Chapter 4.3, Problem 29PPS

To determine

To find: The equation of the straight line in slope-intercept form which passes through the point x-intercept of line $y=-\frac{1}{2}x-4$ and perpendicular to graph of same line.

Answer to Problem 29PPS

Equation of line is $y=2x+16$ .

Explanation of Solution

Given information: The given line is $y=-\frac{1}{2}x-4$ . And, the perpendicular line to this line passes through x-intercept of this line.

Calculation: Slope $\left(m\right)$ of the line perpendicular to line $y=-\frac{1}{2}x-4$ will be $\left(m=2\right)$ .

Now, x-intercept of line $y=-\frac{1}{2}x-4$ can be obtained by plugging $y=0$ in $y=-\frac{1}{2}x-4$ .

  $\begin{array}{c}\left(0\right)=-\frac{1}{2}x-\mathrm{4......}\left(\text{For }y=0\right)\\ \text{or, }\frac{1}{2}x=-4\\ x=-8\end{array}$

Now, x-intercept of line $y=-\frac{1}{2}x-4$ is $\left(-8,0\right)$ . Also the line perpendicular to this line passes through this point $\left(-8,0\right)$ having slope $\left(m=2\right)$ . Therefore, the equation of line perpendicular to line $y=-\frac{1}{2}x-4$ can be written as,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\mathrm{......}\left(\text{Point-slope form}\right)\\ y-\left(0\right)=2\left(x+8\right)\mathrm{......}\left(\begin{array}{l}\left(x_1,y_1\right)\to \left(-8,0\right)\\ \text{and }m=2\end{array}\right)\\ y=2x+16\end{array}$