Chapter 4, Problem 12CR

To determine

An equation of line perpendicular to $3x-2y=7$ and containing point $\left(1,5\right)$

Answer to Problem 12CR

Solution:

The equation of the line perpendicular to line $3x-2y=7$ and containing point $\left(1,5\right)$ is $2x+3y=17$ .

Explanation of Solution

Given information:

The equation of the line is perpendicular to $3x-2y=7$ containing point $\left(1,5\right)$

First, write the equation of line $3x-2y=7$ in the slope-intercept form $y=mx+b$ to find its slope

  $3x-2y=7$

Subtracting $3x$ from both sides

  $\Rightarrow -2y=-3x+7$

Dividing by $-2$ on both sides,

  $\Rightarrow \frac{-2}{-2}y=\frac{-3}{-2}x+\frac{7}{-2}$

  $\Rightarrow y=\frac{3}{2}x-\frac{7}{2}$

Thus, the slope intercept form of the line $3x-2y=7$ is $y=\frac{3}{2}x-\frac{7}{2}$ .

By comparing equation $y=\frac{3}{2}x-\frac{7}{2}$ with $y=mx+b$ , the slope of the line is $\frac{3}{2}$

The product of slopes of perpendicular lines is $-1$

Hence, the slope of the line perpendicular to $3x-2y=7$ is $-\frac{2}{3}$ , since $\frac{3}{2}\left(-\frac{2}{3}\right)=-1$

Now, find the equation of line containing point $\left(1,5\right)$ and has slope $-\frac{2}{3}$ .

Use the point-slope form to get the equation of line

  $y-y_1=m\left(x-x_1\right)$

By substituting values of $m,x_1$ and $y_1$

  $y-5=-\frac{2}{3}\left(x-1\right)$

  $\Rightarrow 3\left(y-5\right)=-2\left(x-1\right)$

  $\Rightarrow 3y-15=-2x+2$

  $\Rightarrow 2x+3y=15+2$

  $\Rightarrow 2x+3y=17$

Thus, the equation of line perpendicular to line $3x-2y=7$ and containing point $\left(1,5\right)$ is $2x+3y=17$