Chapter 4.3, Problem 24PPS

To determine

To Write: An equation of line in slope intercept form that passes through point $\left(-5,2\right)$ and is perpendicular to line $y=\frac{1}{2}x-3$

Answer to Problem 24PPS

An equation of line in slope- intercept form is $y=-2x-8$

Explanation of Solution

Given information:

Line passes through point $\left(-5,2\right)$ and is perpendicular to line $y=\frac{1}{2}x-3$

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 $-1$

Calculation:

The required line is perpendicular to $y=\frac{1}{2}x-3$ .First find the slope of given line by compairing with $y=mx+c$ .

The slope of line $y=\frac{1}{2}x-3$ is $\frac{1}{2}$ .

Let the slope of required line be ‘m’. As the required line is perpendicular to $y=\frac{1}{2}x-3$

Therefore, product of slopes is $-1$

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

The required line passes through $\left(-5,2\right)$ and having slope $-2$ . Using point slope form of equation of line

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(-5,2\right)\&m=-2\\ y-\left(2\right)=-2\left(x-\left(-5\right)\right)\\ y-\left(2\right)=-2\left(x+5\right)\\ y-2=-2x-10\left(\text{Distributive property}\right)\\ \text{Add 2 to both sides}\\ y-2+2=-2x-10+2\\ y=-2x-8\end{array}$

Conclusion:

An equation of line in slope- intercept form is $y=-2x-8$