Chapter 5, Problem 30CR

To determine

To find: equation of line that passes through the point $\left(3,5\right)$ and is perpendicular to the graph of the equation $y=-3x+7$

Answer to Problem 30CR

  $x-3y+12=0$

Explanation of Solution

Given information: Line passes through the point $\left(3,5\right)$ and is perpendicular to the graph of the equation $y=-3x+7$

Formula used: According to slope intercept form, equation of a line is given by $y=mx+c$ where $m$ represents a slope and $c$ represents the y-intercept

According to point slope form, equation of a line that passes through point $\left(x_1,y_1\right)$ and has slope $m_1$ is given by $y-y_1=m_1\left(x-x_1\right)$

Calculation:

Consider the equation $y=-3x+7$

This equation is in slope intercept form $y=mx+c$ where $m$ represents a slope and $c$ represents the y-intercept

Compare the equation $y=-3x+7$ with $y=mx+c$

So, slope of the line is $m=-3$

Now use the fact that if two lines with slopes $m,m_1$ are perpendicular then $m{m}_1=-1$ .

Let $m_1$ denotes slope of the required line.

As the required line is perpendicular to the line $y=-3x+7$ , $m{m}_1=-1$ .

Put $m=-3$

  $\begin{array}{l}-3m_1=-1\\ \Rightarrow m_1=\frac{-1}{-3}=\frac{1}{3}\end{array}$

Also, the line passes through the point $\left(x_1,y_1\right)=\left(3,5\right)$ .

The equation of the line is $y-y_1=m_1\left(x-x_1\right)$ .

Put $\left(x_1,y_1\right)=\left(3,5\right),m_1=\frac{1}{3}$

  $\begin{array}{l}y-5=\frac{1}{3}\left(x-3\right)\\ \Rightarrow 3\left(y-5\right)=x-3\\ \Rightarrow 3y-15=x-3\\ \Rightarrow x-3y-3+15=0\\ \Rightarrow x-3y+12=0\end{array}$