Chapter 4.7, Problem 61STP

To determine

To find: The equation of the line perpendicular to the given graph and passing through the point $\left(2,0\right)$ .

Answer to Problem 61STP

The required equation is $y=-\frac{1}{3}x+\frac{2}{3}$ .

Explanation of Solution

Given information:

The given graph of the line is

The given options are:

A $y=3x-6$

B $y=-3x+6$

C $y=-\frac{1}{3}x+\frac{2}{3}$

D $y=\frac{1}{3}x-\frac{2}{3}$

Formula used:

Point slope form of equation

  $y-y_1=m\left(x-x_1\right)$

The given graph of the line is

Consider any two points from the graph.

  $\left(x_1,y_1\right)=\left(0,4\right)$ and $\left(x_2,y_2\right)=\left(-1.3,0\right)$

Consider the slope of equation

  $m=\frac{y_2-y_1}{x_2-x_1}$

Substitute $0$ for $y_2$

  $4$ for $y_1$ , $-1.3$ for $x_2$ and $0$ for $x_1$ in the above equation.

  $\begin{array}{c}m=\frac{0-4}{-1.3-0}\\ m=3\end{array}$

Slope of the perpendicular is the negative of the reciprocal of $m=3$ .

  $m=-\frac{1}{3}$

Consider the point slope form of an equation

  $y-y_1=m\left(x-x_1\right)$

Substitute $0$ for $y_1$ , $-\frac{1}{3}$ for $m$ in the above equation.

  $\begin{array}{c}y-0=-\frac{1}{3}\left(x-2\right)\\ y=-\frac{1}{3}x+\frac{2}{3}\end{array}$

Hence, the required equation is $y=-\frac{1}{3}x+\frac{2}{3}$ .