Chapter 4.3, Problem 29PPS

To determine

To Write: An equation of line in slope intercept form that passes through

$x-$ intercept of line $y=-\frac{1}{2}x-4$ that is also perpendicular to required line.

Answer to Problem 29PPS

An equation of line in slope- intercept form is $y=2x+16$

Explanation of Solution

Given information:

Line passes through $x-$ intercept and also perpendicular to line $y=-\frac{1}{2}x-4$

Concept and Formula Used:

Slope intercept form $y=mx+c$ where $m$ is slope and $c$ is intercept.

Point slope form of a line: $y-y_1=m\left(x-x_1\right)$

Perpendicular lines: If lines are perpendicular then the product of their slope is $-1$

$x-$ intercept of line is point where $y$ coordinate is zero.

Calculation:

The required line is perpendicular to $y=-\frac{1}{2}x-4$ .First find the slope of given line by compairing with $y=mx+c$ .

The slope of line $y=-\frac{1}{2}x-4$ is $-\frac{1}{2}$ .

Let the slope of required line be ‘m’. As the required line is perpendicular to $y=-\frac{1}{2}x-4$

Therefore, product of slopes is $-1$

  $\begin{array}{l}m\times -\frac{1}{2}=-1\\ m=2\end{array}$

The required line passes through $x-$ intercept of line $y=-\frac{1}{2}x-4$

Putting $y=0$ in

  $\begin{array}{l}y=-\frac{1}{2}x-4\\ 0=-\frac{1}{2}x-4\\ 4=-\frac{1}{2}x\\ x=-8\end{array}$

Therefore point from which line passes is $\left(-8,0\right)$ and having slope $2$

Using point slope form of equation of line

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(-8,0\right)\&m=2\\ y-\left(0\right)=2\left(x-\left(-8\right)\right)\\ y=2\left(x+8\right)\\ y=2x+16\left(\text{Distributive property}\right)\end{array}$

Conclusion:

An equation of line in slope- intercept form is $y=2x+16$