# What Is the Point-Slope Equation?

A straight line is defined by two points, or by one point and the slope of the line. There is an ingenious formula that allows you to find the formula for a straight line using one point and the slope of the line:

Formula

### Point-SlopeEquation

The formula that defines the line with slope $a$ through the point $\left({x}_{1},{y}_{1}\right)$ is

 $y-{y}_{1}=a\left(x-{x}_{1}\right)$

Solve the equation with respect to $y$ and the expression looks like the function for a straight line $y=ax+b$.

Example 1

Find the function for the line through $\left(2,5\right)$ with a slope of 3

Put the numbers into the point-slope equation and solve for $y$: $\begin{array}{llll}\hfill y-5& =3\left(x-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =3x-6+5=3x-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 2

Find the function for the straight line through the points $\left(-3,9\right)$ and $\left(3,-9\right)$

First, you find the slope:

 $a=\frac{-9-9}{3-\left(-3\right)}=-\frac{18}{6}=-3$

Then, select one of the points in the exercise and put it together with the slope into the point-slope equation: $\begin{array}{llll}\hfill y-\left(-9\right)& =-3\left(x-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y+9& =-3x+9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =-3x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Since $b=0$, you know that the line passes through the origin. The function for the straight line is

 $y=-3x$

If you know the function $f\left(x\right)$, you can use the point-slope equation to find the equation of the tangent line at a point on the graph of $f\left(x\right)$. This is because the slope of the tangent is equal to the value of the derivative of the function $f\left(x\right)$ at the same point.

Formula

### TheEquationforanArbitraryTangent

 $y-{y}_{1}={f}^{\prime }\left({x}_{1}\right)\left(x-{x}_{1}\right),$

where $\left({x}_{1},{y}_{1}\right)$ is a point on the tangent (often the point of tangency) and ${f}^{\prime }\left({x}_{1}\right)$ is the slope of the point. When using the formula, you must always solve for $y$— that is, get $y$ alone on one side.

Example 3

Given the function $f\left(x\right)={x}^{2}+3x-2$, find the equation for the tangent at $x=3$

To fill in the equation, you need values for ${y}_{1}$ and ${f}^{\prime }\left({x}_{1}\right)$. You know that ${x}_{1}=3$, so ${y}_{1}=f\left(3\right)$ and ${f}^{\prime }\left({x}_{1}\right)={f}^{\prime }\left(3\right)$. We begin by computing ${f}^{\prime }\left(x\right)$: $\begin{array}{llll}\hfill {f}^{\prime }\left(x\right)& =2x+3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {f}^{\prime }\left({x}_{1}\right)& ={f}^{\prime }\left(3\right)=2\left(3\right)+3=9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {y}_{1}& =f\left({x}_{1}\right)=f\left(3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\left(3\right)}^{2}+3\left(3\right)-2=16\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Now you put this into the equation and get $\begin{array}{llll}\hfill y-16& =9\left(x-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =9x-27+16=9x-11\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 4

Let $g\left(x\right)={e}^{\mathrm{sin}x}$. Find the equation for the tangent at $x=0$.

You need the values of ${f}^{\prime }\left({x}_{1}\right)$ and ${y}_{1}$. First, differentiate the function:

 ${f}^{\prime }\left(x\right)=\mathrm{cos}x\cdot {e}^{\mathrm{sin}x}$

Now you can calculate ${f}^{\prime }\left({x}_{1}\right)$. Since ${x}_{1}=0$, you get $\begin{array}{llll}\hfill {f}^{\prime }\left({x}_{1}\right)& ={f}^{\prime }\left(0\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{cos}0\cdot {e}^{\mathrm{sin}0}=1\cdot {e}^{0}=1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You find ${y}_{1}$ by putting ${x}_{1}$ into $f\left(x\right)$:

 ${y}_{1}=f\left({x}_{1}\right)=f\left(0\right)={e}^{\mathrm{sin}0}={e}^{0}=1$

You put all this into the equation and get $\begin{array}{llll}\hfill y-1& =1\left(x-0\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =x+1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

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