Chapter 5.6, Problem 22PPE

To determine

To write an equation in slope-intercept form of the line that passes through the point $\left(-3,2\right)$ and perpendicular to the equation $x-2y=7$ .

Answer to Problem 22PPE

The required equation of the line in slope-intercept form is $y=-2x-4$ .

Explanation of Solution

Given information :

Givenpoint $\left(-3,2\right)$ and equation of the line $x-2y=7$ .

As given equation $x-2y=7$

  $\begin{array}{l}\Rightarrow 2y=x+7\\ \Rightarrow y=\frac{x}{2}+7\end{array}$

Now let us find slope of the given line $y=\frac{x}{2}+7$ is $m=\frac{1}{2}.$

  $\Rightarrow$ Slope of required line is $m_1=-2$

Thus, the required equation of the line is $y-y_1=m_1\left(x-x_1\right)$

Substitute $\left(x_1,y_1\right)=\left(-3,2\right)$ and $m_1=-2$ in above equation, to get

  $\begin{array}{l}\Rightarrow y-2=-2\left(x+3\right)\\ \Rightarrow y=-2x-6+2\\ \therefore y=-2x-4\end{array}$

Hence, the required equation of the line in slope-intercept form is $y=-2x-4$ .