Chapter 4, Problem 20MCQ

To determine

To Write: An equation of line in slope intercept form that passes through point $\left(3,-4\right)$ and is perpendicular to line $y=-\frac{1}{3}x-5$

Answer to Problem 20MCQ

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

Explanation of Solution

Given information:

Line passes through point $\left(3,-4\right)$ and is perpendicular to line $y=-\frac{1}{3}x-5$ .

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 $y=-\frac{1}{3}x-5$ .First find the slope of given line by comparing with $y=mx+c$ .

The slope of line $y=-\frac{1}{3}x-5$ is $m=-\frac{1}{3}$ .

Let the slope of required line be ‘m’. As the required line is perpendicular to $y=-\frac{1}{3}x-5$

Therefore, product of slopes is $-1$

  $\begin{array}{l}m\times -\frac{1}{3}=-1\\ m=3\end{array}$

The required line passes through $\left(3,-4\right)$ and having slope $3$ . Using point slope form of equation of line

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(3,-4\right)\&m=3\\ y-\left(-4\right)=3\left(x-3\right)\\ y+4=3x-9\left(\text{Distributive property}\right)\\ \text{Subtract 4 from both sides}\\ y+4-4=3x-9-4\\ y=3x-13\end{array}$

Conclusion:

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