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Therefore, the same line can be described in slope-intercept form as y = 1 2 x + 7.

Point-slope form of a linear equation

The point-slope form of a linear equation takes the form

y y 1 = m ( x x 1 )

where m is the slope, x 1   and   y 1 are the x  and  y coordinates of a specific point through which the line passes.

Writing the equation of a line using a point and the slope

The point-slope form is particularly useful if we know one point and the slope of a line. Suppose, for example, we are told that a line has a slope of 2 and passes through the point ( 4 , 1 ) . We know that m = 2 and that x 1 = 4 and y 1 = 1. We can substitute these values into the general point-slope equation.

y y 1 = m ( x x 1 ) y 1 = 2 ( x 4 )

If we wanted to then rewrite the equation in slope-intercept form, we apply algebraic techniques.

y 1 = 2 ( x 4 ) y 1 = 2 x 8 Distribute the  2.         y = 2 x 7 Add 1 to each side .

Both equations, y 1 = 2 ( x 4 ) and y = 2 x 7 , describe the same line. See [link] .

Writing linear equations using a point and the slope

Write the point-slope form of an equation of a line with a slope of 3 that passes through the point ( 6 , –1 ) . Then rewrite it in the slope-intercept form.

Let’s figure out what we know from the given information. The slope is 3, so m = 3. We also know one point, so we know x 1 = 6 and y 1 = −1. Now we can substitute these values into the general point-slope equation.

       y y 1 = m ( x x 1 ) y ( 1 ) = 3 ( x 6 ) Substitute known values .          y + 1 = 3 ( x 6 ) Distribute  1  to find point-slope form .

Then we use algebra to find the slope-intercept form.

y + 1 = 3 ( x 6 ) y + 1 = 3 x 18 Distribute 3 .         y = 3 x 19 Simplify to slope-intercept form .
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Write the point-slope form of an equation of a line with a slope of –2 that passes through the point ( –2 ,   2 ) . Then rewrite it in the slope-intercept form.

y 2 = 2 ( x + 2 ) ; y = 2 x 2

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Writing the equation of a line using two points

The point-slope form of an equation is also useful if we know any two points through which a line passes. Suppose, for example, we know that a line passes through the points ( 0 ,   1 ) and ( 3 ,   2 ) . We can use the coordinates of the two points to find the slope.

m = y 2 y 1 x 2 x 1     = 2 1 3 0     = 1 3

Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let use (0, 1) for our point.

y y 1 = m ( x x 1 ) y 1 = 1 3 ( x 0 )

As before, we can use algebra to rewrite the equation in the slope-intercept form.

y 1 = 1 3 ( x 0 ) y 1 = 1 3 x Distribute the  1 3 .        y = 1 3 x + 1 Add 1 to each side .

Both equations describe the line shown in [link] .

Writing linear equations using two points

Write the point-slope form of an equation of a line that passes through the points (5, 1) and (8, 7). Then rewrite it in the slope-intercept form.

Let’s begin by finding the slope.

m = y 2 y 1 x 2 x 1    = 7 1 8 5    = 6 3    = 2

So m = 2. Next, we substitute the slope and the coordinates for one of the points into the general point-slope equation. We can choose either point, but we will use ( 5 , 1 ) .

y y 1 = m ( x x 1 ) y 1 = 2 ( x 5 )

The point-slope equation of the line is y 2 1 = 2 ( x 2 5 ) . To rewrite the equation in slope-intercept form, we use algebra.

y 1 = 2 ( x 5 ) y 1 = 2 x 10        y = 2 x 9

The slope-intercept equation of the line is y = 2 x 9.

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Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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