Chapter 4.3, Problem 27PPS

To determine

To find the equation of line in slope-intercept form.

Answer to Problem 27PPS

  $y=-\frac{1}{5}x+\frac{43}{5}$

Explanation of Solution

Given:

Line passing through the point $\left(3,8\right)$ and is perpendicular to the line $y=5x-3$ .

Concept used:

Two lines are perpendicular if and only if the product of their slopes is $-1.$

Equation of a line passing through the point $\left(x_1,y_1\right)$ with slope m is $y-y_1=m\left(x-x_1\right)$ .

Calculation:

Comparing $y=5x-3$ with $y=mx+c$ , we get

  $m=\text{slope}=5$

Now let m₁ be the slope of the required line.

Since required line is perpendicular to the line $y=5x-3$ , therefore product of their slopes must be $-1.$

  $\begin{array}{l}{m}_1\cdot m=-1\\ \Rightarrow m_1\cdot 5=-1\\ \Rightarrow m_1=-\frac{1}{5}\end{array}$

Hence equation of the line passing through the point $\left(3,8\right)$ with slope $-\frac{1}{5}$ is,

  $\begin{array}{l}y-8=-\frac{1}{5}\left(x-3\right)\\ \Rightarrow y-8=-\frac{1}{5}x+\frac{3}{5}\\ \Rightarrow y=-\frac{1}{5}x+\frac{3}{5}+8\\ \Rightarrow y=-\frac{1}{5}x+\frac{43}{5}\end{array}$