Chapter 5.6, Problem 24PPE

To determine

To write an equation in slope-intercept form of the line that passes through the point $\left(1,-6\right)$ and perpendicular to the equation $x-2y=4$ .

Answer to Problem 24PPE

The required equation of the line in slope-intercept form is $y=-\frac{x}{2}+\frac{5}{2}$ .

Explanation of Solution

Given information :

Givenpoint $\left(1,-6\right)$ and equation of the line $y=-2x-4$ .

As given equation $x-2y=4$

  $\begin{array}{l}\Rightarrow x-2y=4\\ \Rightarrow 2y=x-4\\ \Rightarrow y=\frac{x}{2}-2\end{array}$

Now let us find slope of the given line $y=\frac{x}{2}-2$ 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 $\left(x_1,y_1\right)=\left(1,-6\right)$ and $m_1=-2$ in above equation, to get

  $\begin{array}{l}\Rightarrow y+6=-2\left(x-1\right)\\ \Rightarrow y=-2x+2-6\\ \therefore y=-2x-4\end{array}$

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