Chapter 5, Problem 31CR

To determine

To find: equation of line that passes through the point $\left(4,10\right)$ and is perpendicular to the graph of the equation $y=8x-1$

Answer to Problem 31CR

  $x+8y-84=0$

Explanation of Solution

Given information: Line passes through the point $\left(4,10\right)$ and is perpendicular to the graph of the equation $y=8x-1$

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 $y=8x-1$

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

Compare the equation $y=8x-1$ with $y=mx+c$

So, slope of the line is $m=8$

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 $y=8x-1$ , $m{m}_1=-1$ .

Put $m=8$

  $\begin{array}{l}8m_1=-1\\ \Rightarrow m_1=\frac{-1}{8}\end{array}$

Also, the line passes through the point $\left(x_1,y_1\right)=\left(4,10\right)$ .

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

Put $\left(x_1,y_1\right)=\left(4,10\right),m_1=\frac{-1}{8}$

  $\begin{array}{l}y-10=\frac{-1}{8}\left(x-4\right)\\ \Rightarrow 8\left(y-10\right)=-\left(x-4\right)\\ \Rightarrow 8\left(y-10\right)=-x+4\\ \Rightarrow 8y-80=-x+4\\ \Rightarrow x+8y-80-4=0\\ \Rightarrow x+8y-84=0\end{array}$