Chapter 4, Problem 21MCQ

To determine

To Write: An equation of line in slope intercept form that passes through point $\left(0,-3\right)$ and is perpendicular to line $y=\frac{1}{2}x-3$

Answer to Problem 21MCQ

An equation of line in slope- intercept form is $y=3x-13$

Explanation of Solution

Given information:

Line passes through point $\left(0,-3\right)$ and is perpendicular to line $y=-2x+4$ .

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 $-1$

Calculation:

The required line is perpendicular to $y=-2x+4$ .First find the slope of given line by comparing with $y=mx+c$ .

The slope of line $y=-2x+4$ is $m=-2$ .

Let the slope of required line be ‘m’. As the required line is perpendicular to $y=-2x+4$

Therefore, product of slopes is $-1$

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

The required line passes through $\left(0,-3\right)$ and having slope $\frac{1}{2}$ . Using point slope form of equation of line

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(0,-3\right)\&m=\frac{1}{2}\\ y-\left(-3\right)=\frac{1}{2}\left(x-0\right)\\ y+3=\frac{1}{2}x\left(\text{Distributive property}\right)\\ \text{Subtract 3 from both sides}\\ y+3-3=\frac{1}{2}x-3\\ y=\frac{1}{2}x-3\end{array}$

Conclusion:

An equation of line in slope- intercept form is $y=\frac{1}{2}x-3$