Chapter 4.3, Problem 51PFA

To determine

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

Answer to Problem 51PFA

An equation of line is $y=3x$ option ‘D’ is correct.

Explanation of Solution

Given information:

Line passes through point $\left(1,3\right)$ 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 $-\frac{1}{3}$ .

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 $\left(1,3\right)$ and having slope $3$ . Using point slope form of equation of line

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

Conclusion:

An equation of line is $y=3x$ option ‘D’ is correct.