex{18}\)

• Graph the circle: \(5 x^{2}+5 y^{2}=125\)
• The circle is centered at \((0,0)\) with a radius of \(5\).

If we expand the equation from Example 11.1.8, \((x+2)^{2}+(y-1)^{2}=9\), the equation of the circle looks very different.

\((x+2)^{2}+(y-1)^{2}=9\)

Square the binomials.

\(x^{2}+4 x+4+y^{2}-2 y+1=9\)

Arrange the terms in descending degree order, and get zero on the right

\(x^{2}+y^{2}+4 x-2 y-4=0\)

This form of the equation is called the general form of the equation of the circle .

Definition \(\PageIndex{5}\)

The general form of the equation of a circle is

\(x^{2}+y^{2}+a x+b y+c=0\)

If we are given an equation in general form, we can change it to standard form by completing the squares in both \(x\) and \(y\). Then we can graph the circle using its center and radius.

Example \(\PageIndex{10}\)

• Graph the circle: \(x^{2}+y^{2}-4 x-6 y+4=0\)

We need to rewrite this general form into standard form in order to find the center and radius.

Exercise \(\PageIndex{19}\)

• Graph the circle: \(x^{2}+y^{2}-6 x-8 y+9=0\).
• The circle is centered at \((3,4)\) with a radius of \(4\).

Exercise \(\PageIndex{20}\)

• Graph the circle: \(x^{2}+y^{2}+6 x-2 y+1=0\)
• The circle is centered at \((-3,1)\) with a radius of \(3\).

In the next example, there is a \(y\)-term and a \(y^{2}\)-term. But notice that there is no \(x\)-term, only an \(x^{2}\)-term. We have seen this before and know that it means \(h\) is \(0\). We will need to complete the square for the \(y\) terms, but not for the \(x\) terms.

Example \(\PageIndex{11}\)

• Graph the circle: \(x^{2}+y^{2}+8 y=0\)

Exercise \(\PageIndex{21}\)

• Graph the circle: \(x^{2}+y^{2}-2 x-3=0\).
• The circle is centered at \((-1,0)\) with a radius of \(2\).

Exercise \(\PageIndex{22}\)

• Graph the circle: \(x^{2}+y^{2}-12 y+11=0\).
• The circle is centered at \((0,6)\) with a radius of \(5\).

Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.

• Distance-Midpoint Formulas and Circles
• Finding the Distance and Midpoint Between Two Points
• Completing the Square to Write Equation in Standard Form of a Circle

Key Concepts

• Circle: A circle is all points in a plane that are a fixed distance from a fixed point in the plane. The given point is called the center, \((h,k)\), and the fixed distance is called the radius, \(r\), of the circle.
• Standard Form of the Equation a Circle: The standard form of the equation of a circle with center, \((h,k)\), and radius, \(r\) , is

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Algebra Topics  - Distance Word Problems

Algebra topics  -, distance word problems, algebra topics distance word problems.

Algebra Topics: Distance Word Problems

Lesson 10: distance word problems.

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What are distance word problems?

Distance word problems are a common type of algebra word problems. They involve a scenario in which you need to figure out how fast , how far , or how long one or more objects have traveled. These are often called train problems because one of the most famous types of distance problems involves finding out when two trains heading toward each other cross paths.

In this lesson, you'll learn how to solve train problems and a few other common types of distance problems. But first, let's look at some basic principles that apply to any distance problem.

The basics of distance problems

There are three basic aspects to movement and travel: distance , rate , and time . To understand the difference among these, think about the last time you drove somewhere.

The distance is how far you traveled. The rate is how fast you traveled. The time is how long the trip took.

The relationship among these things can be described by this formula:

distance = rate x time d = rt

In other words, the distance you drove is equal to the rate at which you drove times the amount of time you drove. For an example of how this would work in real life, just imagine your last trip was like this:

• You drove 25 miles—that's the distance .
• You drove an average of 50 mph—that's the rate .
• The drive took you 30 minutes, or 0 .5 hours—that's the time .

According to the formula, if we multiply the rate and time , the product should be our distance.

And it is! We drove 50 mph for 0.5 hours—and 50 ⋅ 0.5 equals 25 , which is our distance.

What if we drove 60 mph instead of 50? How far could we drive in 30 minutes? We could use the same formula to figure this out.

60 ⋅ 0.5 is 30, so our distance would be 30 miles.

Solving distance problems

When you solve any distance problem, you'll have to do what we just did—use the formula to find distance , rate , or time . Let's try another simple problem.

On his day off, Lee took a trip to the zoo. He drove an average speed of 65 mph, and it took him two-and-a-half hours to get from his house to the zoo. How far is the zoo from his house?

First, we should identify the information we know. Remember, we're looking for any information about distance, rate, or time. According to the problem:

• The rate is 65 mph.
• The time is two-and-a-half hours, or 2.5 hours.
• The distance is unknown—it's what we're trying to find.

You could picture Lee's trip with a diagram like this:

This diagram is a start to understanding this problem, but we still have to figure out what to do with the numbers for distance , rate , and time . To keep track of the information in the problem, we'll set up a table. (This might seem excessive now, but it's a good habit for even simple problems and can make solving complicated problems much easier.) Here's what our table looks like:

We can put this information into our formula: distance = rate ⋅ time .

We can use the distance = rate ⋅ time formula to find the distance Lee traveled.

The formula d = rt looks like this when we plug in the numbers from the problem. The unknown distance is represented with the variable d .

d = 65 ⋅ 2.5

To find d , all we have to do is multiply 65 and 2.5. 65 ⋅ 2.5 equals 162.5 .

We have an answer to our problem: d = 162.5. In other words, the distance Lee drove from his house to the zoo is 162.5 miles.

Be careful to use the same units of measurement for rate and time. It's possible to multiply 65 miles per hour by 2.5 hours because they use the same unit: an hour . However, what if the time had been written in a different unit, like in minutes ? In that case, you'd have to convert the time into hours so it would use the same unit as the rate.

Solving for rate and time

In the problem we just solved we calculated for distance , but you can use the d = rt formula to solve for rate and time too. For example, take a look at this problem:

After work, Janae walked in her neighborhood for a half hour. She walked a mile-and-a-half total. What was her average speed in miles per hour?

We can picture Janae's walk as something like this:

And we can set up the information from the problem we know like this:

The table is repeating the facts we already know from the problem. Janae walked one-and-a-half miles or 1.5 miles in a half hour, or 0.5 hours.

As always, we start with our formula. Next, we'll fill in the formula with the information from our table.

The rate is represented by r because we don't yet know how fast Janae was walking. Since we're solving for r , we'll have to get it alone on one side of the equation.

1.5 = r ⋅ 0.5

Our equation calls for r to be multiplied by 0.5, so we can get r alone on one side of the equation by dividing both sides by 0.5: 1.5 / 0.5 = 3 .

r = 3 , so 3 is the answer to our problem. Janae walked 3 miles per hour.

In the problems on this page, we solved for distance and rate of travel, but you can also use the travel equation to solve for time . You can even use it to solve certain problems where you're trying to figure out the distance, rate, or time of two or more moving objects. We'll look at problems like this on the next few pages.

Two-part and round-trip problems

Do you know how to solve this problem?

Bill took a trip to see a friend. His friend lives 225 miles away. He drove in town at an average of 30 mph, then he drove on the interstate at an average of 70 mph. The trip took three-and-a-half hours total. How far did Bill drive on the interstate?

This problem is a classic two-part trip problem because it's asking you to find information about one part of a two-part trip. This problem might seem complicated, but don't be intimidated!

You can solve it using the same tools we used to solve the simpler problems on the first page:

• The travel equation d = rt
• A table to keep track of important information

Let's start with the table . Take another look at the problem. This time, the information relating to distance , rate , and time has been underlined.

Bill took a trip to see a friend. His friend lives 225 miles away. He drove in town at an average of 30 mph , then he drove on the interstate at an average of 70 mph . The trip took three-and-a-half hours total. How far did Bill drive on the interstate?

If you tried to fill in the table the way we did on the last page, you might have noticed a problem: There's too much information. For instance, the problem contains two rates— 30 mph and 70 mph . To include all of this information, let's create a table with an extra row. The top row of numbers and variables will be labeled in town , and the bottom row will be labeled interstate .

We filled in the rates, but what about the distance and time ? If you look back at the problem, you'll see that these are the total figures, meaning they include both the time in town and on the interstate. So the total distance is 225 . This means this is true:

Interstate distance + in-town distance = Total distance

Together, the interstate distance and in-town distance are equal to the total distance. See?

In any case, we're trying to find out how far Bill drove on the interstate , so let's represent this number with d . If the interstate distance is d , it means the in-town distance is a number that equals the total, 225 , when added to d . In other words, it's equal to 225 - d .

We can fill in our chart like this:

We can use the same technique to fill in the time column. The total time is 3.5 hours . If we say the time on the interstate is t , then the remaining time in town is equal to 3.5 - t . We can fill in the rest of our chart.

Now we can work on solving the problem. The main difference between the problems on the first page and this problem is that this problem involves two equations. Here's the one for in-town travel :

225 - d = 30 ⋅ (3.5 - t)

And here's the one for interstate travel :

If you tried to solve either of these on its own, you might have found it impossible: since each equation contains two unknown variables, they can't be solved on their own. Try for yourself. If you work either equation on its own, you won't be able to find a numerical value for d . In order to find the value of d , we'll also have to know the value of t .

We can find the value of t in both problems by combining them. Let's take another look at our travel equation for interstate travel.

While we don't know the numerical value of d , this equation does tell us that d is equal to 70 t .

Since 70 t and d are equal , we can replace d with 70 t . Substituting 70 t for d in our equation for interstate travel won't help us find the value of t —all it tells us is that 70 t is equal to itself, which we already knew.

But what about our other equation, the one for in-town travel?

When we replace the d in that equation with 70 t , the equation suddenly gets much easier to solve.

225 - 70t = 30 ⋅ (3.5 - t)

Our new equation might look more complicated, but it's actually something we can solve. This is because it only has one variable: t . Once we find t , we can use it to calculate the value of d —and find the answer to our problem.

To simplify this equation and find the value of t , we'll have to get the t alone on one side of the equals sign. We'll also have to simplify the right side as much as possible.

Let's start with the right side: 30 times (3.5 - t ) is 105 - 30 t .

225 - 70t = 105 - 30t

Next, let's cancel out the 225 next to 70 t . To do this, we'll subtract 225 from both sides. On the right side, it means subtracting 225 from 105 . 105 - 225 is -120 .

- 70t = -120 - 30t

Our next step is to group like terms—remember, our eventual goal is to have t on the left side of the equals sign and a number on the right . We'll cancel out the -30 t on the right side by adding 30 t to both sides. On the right side, we'll add it to -70 t . -70 t + 30 t is -40 t .

- 40t = -120

Finally, to get t on its own, we'll divide each side by its coefficient: -40. -120 / - 40 is 3 .

So t is equal to 3 . In other words, the time Bill traveled on the interstate is equal to 3 hours . Remember, we're ultimately trying to find the distanc e Bill traveled on the interstate. Let's look at the interstate row of our chart again and see if we have enough information to find out.

It looks like we do. Now that we're only missing one variable, we should be able to find its value pretty quickly.

To find the distance, we'll use the travel formula distance = rate ⋅ time .

We now know that Bill traveled on the interstate for 3 hours at 70 mph , so we can fill in this information.

d = 3 ⋅ 70

Finally, we finished simplifying the right side of the equation. 3 ⋅ 70 is 210 .

So d = 210 . We have the answer to our problem! The distance is 210 . In other words, Bill drove 210 miles on the interstate.

Solving a round-trip problem

It might have seemed like it took a long time to solve the first problem. The more practice you get with these problems, the quicker they'll go. Let's try a similar problem. This one is called a round-trip problem because it describes a round trip—a trip that includes a return journey. Even though the trip described in this problem is slightly different from the one in our first problem, you should be able to solve it the same way. Let's take a look:

Eva drove to work at an average speed of 36 mph. On the way home, she hit traffic and only drove an average of 27 mph. Her total time in the car was 1 hour and 45 minutes, or 1.75 hours. How far does Eva live from work?

If you're having trouble understanding this problem, you might want to visualize Eva's commute like this:

As always, let's start by filling in a table with the important information. We'll make a row with information about her trip to work and from work .

1.75 - t to describe the trip from work. (Remember, the total travel time is 1.75 hours , so the time to work and from work should equal 1.75 .)

From our table, we can write two equations:

• The trip to work can be represented as d = 36 t .
• The trip from work can be represented as d = 27 (1.75 - t ) .

In both equations, d represents the total distance. From the diagram, you can see that these two equations are equal to each other—after all, Eva drives the same distance to and from work .

Just like with the last problem we solved, we can solve this one by combining the two equations.

We'll start with our equation for the trip from work .

d = 27 (1.75 - t)

Next, we'll substitute in the value of d from our to work equation, d = 36 t . Since the value of d is 36 t , we can replace any occurrence of d with 36 t .

36t = 27 (1.75 - t)

Now, let's simplify the right side. 27 ⋅(1.75 - t ) is 47.25 .

36t = 47.25 - 27t

Next, we'll cancel out -27 t by adding 27 t to both sides of the equation. 36 t + 27 t is 63 t .

63t = 47.25

Finally, we can get t on its own by dividing both sides by its coefficient: 63 . 47.25 / 63 is .75 .

t is equal to .75 . In other words, the time it took Eva to drive to work is .75 hours . Now that we know the value of t , we'll be able to can find the distance to Eva's work.

If you guessed that we were going to use the travel equation again, you were right. We now know the value of two out of the three variables, which means we know enough to solve our problem.

First, let's fill in the values we know. We'll work with the numbers for the trip to work . We already knew the rate : 36 . And we just learned the time : .75 .

d = 36 ⋅ .75

Now all we have to do is simplify the equation: 36 ⋅ .75 = 27 .

d is equal to 27 . In other words, the distance to Eva's work is 27 miles . Our problem is solved.

Intersecting distance problems

An intersecting distance problem is one where two things are moving toward each other. Here's a typical problem:

Pawnee and Springfield are 420 miles apart. A train leaves Pawnee heading to Springfield at the same time a train leaves Springfield heading to Pawnee. One train is moving at a speed of 45 mph, and the other is moving 60 mph. How long will they travel before they meet?

This problem is asking you to calculate how long it will take these two trains moving toward each other to cross paths. This might seem confusing at first. Even though it's a real-world situation, it can be difficult to imagine distance and motion abstractly. This diagram might help you get a sense of what this situation looks like:

If you're still confused, don't worry! You can solve this problem the same way you solved the two-part problems on the last page. You'll just need a chart and the travel formula .

Pawnee and Springfield are 420 miles apart. A train leaves Pawnee heading toward Springfield at the same time a train leaves Springfield heading toward Pawnee. One train is moving at a speed of 45 mph , and the other is moving 60 mph . How long will they travel before they meet?

Let's start by filling in our chart. Here's the problem again, this time with the important information underlined. We can start by filling in the most obvious information: rate . The problem gives us the speed of each train. We'll label them fast train and slow train . The fast train goes 60 mph . The slow train goes only 45 mph .

We can also put this information into a table:

We don't know the distance each train travels to meet the other yet—we just know the total distance. In order to meet, the trains will cover a combined distance equal to the total distance. As you can see in this diagram, this is true no matter how far each train travels.

This means that—just like last time—we'll represent the distance of one with d and the distance of the other with the total minus d. So the distance for the fast train will be d , and the distance for the slow train will be 420 - d .

Because we're looking for the time both trains travel before they meet, the time will be the same for both trains. We can represent it with t .

The table gives us two equations: d = 60 t and 420 - d = 45 t . Just like we did with the two-part problems, we can combine these two equations.

The equation for the fast train isn't solvable on its own, but it does tell us that d is equal to 60 t .

The other equation, which describes the slow train, can't be solved alone either. However, we can replace the d with its value from the first equation.

420 - d = 45t

Because we know that d is equal to 60 t , we can replace the d in this equation with 60 t . Now we have an equation we can solve.

420 - 60t = 45t

To solve this equation, we'll need to get t and its coefficients on one side of the equals sign and any other numbers on the other. We can start by canceling out the -60 t on the left by adding 60 t to both sides. 45 t + 60 t is 105 t .

Now we just need to get rid of the coefficient next to t . We can do this by dividing both sides by 105 . 420 / 105 is 4 .

t = 4 . In other words, the time it takes the trains to meet is 4 hours . Our problem is solved!

If you want to be sure of your answer, you can check it by using the distance equation with t equal to 4 . For our fast train, the equation would be d = 60 ⋅ 4 . 60 ⋅ 4 is 240 , so the distance our fast train traveled would be 240 miles. For our slow train, the equation would be d = 45 ⋅ 4 . 45 ⋅ 4 is 180 , so the distance traveled by the slow train is 180 miles . Remember how we said the distance the slow train and fast train travel should equal the total distance? 240 miles + 180 miles equals 420 miles , which is the total distance from our problem. Our answer is correct.

Practice problem 1

Here's another intersecting distance problem. It's similar to the one we just solved. See if you can solve it on your own. When you're finished, scroll down to see the answer and an explanation.

Jon and Dani live 270 miles apart. One day, they decided to drive toward each other and hang out wherever they met. Jon drove an average of 65 mph, and Dani drove an average of 70 mph. How long did they drive before they met up?

Problem 1 answer

Here's practice problem 1:

Jon and Dani live 270 miles apart. One day, they decided to drive toward each other and hang out wherever they met. Jon drove an average of 65 mph, and Dani drove 70 mph. How long did they drive before they met up?

Answer: 2 hours .

Let's solve this problem like we solved the others. First, try making the chart. It should look like this:

Here's how we filled in the chart:

• Distance: Together, Dani and Jon will have covered the total distance between them by the time they meet up. That's 270 . Jon's distance is represented by d , so Dani's distance is 270 - d .
• Rate: The problem tells us Dani and Jon's speeds. Dani drives 65 mph , and Jon drives 70 mph .
• Time: Because Jon and Dani drive the same amount of time before they meet up, both of their travel times are represented by t .

Now we have two equations. The equation for Jon's travel is d = 65 t . The equation for Dani's travel is 270 - d = 70 t . To solve this problem, we'll need to combine them.

The equation for Jon tells us that d is equal to 65 t . This means we can combine the two equations by replacing the d in Dani's equation with 65 t .

270 - 65t = 70t

Let's get t on one side of the equation and a number on the other. The first step to doing this is to get rid of -65 t on the left side. We'll cancel it out by adding 65 t to both sides: 70 t + 65 t is 135 t .

All that's left to do is to get rid of the 135 next to the t . We can do this by dividing both sides by 135 : 270 / 135 is 2 .

That's it. t is equal to 2 . We have the answer to our problem: Dani and Jon drove 2 hours before they met up.

Overtaking distance problems

The final type of distance problem we'll discuss in this lesson is a problem in which one moving object overtakes —or passes —another. Here's a typical overtaking problem:

The Hill family and the Platter family are going on a road trip. The Hills left 3 hours before the Platters, but the Platters drive an average of 15 mph faster. If it takes the Platter family 13 hours to catch up with the Hill family, how fast are the Hills driving?

You can picture the moment the Platter family left for the road trip a little like this:

The problem tells us that the Platter family will catch up with the Hill family in 13 hours and asks us to use this information to find the Hill family's rate . Like some of the other problems we've solved in this lesson, it might not seem like we have enough information to solve this problem—but we do. Let's start making our chart. The distance can be d for both the Hills and the Platters—when the Platters catch up with the Hills, both families will have driven the exact same distance.

Filling in the rate and time will require a little more thought. We don't know the rate for either family—remember, that's what we're trying to find out. However, we do know that the Platters drove 15 mph faster than the Hills. This means if the Hill family's rate is r , the Platter family's rate would be r + 15 .

Now all that's left is the time. We know it took the Platters 13 hours to catch up with the Hills. However, remember that the Hills left 3 hours earlier than the Platters—which means when the Platters caught up, they'd been driving 3 hours more than the Platters. 13 + 3 is 16 , so we know the Hills had been driving 16 hours by the time the Platters caught up with them.

Our chart gives us two equations. The Hill family's trip can be described by d = r ⋅ 16 . The equation for the Platter family's trip is d = ( r + 15) ⋅ 13 . Just like with our other problems, we can combine these equations by replacing a variable in one of them.

The Hill family equation already has the value of d equal to r ⋅ 16. So we'll replace the d in the Platter equation with r ⋅ 16 . This way, it will be an equation we can solve.

r ⋅ 16 = (r + 15) ⋅ 13

First, let's simplify the right side: r ⋅ 16 is 16 r .

16r = (r + 15) ⋅ 13

Next, we'll simplify the right side and multiply ( r + 15) by 13 .

16r = 13r + 195

We can get both r and their coefficients on the left side by subtracting 13 r from 16 r : 16 r - 13 r is 3 r .

Now all that's left to do is get rid of the 3 next to the r . To do this, we'll divide both sides by 3: 195 / 3 is 65 .

So there's our answer: r = 65. The Hill family drove an average of 65 mph .

You can solve any overtaking problem the same way we solved this one. Just remember to pay special attention when you're setting up your chart. Just like the Hill family did in this problem, the person or vehicle who started moving first will always have a greater travel time.

Practice problem 2

Try solving this problem. It's similar to the problem we just solved. When you're finished, scroll down to see the answer and an explanation.

A train moving 60 mph leaves the station at noon. An hour later, a train moving 80 mph leaves heading the same direction on a parallel track. What time does the second train catch up to the first?

Problem 2 answer

Here's practice problem 2:

Answer: 4 p.m.

To solve this problem, start by making a chart. Here's how it should look:

Here's an explanation of the chart:

• Distance: Both trains will have traveled the same distance by the time the fast train catches up with the slow one, so the distance for both is d .
• Rate: The problem tells us how fast each train was going. The fast train has a rate of 80 mph , and the slow train has a rate of 60 mph .
• Time: We'll use t to represent the fast train's travel time before it catches up. Because the slow train started an hour before the fast one, it will have been traveling one hour more by the time the fast train catches up. It's t + 1 .

Now we have two equations. The equation for the fast train is d = 80 t . The equation for the slow train is d = 60 ( t + 1) . To solve this problem, we'll need to combine the equations.

The equation for the fast train says d is equal to 80 t . This means we can combine the two equations by replacing the d in the slow train's equation with 80 t .

80t = 60 (t + 1)

First, let's simplify the right side of the equation: 60 ⋅ ( t + 1) is 60 t + 60 .

80t = 60t + 60

To solve the equation, we'll have to get t on one side of the equals sign and a number on the other. We can get rid of 60 t on the right side by subtracting 60 t from both sides: 80 t - 60 t is 20 t .

Finally, we can get rid of the 20 next to t by dividing both sides by 20 . 60 divided by 20 is 3 .

So t is equal to 3 . The fast train traveled for 3 hours . However, it's not the answer to our problem. Let's look at the original problem again. Pay attention to the last sentence, which is the question we're trying to answer.

Our problem doesn't ask how long either of the trains traveled. It asks what time the second train catches up with the first.

The problem tells us that the slow train left at noon and the fast one left an hour later. This means the fast train left at 1 p.m . From our equations, we know the fast train traveled 3 hours . 1 + 3 is 4 , so the fast train caught up with the slow one at 4 p.m . The answer to the problem is 4 p.m .

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"Distance" Word Problems

Explanation More Examples

What is a "distance" word problem?

"Distance" word problems, often also called "uniform rate" problems, involve something travelling at some fixed and steady ("uniform") pace ("rate", "velocity", or "speed"), or else you are told to regard to object as moving at some average speed.

Content Continues Below

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Distance Word Problems

Whenever you read a problem that involves "how fast", "how far", or "for how long", you should think of the distance equation, d = rt , where d stands for distance, r stands for the (constant or average) rate of speed, and t stands for time.

Make sure that the units for time and distance agree with the units for the rate. For instance, if they give you a rate of feet per second, then your time must be in seconds and your distance must be in feet. Sometimes they try to trick you by using mis-matched units, and you have to catch this and convert to the correct units.

In case you're wondering, this type of exercise requires that the rate be fixed and steady (that is, unchanging) for the d = rt formula to work. The only way you can deal with a speed that might be changing over time is to take the average speed over the time or distance in question. Working directly with changing speeds will be something you'll encounter in calculus, as it requires calculus-based (or more advanced) methods.

What is the difference between a fixed speed and an average speed?

A fixed-speed exercise is one in which the car, say, is always going exactly sixty miles an hour; in three hours, the car, on cruise-control, will have gone 180 miles. An average-speed exercise is one in which the car, say, averaged forty miles an hour, but this average includes the different speeds related to stop lights, highways, and back roads; in three hours the car went 120 miles, though the car's speed was not constant. Most of the exercises you'll see will be fixed-speed exercises, but obviously they're not very "real world". It's a simplification they do in order to make the situation feasible using only algebraic methods.

What is an example of a "distance" word problem?

• A 555 -mile, 5 -hour plane trip was flown at two speeds. For the first part of the trip, the average speed was 105 mph. Then the tailwind picked up, and the remainder of the trip was flown at an average speed of 115 mph. For how long did the plane fly at each speed?

There is a method for setting up and solving these exercises that I first encountered well after I'd actually been doing them while taking a class as an undergraduate. But, as soon as I was introduced to the method, I switched over, because it is *so* way easier.

First I set up a grid, with the columns being labelled with the variables from the "distance" formula, and the rows being labelled with the "parts" involved:

In the first part, the plane covered some distance. I don't know how much, so I'll need a variable to stand for this unknown value. I'll use the variable they give me in the distance equation:

They gave me the speed, or rate, for this part, so I'll add this to my table:

The plane flew for some amount of time during this first part, but I don't know how long that was. So I need a variable to stand for this unknown value; I'll use the variable from the distance equation:

For the second part, the plane travelled the rest of the total distance. I don't know the exact distance that was flown during this second part, but I do know that it was "however much was left of the 555 miles, after the first d miles were flown in the first part. "How much was left after [some amount] was taken out" is expressed with subtraction: I take the amount that has been taken care of already, and subtract this from whatever was the total. Adding this to my table, I get:

They've given me the speed, or rate, for the second part, and I can use the same "How much is left?" construction for whatever was the time for this second part. So now my table looks like this:

For the "total" row, I add down (or take info from the exercise statement):

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Why did I not add down in the "rate" column? Because I cannot add rates! In this exercise, adding the rates would have said that the average rate for the entire trip was 105 + 115 = 220 miles per hour. But obviously this makes no sense.

The genius of this table-based method of set-up is that I can now create equations from the rows and columns. In this exercise, there is more than one way to proceed. I'll work with the "distance" equation to create expressions for the distances covered in each part.

Multiplying across, the first row tells me that the distance covered in the first part of the flight was:

1st part distance: 105 t

Again multiplying across, the second row tells me that the distance covered in the second part of the flight was:

2nd part distance: 115(5 − t )

I can add these two partial-distance expressions, and set them equal to the known total distance:

105 t + 115(5 − t ) = 555

This is an equation in one variable, which I can solve:

105 t + 115(5 − t ) = 555 105 t + 575 − 115 t = 555 575 − 10 t = 555 20 = 10 t 2 = t

Looking back at my table, I see the I had defined t to be the time that the plane spent in the air on the first part of its journey. Looking back at the original exercise, I see that they want to know the times that the plane spent at each of the two speeds.

I now have the time for the first part of the flight; the time was two hours. The exercise said that the entire trip was five hours, so the second part must have taken three hours (found by subtracting the first-part time from the total time). They haven't asked for the partial distances, so I now have all the information I need; no further computations are necessary. My answer is:

first part: 2 hours second part: 3 hours

When I was setting up my equation, I mentioned that there was more than one way to proceed. What was the other way? I could have used the table to create an expression for each of the two partial times, added, set the result equal to the given total, and solved for the variable d . Since the distance equation is d  =  rt , then the expressions for the partial times would be created by solving the equation for t  = . My work would have looked like this:

first part: d /105

second part: (555 − d )/115

adding: d /105 + (555 − d )/115 = 5 23 d + 11,655 − 21 d = 12,075 2 d = 420 d = 210

Looking back at my table, I would have seen that this gives me the distance covered in the first half of the flight. Looking back at the exercise, I would have seeing that they are wanting times, not distances. So I would have back-solved for the time for the first part, and then done the subtraction to find the time for the second part. My work would have had more steps, but my answer would have been the same.

There are three things that I hope you take from the above example:

• Using a table or grid to keep track of what you're doing can be incredibly helpful.
• It is important to clearly define your variables, so you know (by the end) what you'd meant (back in the beginning), so you can apply your results correctly.
• You should always check the original exercise, so you can be sure that you're answering the question that they'd actually asked.

(My value for the distance, found above, is correct, but was not what they'd asked for.) But even more important to understand is this:

NEVER TRY TO ADD RATES! Think about it: If you drive 20 mph on one street, and 40 mph on another street, does that mean you averaged 60 mph? Of course not.

Can I even average the rates? If I drove at 20 mph for one hour, and then drove 60 mph for two hours, then I would have travelled 140 miles in three hours, or a little under 47 mph. But 47 is not the average of 20 and 60 .

As you can see, the actual math involved in solving this type of exercise is often quite simple. It's the set-up that's the hard part. So what follows are some more examples, but with just the set-up displayed. Try your hand at solving, and click on the links to get pop-ups from which to check your equations and solutions.

• An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at an average speed of 60 mph. The entire distance was 150 miles; the entire trip took three hours. Find the distance from the airport to the corporate offices.

I will start in the usual way, by setting up my table:

I have labelled my rows so it's clear how they relate to the exercise. Now I need to fill in the rows. As before, I don't know the distance or the time for the part where the executive was driving, so I'll use variables for these unknowns, along with the given rate.

For the flying portion of the trip, I'll use the "how much is left" construction, along with the given rate, to fill in my second row. I'll also fill in the totals.

The first row gives me the equation d  = 30 t . The second row is messier, giving me the equation:

150 − d = 60(3 − t )

There are various ways I can go from here; I think I'll solve this second equation for the variable d , and then set the results equal to each other.

150 − d = 60(3 − t ) 150 − 60(3 − t ) = d

Setting equal these two expressions for d , I get:

30 t = 150 − 60(3 − t )

Solve for t ; interpret the value; state the final answer.

• A car and a bus set out at 2 p.m. from the same point, headed in the same direction. The average speed of the car is 30 mph slower than twice the speed of the bus. In two hours, the car is 20 miles ahead of the bus. Find the rate of the car.

Both vehicles travelled for the same amount of time.

The car's values are expressed in terms of the bus' values, so I'll use variables for the bus' unknowns, and then define the car in terms of the bus' variables. This gives me:

(As it turns out, I won't need the "total" row this time.) The car's row gives me:

d + 20 = 2(2 r − 30)

This is not terribly helpful. The second row gives me:

I'll use the second equation to simplify the first equation by substituting " 2 r " from the second equation in for the " d " in the first equation. Then I'll solve the equation for the value of " r ". Finally, I'll need to interpret this value within the context of the exercise, and then I'll state the final answer.

(Remember that the expression for the car's speed, from the table, was 2 r  − 30 , so all you need to do is find the numerical value of this expression. Just evaluate; don't try to solve — again — for the value of r .)

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• Problems on Distance Formula

We will discuss here how to solve the problems on distance formula.

The distance between two points A (x\(_{1}\), y\(_{1}\)) and B (x\(_{2}\), y\(_{2}\)) is given by the formula

AB = \(\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\)

1.  If the distance between the points (5, - 2) and (1, a) is 5, find the values of a.

We know, the distance between (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\))

is \(\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\)

Here, the distance = 5, x\(_{1}\) = 5, x\(_{2}\) = 1, y\(_{1}\) = -2 and y\(_{2}\) = a

Therefore, 5 = \(\sqrt{(5 - 1)^{2} + (-2 - a)^{2}}\)

⟹ 25 = 16 + (2 + a)\(^{2}\)

⟹ (2 + a)\(^{2}\) = 25 - 16

⟹ (2 + a)\(^{2}\) = 9

Taking square root, 2 + a = ±3

⟹ a = -2 ± 3

⟹ a = 1, -5

2. The co-ordinates of points on the x-axis which are at a distance of 5 units from the point (6, -3).

Let the co-ordinates of the point on the x-axis be (x, 0)

Since, distance = \(\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}\)

Now taking (6, -3) = (x\(_{1}\), y\(_{1}\)) and (x, 0) = (x\(_{2}\), y\(_{2}\)), we get

5 = \(\sqrt{(x - 6)^{2} + (0 + 3)^{2}}\)

Squaring both sides we get

⟹ 25 = (x – 6)\(^{2}\) + 3\(^{2}\)

⟹ 25 = x\(^{2}\) – 12x + 36 + 9

⟹ 25 = x\(^{2}\) – 12x + 45

⟹ x\(^{2}\) – 12x + 45 – 25 = 0

⟹ x\(^{2}\) – 12x + 20 = 0

⟹ (x – 2)(x – 10) = 0

⟹ x = 2 or x = 10

Therefore, the required points on the x-axis are (2, 0) and (10, 0).

3. Which point on the y-axis is equidistance from the points (12, 3) and (-5, 10)?

Let the required point on the y-axis (0, y).

Given (0, y) is equidistance from (12, 3) and (-5, 10)

i.e., distance between (0, y) and (12, 3) = distance between (0, y) and (-5, 10)

⟹ \(\sqrt{(12 - 0)^{2} + (3 - y)^{2}}\) = \(\sqrt{(-5 - 0)^{2} + (10 - y)^{2}}\)

⟹ 144 + 9 + y\(^{2}\) – 6y = 25 + 100 + y\(^{2}\) – 20y

⟹ 14y = -28

Therefore, the required point on the y-axis = (0, -2)

4.  Find the values of a such that PQ = QR, where P, Q and R are the points whose coordinates are (6, - 1), (1, 3) and (a, 8) respectively.

PQ = \(\sqrt{(6 - 1)^{2} + (-1 - 3)^{2}}\)

     = \(\sqrt{5^{2} + (-4)^{2}}\)

     = \(\sqrt{25 + 16}\)

     = \(\sqrt{41}\)

QR = \(\sqrt{(1 - a)^{2} + (3 - 8)^{2}}\)

     = \(\sqrt{(1 - a)^{2} + (-5)^{2}}\)

     = \(\sqrt{(1 - a)^{2} + 25}\)

Therefore, PQ = QR

⟹ \(\sqrt{41}\) = \(\sqrt{(1 - a)^{2} + 25}\)

⟹ 41 = (1 - a)\(^{2}\) + 25

⟹ (1 - a)\(^{2}\) = 41 - 25

⟹ (1 - a)\(^{2}\) = 16

⟹ 1 - a = ±4

⟹ a = -3, 5

5.  Find the points on the y-axis, each of which is at a distance of 13 units from the point (-5, 7).

Let A (-5, 7) be the given point and let P (0, y) be the required point on the y-axis. Then,

PA = 13 units

⟹ PA\(^{2}\) = 169

⟹ (0 + 5)\(^{2}\) + (y - 7)\(^{2}\) = 169

⟹ 25 + y\(^{2}\) - 14y + 49 = 169

⟹ y\(^{2}\) – 14y + 74 = 169

⟹ y\(^{2}\) – 14y – 95 = 0

⟹ (y - 19)(y + 5) = 0

⟹ y – 19 = 0 or, y + 5 = 0

⟹ y = 19 or, y = -5

Hence, the required points are (0, 19) and (0, -5)

●   Distance and Section Formulae

• Distance Formula
• Distance Properties in some Geometrical Figures
• Conditions of Collinearity of Three Points
• Distance of a Point from the Origin
• Distance Formula in Geometry
• Section Formula
• Midpoint Formula
• Centroid of a Triangle
• Worksheet on Distance Formula
• Worksheet on Collinearity of Three Points
• Worksheet on Finding the Centroid of a Triangle
• Worksheet on Section Formula

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Algebra: Distance Problems

In these lessons, we will learn to solve word problems involving distance, rate (speed) and time.

Related Pages Rate Distance Time Word Problems Distance Word Problems Average Speed Problems

What Are Distance Word Problems Or Distance Rate Time Problems?

Distance problems are word problems that involve the distance an object will travel at a certain average rate for a given period of time .

The formula for distance problems is: distance = rate × time or d = r × t

Things to watch out for: Make sure that you change the units when necessary. For example, if the rate is given in miles per hour and the time is given in minutes then change the units appropriately.

It would be helpful to use a table to organize the information for distance problems. A table helps you to think about one number at a time instead being confused by the question.

The following diagrams give the steps to solve Distance-Rate-Time Problems. Scroll down the page for examples and solutions.

We will show you how to solve distance problems by the following examples:

• Traveling At Different Rates
• Traveling In Different Directions
• Given Total Time
• Wind and Current Problems

How To Solve Distance Problems: Traveling At Different Rates

Example: A bus traveling at an average rate of 50 kilometers per hour made the trip to town in 6 hours. If it had traveled at 45 kilometers per hour, how many more minutes would it have taken to make the trip?

Solution: Step 1: Set up a rtd table.

Step 2: Fill in the table with information given in the question.

A bus traveling at an average rate of 50 kilometers per hour made the trip to town in 6 hours. If it had traveled at 45 kilometers per hour, how many more minutes would it have taken to make the trip?

Let t = time to make the trip in Case 2.

Step 3: Fill in the values for d using the formula d = rt

Step 4: Since the distances traveled in both cases are the same, we get the equation:

Step 5: Beware - the question asked for “how many more minutes would it have taken to make the trip”, so we need to deduct the original 6 hours taken.

Answer: The time taken would have been 40 minutes longer.

How To Solve Distance Problems: Two Objects Traveling In The Same Direction

Example: This motion problem (or distance rate time problem or uniform rate problem) involves traveling in the same direction, solving for “how long” one moving object traveling until it meets up with the second moving object.

It uses d = rt (distance equals rate times time).

Car 1 starts from point A and heads for point B at 60 mph. Fifteen minutes later, car 2 leaves the same point A and heads for point B at 75 mph. How long before car 2 overtakes car 1?

How To Solve Distance Problems: Two Objects Traveling In The Opposite Directions

Example: A bus and a car leave the same place and traveled in opposite directions. If the bus is traveling at 50 mph and the car is traveling at 55 mph, in how many hours will they be 210 miles apart?

If the bus is traveling at 50 mph and the car is traveling at 55 mph, in how many hours will they be 210 miles apart?

Let t = time when they are 210 miles apart.

Step 4: Since the total distance is 210, we get the equation:

50t + 55t = 210 105t = 210

Answer: They will be 210 miles apart in 2 hours.

Objects Traveling At Opposite Directions, Calculate How Long It Takes For Them To Be A Given Distance Apart

This motion problem (or distance rate time problem or uniform rate problem) involves one object traveling in one direction and the other in the opposite direction, solving for “how long” (or the amount of time) two moving objects traveling until they are certain distance apart.

Example: Two planes leave the same point at 8 AM. Plane 1 heads East at 600 mph and Plane 2 heads West at 450 mph. How long will they be 1400 miles apart? At what time will they be 1400 miles apart? How far has each plane traveled?

How To Solve Distance Problems: Given The Total Time

Example: John took a drive to town at an average rate of 40 mph. In the evening, he drove back at 30 mph. If he spent a total of 7 hours traveling, what is the distance traveled by John?

John took a drive to town at an average rate of 40 mph. In the evening, he drove back at 30 mph. If he spent a total of 7 hours traveling, what is the distance traveled by John?

Let t = time to travel to town.

7 – t = time to return from town.

40t = 30(7 – t) Use distributive property 40t = 210 – 30t

Step 5: The distance traveled by John to town is

40t = 120 The distance traveled by John to go back is also 120 So, the total distance traveled by John is 240

Answer: The distance traveled by John is 240 miles.

How To Find The Total Distance Given Total Time And Two Rates?

Example: Roy took 5 hours to complete a journey. For the first 2 hours, he traveled at an average speed of 65 km/h. For the rest of the journey, he traveled at an average speed of 78 km/h. What was the total distance of the journey?

How To Solve Wind And Current Word Problems?

There is another group of distance-time problems that involves the speed of the water current or the speed of wind affecting the speed of the vehicle. The following video shows an example of such a problem.

How to solve Wind Word Problems?

Example: Into the headwind, the plane flew 2000 miles in 5 hours. With a tailwind, the return trip took 4 hours. Find the speed of the plane in still air and the speed of the wind.

How to find the speed of the current of a stream?

Example: The speed of a boat in still water is 10 mph. It travels 24 miles upstream and 24 miles downstream in 5 hours. What is the speed of the current?

How to solve Current Word Problems?

Example: Traveling downstream, Elmo can go 6 km in 45 minutes. On the return trip, it takes him 1.5 hours. What is the boat’s speed in still water and what is the rate of the current?

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Module 9: Multi-Step Linear Equations

Using the distance, rate, and time formula, learning outcomes.

• Use the problem-solving method to solve problems using the distance, rate, and time formula

One formula you’ll use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant speed. The basic idea is probably already familiar to you. Do you know what distance you traveled if you drove at a steady rate of [latex]60[/latex] miles per hour for [latex]2[/latex] hours? (This might happen if you use your car’s cruise control while driving on the Interstate.) If you said [latex]120[/latex] miles, you already know how to use this formula!

The math to calculate the distance might look like this:

[latex]\begin{array}{}\\ \text{distance}=\left(\Large\frac{60\text{ miles}}{1\text{ hour}}\normalsize\right)\left(2\text{ hours}\right)\hfill \\ \text{distance}=120\text{ miles}\hfill \end{array}[/latex]

In general, the formula relating distance, rate, and time is

[latex]\text{distance}\text{=}\text{rate}\cdot \text{time}[/latex]

Distance, Rate, and Time

For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula

[latex]d=rt[/latex]

where [latex]d=[/latex] distance, [latex]r=[/latex] rate, and [latex]t=[/latex] time.

Notice that the units we used above for the rate were miles per hour, which we can write as a ratio [latex]\Large\frac{miles}{hour}[/latex]. Then when we multiplied by the time, in hours, the common units “hour” divided out. The answer was in miles.

Jamal rides his bike at a uniform rate of [latex]12[/latex] miles per hour for [latex]3\Large\frac{1}{2}[/latex] hours. How much distance has he traveled?

In the following video we provide another example of how to solve for distance given rate and time.

Rey is planning to drive from his house in San Diego to visit his grandmother in Sacramento, a distance of [latex]520[/latex] miles. If he can drive at a steady rate of [latex]65[/latex] miles per hour, how many hours will the trip take?

Show Solution

In the following video we show another example of how to find rate given distance and time.

• Question ID 145550, 145553,145619,145620. Authored by : Lumen Learning. License : CC BY: Attribution
• Ex: Find the Rate Given Distance and Time. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/3rYh32ErDaE . License : CC BY: Attribution
• Example: Solve a Problem using Distance = Rate x Time. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/lMO1L_CvH4Y . License : CC BY: Attribution
• Prealgebra. Provided by : OpenStax. License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

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High school geometry

Course: high school geometry   >   unit 6.

• Getting ready for analytic geometry
• Distance formula

Distance between two points

• Midpoint formula
• Distance formula review
• Midpoint formula review
• (Choice A)   9 ‍   A 9 ‍  
• (Choice B)   8 ‍   B 8 ‍  
• (Choice C)   45 ‍   C 45 ‍  
• (Choice D)   27 ‍   D 27 ‍  

Distance Formula

Distance formula calculator

How it works:  Type the two x coordinates and two y coordinates into the boxes below and it will automatically calculate the distance between those 2 points and show you step by step.

Distance between two points

In a Cartesian grid, to measure a line segment that is either vertical or horizontal is simple enough. You can count the distance either up and down the y-axis or across the x-axis.

But what about diagonal lines? How can you know precisely how long the line segment is if it cuts across those tiny boxes? See this example:

You can use the distance formula to calculate any line segment if you know the coordinates of the two endpoints. You will be mentally constructing a right triangle, using the diagonal as if it were a hypotenuse.

Distance formula

The  Distance Formula  squares the differences between the two x coordinates and two y coordinates, then adds those squares, and finally takes their square root to get the total distance along the diagonal line:

The expression ( x 2 − x 1 ) ({x}_{2}-{x}_{1}) ( x 2 ​ − x 1 ​ )  is read as  the change in x  and  ( y 2 − y 1 ({y}_{2}-{y}_{1} ( y 2 ​ − y 1 ​  is  the change in y .

How to use the distance formula

What this is really doing is calculating the distance horizontally between x- values , as if a line segment was forming a side of a right triangle, and then doing that again with the y -values , as if a vertical line segment was the second side of a right triangle.

That leaves a calculation about the hypotenuse, your given diagonal. The distance formula is a special application of the Pythagorean theorem .

All you need to do is plug the coordinates in very carefully. Let's use our line's endpoints,  (1, 3)  and  (7, 6) :

Distance formula examples

You need not even have a coordinate grid in front of you to use the Distance Formula, so long as you have both sets of coordinate points. So, try these three practice problems!

( − 2 , 4 ) (-2, 4) ( − 2 , 4 ) and ( 10 , 1 ) (10, 1) ( 10 , 1 )

( 5 , 5 ) (5, 5) ( 5 , 5 ) and ( 10 , 1 ) (10, 1) ( 10 , 1 )

( − 11 , − 2 ) (-11, -2) ( − 11 , − 2 ) and ( − 21 , − 6 ) (-21, -6) ( − 21 , − 6 )

We will not leave you hanging out on a diagonal. Here are the beginning steps, to help you get started:

D = ( 10 − ( − 2 ) ) 2 + ( 1 − 4 ) 2 D=\sqrt{{(10-(-2))}^{2}+{(1-4)}^{2}} D = ( 10 − ( − 2 )) 2 + ( 1 − 4 ) 2 ​

D = ( 2 − 5 ) 2 + ( 10 − 5 ) 2 D=\sqrt{{(2-5)}^{2}+{(10-5)}^{2}} D = ( 2 − 5 ) 2 + ( 10 − 5 ) 2 ​

D = ( − 21 − ( − 11 ) ) 2 + ( − 6 − ( − 2 ) ) 2 D=\sqrt{{(-21-(-11))}^{2}+{(-6-(-2))}^{2}} D = ( − 21 − ( − 11 )) 2 + ( − 6 − ( − 2 )) 2 ​

What comes next?

D = ( 12 ) 2 + ( − 3 ) 2 D=\sqrt{{\left(12\right)}^{2}+{\left(-3\right)}^{2}} D = ( 12 ) 2 + ( − 3 ) 2 ​

D = ( − 3 ) 2 + ( 5 ) 2 D=\sqrt{{\left(-3\right)}^{2}+{\left(5\right)}^{2}} D = ( − 3 ) 2 + ( 5 ) 2 ​

D = ( − 10 ) 2 + ( − 4 ) 2 D=\sqrt{{\left(-10\right)}^{2}+{\left(-4\right)}^{2}} D = ( − 10 ) 2 + ( − 4 ) 2 ​

You really should be able to take the last few steps by yourself. See if you got these answers:

153 ≈ 12.369 153\approx 12.369 153 ≈ 12.369

34 ≈ 5.8309 34\approx 5.8309 34 ≈ 5.8309

116 ≈ 10.7703 116\approx 10.7703 116 ≈ 10.7703

The distance formula gets its precision and perfection from the concept of using the angled line segment as if it were the hypotenuse of a right triangle formed on the grid.

You need not construct the other two sides to apply the Distance Formula, but you can see those two "sides" in the differences (distances) between x-values (a horizontal line) and y- values (a vertical line).

Lesson summary

Now that you have worked through the lesson and practice, you are able to apply the Distance Formula to the endpoints of any diagonal line segment appearing in a coordinate, or Cartesian, grid. You are also able to relate the Distance Formula to the Pythagorean Theorem.

Pythagoras was a generous and brilliant mathematician, no doubt, but he did not make the great leap to applying the Pythagorean theorem to coordinate grids. To take us from his theorem of the relationships among sides of right triangles to coordinate grids, the mathematical world had to wait for René Descartes.

His Cartesian grid combines geometry and algebra. You can use formulas, including the Distance Formula, to get precise measurements of line segments on the grid.

Solving Problems Involving Distance, Rate, and Time

• Pre Algebra & Algebra
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• Worksheets By Grade

In math, distance, rate, and time are three important concepts you can use to solve many problems if you know the formula. Distance is the length of space traveled by a moving object or the length measured between two points. It is usually denoted by d in math problems .

The rate is the speed at which an object or person travels. It is usually denoted by  r  in equations . Time is the measured or measurable period during which an action, process, or condition exists or continues. In distance, rate, and time problems, time is measured as the fraction in which a particular distance is traveled. Time is usually denoted by t in equations. 

Solving for Distance, Rate, or Time

When you are solving problems for distance, rate, and time, you will find it helpful to use diagrams or charts to organize the information and help you solve the problem. You will also apply the formula that solves distance, rate, and time, which is  distance = rate x tim e. It is abbreviated as:

There are many examples where you might use this formula in real life. For example, if you know the time and rate a person is traveling on a train, you can quickly calculate how far he traveled. And if you know the time and distance a passenger traveled on a plane, you could quickly figure the distance she traveled simply by reconfiguring the formula.

Distance, Rate, and Time Example

You'll usually encounter a distance, rate, and time question as a word problem in mathematics. Once you read the problem, simply plug the numbers into the formula.

For example, suppose a train leaves Deb's house and travels at 50 mph. Two hours later, another train leaves from Deb's house on the track beside or parallel to the first train but it travels at 100 mph. How far away from Deb's house will the faster train pass the other train?

To solve the problem, remember that d represents the distance in miles from Deb's house and t  represents the time that the slower train has been traveling. You may wish to draw a diagram to show what is happening. Organize the information you have in a chart format if you haven't solved these types of problems before. Remember the formula:

distance = rate x time

When identifying the parts of the word problem, distance is typically given in units of miles, meters, kilometers, or inches. Time is in units of seconds, minutes, hours, or years. Rate is distance per time, so its units could be mph, meters per second, or inches per year.

Now you can solve the system of equations:

50t = 100(t - 2) (Multiply both values inside the parentheses by 100.) 50t = 100t - 200 200 = 50t (Divide 200 by 50 to solve for t.) t = 4

Substitute t = 4 into train No. 1

d = 50t = 50(4) = 200

Now you can write your statement. "The faster train will pass the slower train 200 miles from Deb's house."

Sample Problems

Try solving similar problems. Remember to use the formula that supports what you're looking for—distance, rate, or time.

d = rt (multiply) r = d/t (divide) t = d/r (divide)

Practice Question 1

A train left Chicago and traveled toward Dallas. Five hours later another train left for Dallas traveling at 40 mph with a goal of catching up with the first train bound for Dallas. The second train finally caught up with the first train after traveling for three hours. How fast was the train that left first going?

Remember to use a diagram to arrange your information. Then write two equations to solve your problem. Start with the second train, since you know the time and rate it traveled:

Second train t x r = d 3 x 40 = 120 miles First train t x r = d 8 hours x r = 120 miles Divide each side by 8 hours to solve for r. 8 hours/8 hours x r = 120 miles/8 hours r = 15 mph

Practice Question 2

One train left the station and traveled toward its destination at 65 mph. Later, another train left the station traveling in the opposite direction of the first train at 75 mph. After the first train had traveled for 14 hours, it was 1,960 miles apart from the second train. How long did the second train travel? First, consider what you know:

First train r = 65 mph, t = 14 hours, d = 65 x 14 miles Second train r = 75 mph, t = x hours, d = 75x miles

Then use the d = rt formula as follows:

d (of train 1) + d (of train 2) = 1,960 miles 75x + 910 = 1,960 75x = 1,050 x = 14 hours (the time the second train traveled)

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