# 4.1 Linear functions  (Page 13/27)

 Page 13 / 27
$f\left(x\right)=2x+4$

The slope of the line is 2, and its negative reciprocal is $\text{\hspace{0.17em}}-\frac{1}{2}.\text{\hspace{0.17em}}$ Any function with a slope of $\text{\hspace{0.17em}}-\frac{1}{2}\text{\hspace{0.17em}}$ will be perpendicular to $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ So the lines formed by all of the following functions will be perpendicular to $\text{\hspace{0.17em}}f\left(x\right).$

$\begin{array}{ccc}\hfill g\left(x\right)& =& -\frac{1}{2}x+4\hfill \\ \hfill h\left(x\right)& =& -\frac{1}{2}x+2\hfill \\ \hfill p\left(x\right)& =& -\frac{1}{2}x-\frac{1}{2}\hfill \end{array}$

As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ and passes through the point $\text{\hspace{0.17em}}\left(4,\text{0}\right).\text{\hspace{0.17em}}$ We already know that the slope is $\text{\hspace{0.17em}}-\frac{1}{2}.\text{\hspace{0.17em}}$ Now we can use the point to find the y -intercept by substituting the given values into the slope-intercept form of a line and solving for $\text{\hspace{0.17em}}b.$

$\begin{array}{ccc}\hfill g\left(x\right)& =& mx+b\hfill \\ \hfill 0& =& -\frac{1}{2}\left(4\right)+b\hfill \\ \hfill 0& =& -2+b\hfill \\ \hfill 2& =& b\hfill \\ \hfill b& =& 2\hfill \end{array}$

The equation for the function with a slope of $\text{\hspace{0.17em}}-\frac{1}{2}\text{\hspace{0.17em}}$ and a y- intercept of 2 is

$g\left(x\right)=-\frac{1}{2}x+2$

So $\text{\hspace{0.17em}}g\left(x\right)=-\frac{1}{2}x+2\text{\hspace{0.17em}}$ is perpendicular to $\text{\hspace{0.17em}}f\left(x\right)=2x+4\text{\hspace{0.17em}}$ and passes through the point $\text{\hspace{0.17em}}\left(4,\text{0}\right).\text{\hspace{0.17em}}$ Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.

A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not –1. Doesn’t this fact contradict the definition of perpendicular lines?

No. For two perpendicular linear functions, the product of their slopes is –1. However, a vertical line is not a function so the definition is not contradicted.

Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.

1. Find the slope of the function.
2. Determine the negative reciprocal of the slope.
3. Substitute the new slope and the values for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ from the coordinate pair provided into $\text{\hspace{0.17em}}g\left(x\right)=mx+b.$
4. Solve for $\text{\hspace{0.17em}}b.$
5. Write the equation of the line.

## Finding the equation of a perpendicular line

Find the equation of a line perpendicular to $\text{\hspace{0.17em}}f\left(x\right)=3x+3\text{\hspace{0.17em}}$ that passes through the point $\text{\hspace{0.17em}}\left(3,\text{0}\right).$

The original line has slope $\text{\hspace{0.17em}}m=3,\text{\hspace{0.17em}}$ so the slope of the perpendicular line will be its negative reciprocal, or $\text{\hspace{0.17em}}-\frac{1}{3}.\text{\hspace{0.17em}}$ Using this slope and the given point, we can find the equation of the line.

$\begin{array}{ccc}\hfill g\left(x\right)& =& –\frac{1}{3}x+b\hfill \\ \hfill 0& =& –\frac{1}{3}\left(3\right)+b\hfill \\ \hfill 1& =& b\hfill \\ \hfill b& =& 1\hfill \end{array}$

The line perpendicular to $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ that passes through $\text{\hspace{0.17em}}\left(3,\text{0}\right)\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}g\left(x\right)=-\frac{1}{3}x+1.$

Given the function $\text{\hspace{0.17em}}h\left(x\right)=2x-4,$ write an equation for the line passing through $\text{\hspace{0.17em}}\left(0,0\right)\text{\hspace{0.17em}}$ that is

1. parallel to $\text{\hspace{0.17em}}h\left(x\right)$
2. perpendicular to $\text{\hspace{0.17em}}h\left(x\right)$

a. $\text{\hspace{0.17em}}f\left(x\right)=2x;$ b. $\text{\hspace{0.17em}}g\left(x\right)=-\frac{1}{2}x$

Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.

1. Determine the slope of the line passing through the points.
2. Find the negative reciprocal of the slope.
3. Use the slope-intercept form or point-slope form to write the equation by substituting the known values.
4. Simplify.

## Finding the equation of a line perpendicular to a given line passing through a point

A line passes through the points $\text{\hspace{0.17em}}\left(-2,\text{6}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(4,5\right).\text{\hspace{0.17em}}$ Find the equation of a perpendicular line that passes through the point $\text{\hspace{0.17em}}\left(4,5\right).$

From the two points of the given line, we can calculate the slope of that line.

$\begin{array}{ccc}\hfill {m}_{1}& =& \frac{5-6}{4-\left(-2\right)}\hfill \\ & =& \frac{-1}{6}\hfill \\ & =& -\frac{1}{6}\hfill \end{array}$

Find the negative reciprocal of the slope.

$\begin{array}{ccc}\hfill {m}_{2}& =& \frac{-1}{-\frac{1}{6}}\hfill \\ & =& -1\left(-\frac{6}{1}\right)\hfill \\ & =& 6\hfill \end{array}$

We can then solve for the y- intercept of the line passing through the point $\text{\hspace{0.17em}}\left(4,5\right).$

$\begin{array}{ccc}\hfill g\left(x\right)& =& 6x+b\hfill \\ \hfill 5& =& 6\left(4\right)+b\hfill \\ \hfill 5& =& 24+b\hfill \\ \hfill -19& =& b\hfill \\ \hfill b& =& -19\hfill \end{array}$

The equation for the line that is perpendicular to the line passing through the two given points and also passes through point $\text{\hspace{0.17em}}\left(4,5\right)\text{\hspace{0.17em}}$ is

$y=6x-19$

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