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5.2: Linear Equations in Point-Slope Form

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Learning Objectives

  • Write an equation in point-slope form.
  • Graph an equation in point-slope form.
  • Write a linear function in point-slope form.
  • Solve real-world problems using linear models in point-slope form.

Introduction

In the last lesson, we saw how to write the equation of a straight line in slope-intercept form. We can rewrite this equation in another way that sometimes makes solving the problem easier. The equation of a straight line that we are going to talk about is called point-slope form.

y-y_0 =m(x-x_0 )

Here m is the slope and (x_0 , y_0 ) is a point on the line. Let’s see how we can use this form of the equation in the three cases that we talked about in the last section.

Case 1: You know the slope of the line and the y-intercept.

Case 2: You know the slope of the line and a point on the line.

Case 3: You know two points on the line.

Write an Equation in Point-Slope Form

Case 1 You know the slope and the y-intercept.

  1. Start with the equation in point-slope form y-y_0 =m(x-x_0 ).
  2. Plug in the value of the slope.
  3. Plug in 0 for x_0 and b for y_0.

Example 1

Write the equation of the line in point-slope form, given that the \text{slope} = -5 and the y-intercept = 4.

Solution:

  1. Start with the equation in point-slope form. y-y_0 =m(x-x_0 )
  2. Plug in the value of the slope. y-y_0 =-5(x-x_0 )
  3. Plug in 0 for x_0 and 4 for y_0. y-(-4) =-5(x-(0) )

Therefore, the equation is y+4 = -5x

Case 2 You know the slope and a point on the line.

  1. Start with the equation in point-slope form y-y_0 =m(x-x_0 ).
  2. Plug in the value of the slope.
  3. Plug in the x and y values in place of x_0 and y_0.

Example 2

Write the equation of the line in point-slope form, given that the \text{slope} = \frac{3}{5} and the point (2,6) is on the line.

Solution:

  1. Start with the equation in point-slope form. y-y_0 =m(x-x_0 )
  2. Plug in the value of the slope. y-y_0 = \frac {3}{5}(x-x_0)
  3. Plug in 2 for x_0 and 6 for y_0. y-(6) = \frac {3}{5}(x-(2))

The equation is  y-6 = \frac {3}{5}(x-2)

Notice that the equation in point-slope form is not solved for y.

Case 3 You know two points on the line.

  1. Start with the equation in point-slope form y-y_0 =m(x-x_0 ).
  2. Find the slope using the slope formula.  m = \frac {y_2-y_1}{x_2-x_1}
  3. Plug in the value of the slope.
  4. Plug in the x and y values of one of the given points in place of x_0 and y_0

Example 3

Write the equation of the line in point-slope form, given that the line contains points (-4, -2) and (8, 12).

Solution

1. Start with the equation in point-slope form. y-y_0 =m(x-x_0 )

2. Find the slope using the slope formula.  m = \frac {12-(-2)}{8-(-4)}=\frac{14}{12}=\frac {7}{6}

3. Plug in the value of the slope.  y-y_0 = \frac {7}{6}(x-x_0)

4. Plug in -4 for x_0 and -2 for y_0.  y-(-2)= \frac {7}{6}(x-(-4))

Therefore, the equation is  y + 2 = \frac {7}{6}(x + 4) Answer 1

In the last example, you were told that for the last step you could choose either of the points you were given to plug in for the point (x_0 , y_0 ) but it might not seem like you would get the same answer if you plug the second point in instead of the first. Let’s redo Step 4.

4. Plug in 8 for x_0 and 12 for y_0 \ y-12 = \frac{7}{6}(x-8) Answer 2

This certainly does not see like the same answer as we got by plugging in the first point. What is going on?

Notice that the equation in point-slope form is not solved for y. Let’s change both answers into slope-intercept form by solving for y.

& \text{Answer}\ 1 && \text{Answer}\ 2\\& y+2 = \frac {7}{6}(x+4) && y-12 = \frac {7}{6}(x-8) \\& y+2 = \frac {7}{6}x+\frac {28}{6} && y-12 = \frac {7}{6}x-\frac {56}{6} \\& y = \frac {7}{6}x+\frac {14}{3}-2 && y = \frac {7}{6}x-\frac {28}{3}+12 \\& y = \frac {7}{6}x+\frac {8}{3} && y = \frac {7}{6}x+\frac {8}{3}

Now that the two answers are solved for y, you can see that they simplify to the same thing. In point-slope form you can get an infinite number of right answers, because there are an infinite number of points on a line. The slope of the line will always be the same but the answer will look different because you can substitute any point on the line for (x_0 , y_0 ). However, regardless of the point you pick, the point-slope form should always simplify to the same slope-intercept equation for points that are on the same line.

In the last example you saw that sometimes we need to change between different forms of the equation. To change from point-slope form to slope-intercept form, we just solve for y.

Example 4

Re-write the following equations in slope-intercept form.

a) y-5= 3(x- 2)

b) y+ 7 = -(x + 4)

Solution

a) To re-write in slope-intercept form, solve for y.

y-5 & = 3(x- 2)\\-5 & = 3x- 6\\y & = 3x- 1

b) To re-write in slope-intercept form, solve for y.

y + 7 & = -(x + 4)\\y + 7 & = -x-4\\y & = -x- 11

Graph an Equation in Point-Slope Form

If you are given an equation in point-slope form, it is not necessary to re-write it in slope-intercept form in order to graph it. The point-slope form of the equation gives you enough information so you can graph the line y-y_0 = m(x-x_0 ). From this equation, we know a point on the line (x_0 , y_0 ) and the slope of the line.

To graph the line, you first plot the point (x_0 , y_0). Then the slope tells you how many units you should go up or down and how many units you should go to the right to get to the next point on the line. Let’s demonstrate this method with an example.

Example 5

Make a graph of the line given by the equation  y-2=\frac {2}{3}(x+2)

Solution

Let’s rewrite the equation  y-(2)=\frac {2}{3}(x+2).

Now we see that point (-2, 2) is on the line and that the \text{slope} = \frac{2}{3}.

First plot point (-2, 2) on the graph.

A slope of \frac{2}{3} tells you that from your point you should move. 2 units up and 3 units to the right and draw another point.

Now draw a line through the two points and extend the line in both directions.

Write a Linear Function in Point-Slope Form

The functional notation for the point-slope form of the equation of a line is:

f(x) -f(x_0) = m(x-x_0)

Note that we replaced the each y with the f(x)

y = f(x) and y_0 = f(x_0)

That tells us more clearly that we find values of y by plugging in values of x into the function defined by the equation of the line. Let’s use the functional notation to solve some examples.

Example 6

Write the equation of the following linear functions in point-slope form.

a) m = 25 and f(0) = 250

b) m = 9.8 and f(5.5) = 12.5

c) f(32) = 0 and f(77) = 25

a) Here we are given the \text{slope} = 25 and the point on the line gives x_0 =0, f(x_0 )=250

  1. Start with the equation in point-slope form. f(x) - f(x_0 )=m(x-x_0)
  2. Plug in the value of the slope. f(x)-f(x_0 )=25(x-x_0 )
  3. Plug in 0 for x_0 and 250 for f(x_0). f(x)-250 = 25(x- 0)

Solution

The linear function is f(x)-250=25x.

b) Here we are given that \text{slope} = 9.8 and the point on the line gives x_0 =5.5, f(x_0) = 12.5

  1. Start with the equation in point-slope form. f(x) -f(x_0 ) = m(x-x_0 )
  2. Plug in the value of the slope. f(x)-f(x_0 ) = 9.8(x-x_0 )
  3. Plug in 5.5 for x_0 and 12.5 for f(x_0). f(x)- 12.5 = 9.8(x-5.5)

Solution

The linear function is f(x)-12.5= 9.8(x-5.5).

c) Here we are given two points (32, 0) and (77, 25).

  1. Start with the equation in point-slope form. f(x)-f(x_0) = m(x-x_0)
  2. Find the value of the slope. m = \frac {25-0}{77-32}=\frac {25}{45}=\frac {5}{9}
  3. Plug in the value of the slope. f(x)-f(x_0) = \frac {5}{9}(x-x_0)
  4. Plug in 32 for and 0 for f(x_0). f(x)-0 = \frac {5}{9}(x-32)

Solution

The linear function is f(x)-0 = \frac {5}{9}(x-32).

Solve Real-World Problems Using Linear Models in Point-Slope Form

Let’s solve some word problems where we need to write the equation of a straight line in point-slope form.

Example 7

Marciel rented a moving truck for the day. Marciel only remembers that the rental truck company charges $40 per day and some amount of cents per mile. Marciel drives 46 miles and the final amount of the bill (before tax) is $63. What is the amount per mile the truck rental company charges per day? Write an equation in point-slope form that describes this situation. How much would it cost to rent this truck if Marciel drove 220 miles?

Let’s define our variables:

x = distance in miles

y = cost of the rental truck in dollars

We see that we are given the y-intercept and the point (46, 63).

Peter pays a flat fee of $40 for the day. This is the y-intercept.

He pays $63 for 46 miles – this is the coordinate point (46, 63).

\text{Start with the point-slope form of the line}. & & (y -y_0 ) & = m(x-x_0)\\\text{Plug in the coordinate point}. & & 63-y_0 & = m(46-x_0)\\\text{Plug in point} \ (0, 40). & & 63 -40 & = m(46 - 0)\\\text{Solve for the slope}. & & 23 & = m (46) \rightarrow m=\frac {23}{46}=0.5\\\text{The slope is}: & & & 0.5 \ \text{dollars per mile}\\\text{So, the truck company charges} \ 50 \ \text{cents per mile}.& & (\$0.5 & = 50 \ \text{cents})\text{Equation of line is}: & & y & = 0.5x + 40

To answer the question of 220 miles we plug in x=220.

Solution

y-40=0.5(220)\Rightarrow y=\$150

Example 8

Anne got a job selling window shades. She receives a monthly base salary and a $6 commission for each window shade she sells. At the end of the month, she adds up her sales and she figures out that she sold 200 window shades and made $2500. Write an equation in point-slope form that describes this situation. How much is Anne’s monthly base salary?

Let’s define our variables

x = number of window shades sold

y = Anne’s monthly salary in dollars

We see that we are given the slope and a point on the line:

Anne gets $6 for each shade, so the \text{slope} = 6 dollars/shade.

She sold 200 shades and made $2500, so the point is (200, 2500).

\text{Start with the point-slope form of the line}. & & y -y_0 & = m(x -x_0)\\\text{Plug in the coordinate point}. & & y-y_0 & = 6(x-x_0)\\\text{Plug in point} \ (200, 2500). & & y- 2500 & = 6(x- 200)

Anne’s base salary is found by plugging in x=0. We obtain  y-2500 = -1200 \Rightarrow y = \$1300

Solution

Anne’s monthly base salary is $1300.

Lesson Summary

  • The point-slope form of an equation for a line is: y-y_0=m(x-x_0).
  • If you are given the slope and a point on the line:
    1. Simply plug the point and the slope into the equation.
  • If you are given the slope and y-intercept of a line:
    1. Plug the value of m into the equation
    2. Plug the y-intercept point into the equation y_0=y-intercept and x_0 = 0.
  • If you are given two points on the line:
    1. Use the two points to find the slope using the slope formula  m = \frac {y_2-y_1}{x_2-x_1}.
    2. Plug the value of m into the equation.
    3. Plug either of the points into the equation as (x_0, y_0).
  • The functional notation of point-slope form is f(x)-f(x_0)=m(x-x_0).

Review Questions

Write the equation of the line in point-slope form.

  1. The line has slope  -\frac {1}{10} and goes through point (10, 2).
  2. The line has slope -75 and goes through point (0, 125).
  3. The line has slope 10 and goes through point (8, -2).
  4. The line goes through the points (-2, 3) and (-1, -2).
  5. The line contains points (10, 12) and (5, 25).
  6. The line goes through points (2, 3) and (0, 3).
  7. The line has a slope \frac {3}{5} and a y-intercept -3.
  8. The line has a slope -6 and a y-intercept 0.5.

Write the equation of the linear function in point-slope form.

  1. m=-\frac {1}{5} and f(0) = 7
  2. m = -12 and f(-2) = 5
  3. f(-7) = 5 and f(3) = -4
  4. f(6) = 0 and f(0) = 6
  5. m = 3 and f(2) = -9
  6.  m=-\frac {9}{5} and f(0) = 32
  7. Nadia is placing different weights on a spring and measuring the length of the stretched spring. She finds that for a 100 gram weight the length of the stretched spring is 20 cm and for a 300 gram weight the length of the stretched spring is 25 cm. Write an equation in point-slope form that describes this situation. What is the unstretched length of the spring?
  8. Andrew is a submarine commander. He decides to surface his submarine to periscope depth. It takes him 20 minutes to get from a depth of 400 feet to a depth of 50 feet. Write an equation in point-slope form that describes this situation. What was the submarine’s depth five minutes after it started surfacing?

Review Answers

  1. y-2 = -\frac{1}{10}(x-10)
  2. y- 125 = -75x
  3. y + 2 = 10(x-8)
  4. y + 2= -5(x + 1) or y- 3 = -5(x + 2)
  5. y-25 = -\frac{13}{5}(x- 5) or y- 12 = -\frac{13}{5}(x- 10)
  6. y-3 = 0
  7. y+ 3 = \frac{3}{5}x
  8. y- 0.5 = -6x
  9. f(x)- 7 = -\frac{1}{5}x
  10. f(x) - 5 = -12(x + 2)
  11. f(x)- 5 = -\frac{9}{10}(x + 7) or f(x) +4 = -\frac{9}{10}(x - 3)
  12. f(x) = -x(x-6) or f(x) -6 = -x
  13. f(x) + 9 = 3(x-2)
  14. f(x)-32 = \frac{9}{5}x
  15. y- 20 = \frac{1}{40}(x- 100) unstretched length = 17.5 \ cm
  16. y- 50 = -17.5(x- 20) or y- 400 = -17.5x depth = 312.5 \ feet

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