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Writing Linear Equations in Point-Slope Form

The equation of a line can be written in slope-intercept form or standard form. However, when the slope and a point are given, point-slope form is another alternative.

yy1=m(xx1)y-y_1 = m(x-x_1)

Here, mm is the slope and (x1,y1)(x_1,y_1) is a point on the line. By simplifying and solving for y,y, it's possible to rewrite a point-slope equation in slope-intercept form: y=mx+b.y = mx + b.

Writing a Line in Point-Slope Form

As its name suggests, point-slope form requires a point on and the slope of the line.

To write the point-slope form for the line that passes through the points (-1,5)and(1,1), (\text{-}1,5) \quad \text{and} \quad (1,1), find the slope, then use either point to write the equation.


Find the slope
First, it can be helpful to determine the slope of the line, m.m. Depending on what information about the line is given, finding the slope could be done both algebraically and graphically. In this case, two points are given, so the slope formula is preferable.
m=y2y1x2x1m = \dfrac{y_2-y_1}{x_2-x_1}
m=151(-1)m = \dfrac{{\color{#0000FF}{1}}-{\color{#009600}{5}}}{{\color{#0000FF}{1}}-\left( {\color{#009600}{\text{-}1}} \right)}
The slope of the line that passes through the given points is m=-2.m=\text{-} 2.


Choose one point on the line

The next step is to choose one point on the line to use. If the line is given as a graph, find a point whose coordinates are easy to identify. In this case, two points are given. Either can be used. For simplicity, here, use (1,1).(1,1).


Substitute values

Now that the slope has been found and a point has been chosen, the equation of the line can be written by substituting the corresponding values. Here, substitute m=-2m={\color{#0000FF}{\text{-} 2}} and (x1,y1)=(1,1).(x_1,y_1)=({\color{#009600}{1}},{\color{#009600}{1}}). yy1=m(xx1)y1=-2(x1). y-y_1=m(x-x_1) \quad \Rightarrow \quad y-{\color{#009600}{1}}={\color{#0000FF}{\text{-}2}}(x-{\color{#009600}{1}}).


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

Show Solution

To write the equation of the line in point-slope form, yy1=m(xx1),y-y_1=m(x-x_1), we need the slope and a point.


The slope

Since we're given the line as a graph we can use the rise and run to find the slope. We'll find the rise and run between two arbitratily chosen points, the yy-intercept, (0,-2),(0,\text{-} 2), and (2,1).(2,1).

Here, the run is 22 and the rise is 3,3, which gives a slope of m=32. m=\dfrac{3}{2}.


Choosing a point

Next, any point on the line can be used for (x1,y1).(x_1,y_1). Let's use the same point as above, (2,1).(2,1).


Writing the equation

Lastly, substitute the found values of mm and (x1,y1)(x_1,y_1) into yy1=m(xx1).y-y_1=m(x-x_1). Here, m=32m={\color{#0000FF}{\frac{3}{2}}} and (x1,y1)=(2,1)(x_1,y_1)=({\color{#009600}{2}},{\color{#009600}{1}}) will be substituted. y1=32(x2). y-{\color{#009600}{1}}={\color{#0000FF}{\dfrac{3}{2}}}(x-{\color{#009600}{2}}).


All points in the table lie on the same line.

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

Show Solution

To write an equation in the form yy1=m(xx1), y-y_1=m(x-x_1), the slope of the line, m,m, and any point on the line, (x1,y1),(x_1,y_1), must be known. We'll begin by finding the slope between consecutive points. It can be seen that the difference between each xx-value is 1.1.

In the right column, we see that the difference between each value is 2.2.

Therefore, m=2.m=2. Next, we need any point on the line. We know that all points in the table are on the line, so we can choose any of them. Let's use (-5,-1)(\text{-}5,\text{-}1) to be (x1,y1.)(x_1,y_1.) Substituting this point and m=2m=2 gives us the line.
The equation of the line is y+1=2(x+5).y+1=2(x+5).
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