
## Section6.5The Point-Slope Form of the Equation of a Line

Consider the following slope equation.

\begin{equation*} m=\frac{y_2-y_1}{x_2-x_1}\text{.} \end{equation*}

If we multiply both sides of the equation by $x_2-x_1$ the resultant equation is $m(x_2-x_1)=y_2-y_1$ which is equivalent to

\begin{equation*} y_2-y_1=m(x_2-x_1)\text{.} \end{equation*}

In the original form of the slope-equation, both $(x_1,y_1)$ and $(x_2,y_2)$ are specific (and distinct) points on the line. In the new form of the equation, it is useful to still consider $(x_1,y_1)$ as a specific point on the line, but it is beneficial to think of $(x_2,y_2)$ as the variable (movable) point $(x,y)\text{.}$

The resultant equation is called the point-slope form of the equation of the line. The point-slope form can be useful when determining the equation of a line. Specifically, if a non-vertical line has a slope of $m$ and the line passes through the point $(x_1,y_1)\text{,}$ then the equation of the line can be determined using the point-slope template

\begin{equation*} y-y_1=m(x-x_1)\text{.} \end{equation*}

For example, let's determine the equation of the line that passes through the points $(-5,20)$ and $(15,12)\text{.}$

We begin by determining the slope of the line. Let $(x_1,y_1)$ be the ordered pair $(-5,20)$ and $(x_2,y_2)$ be the point $(15,12)\text{.}$ We determine the slope as follows.

\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{12-20}{15-(-5)}\\ \amp=\frac{-8}{20}\\ \amp=-\frac{2}{5} \end{align*}

Using that slope along with the ordered pair $(-5,20)$ in the point slope-slope equation we determine the equation of the line as follows.

\begin{align*} y-y_1\amp=m(x-x_1)\\ y-20\amp=-\frac{2}{5}(x-(-5)) \end{align*}

It is most common to state the equation of a line in slope-intercept form, so to that end let's pick up where we left off.

So we conclude that the slope-intercept equation of the line that passes through the points $(-5,20)$ and $(15,12)$ is

\begin{equation*} y=-\frac{2}{5}x+18\text{.} \end{equation*}

### Subsection6.5.1Exercises

Use the point-slope form of the equation of the line to determine the equation of the line that passes through each of the following pairs of points. In each case state the equation in slope-intercept form.

###### 1

$(1,-5)$ and $(-2,-32)$

Solution

We begin by determining the slope of the line.

\begin{equation*} (x_1,y_1) \text{ is } (1,-5) \text{ and } (x_2,y_2) \text{ is } (-2,-32) \end{equation*}
\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{-32-(-5)}{-2-1}\\ \amp=\frac{-27}{-3}\\ \amp=9 \end{align*}

We can use the slope of $9$ along with the ordered pair $(1,-5)$ in the equation $y-y_1=m(x-x_1)$ to determine the equation of the line.

\begin{align*} y-(-5)\amp=9(x-1)\\ y+5\amp=9x-9\\ y+5\subtractright{5}\amp=9x-9\subtractright{5}\\ y\amp=9x-14 \end{align*}

he equation of the line is $y=9x-14\text{.}$ Both points check.

###### 2

$(36,-1)$ and $(-12,19)$

Solution

We begin by determining the slope of the line.

\begin{equation*} (x_1,y_1) \text{ is } (36,-1) \text{ and } (x_2,y_2) \text{ is } (-12,19) \end{equation*}
\begin{align*} m\amp=\frac{y_2-y_1}{x_1-x_1}\\ \amp=\frac{19-(-1)}{-12-36}\\ \amp=\frac{20}{-48}\\ \amp=-\frac{5}{12} \end{align*}

We can use the slope of $-\frac{5}{12}$ along with the ordered pair $(36,-1)$ in the equation $y-y_1=m(x-x_1)$ to determine the equation of the line.

\begin{align*} y-(-1)\amp=-\frac{5}{12}(x-36)\\ y+1\amp=-\frac{5}{12}x+15\\ y+1\subtractright{1}\amp=-\frac{5}{12}x+15\subtractright{1}\\ y\amp=-\frac{5}{12}+14 \end{align*}

So the equation of the line is $y=-\frac{5}{12}x+14\text{.}$ Checks.

###### 3

$(5,-6)$ and $(-7,-2)$

Solution

We begin by determining the slope of the line.

\begin{equation*} (x_1,y_1) \text{ is } (5,-6) \text{ and } (x_2,y_2) \text{ is } (-7,-2) \end{equation*}
\begin{align*} m\amp=\frac{y_2-y_1}{x_1-x_1}\\ \amp=\frac{-2-(-6)}{-7-5}\\ \amp=\frac{4}{-12}\\ \amp=-\frac{1}{3} \end{align*}

We can use the slope of $-\frac{1}{3}$ along with the ordered pair $(5,-6)$ in the equation $y-y_1=m(x-x_1)$ to determine the equation of the line.

\begin{align*} y-(-6)\amp=-\frac{1}{3}(x-5)\\ y+6\amp=-\frac{1}{3}x+\frac{5}{3}\\ y+6\subtractright{6}\amp=-\frac{1}{3}x+\frac{5}{3}\subtractright{\frac{18}{3}}\\ y\amp=-\frac{1}{3}x-\frac{13}{3} \end{align*}

So the equation of the line seems to be $y=-\frac{1}{3}x-\frac{13}{3}\text{.}$ Hmmm ... let's check the point $(-7,-2)\text{.}$

\begin{align*} -2\amp=-\frac{1}{3} \cdot -7-\frac{13}{3}?\\ -2\amp=\frac{7}{3}-\frac{13}{3}?\\ -2\amp=-\frac{6}{3} \checkmark \end{align*}

Well I'll be darned, the equation of the line really is $y=-\frac{1}{3}x-\frac{13}{3}\text{.}$