Chapter 4, Problem 36SGR

To determine

To Write: An equation of line in slope intercept form that passes through point $\left(-5,2\right)$ and is parallel to line $y=\frac{1}{3}x+\frac{11}{3}$

Answer to Problem 36SGR

An equation of line in slope- intercept form is $y=-\frac{1}{2}x-\frac{7}{2}$

Explanation of Solution

Given information:

Line passes through point $\left(-5,2\right)$ and is parallel to line $y=\frac{1}{3}x+4$

Concept and Formula Used:

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

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

Parallellines: If lines are parallel then the slopes of lines are equal

Calculation:

The required line is parallel to

  $y=\frac{1}{3}x+4$

Now, find the slope of given line by compairing it with $y=mx+c$ .

The slope of line $y=\frac{1}{3}x+4$ is $m=\frac{1}{3}$ .

Let the slope of required line be ‘m’. As the required line is parallel to $y=\frac{1}{3}x+4$

Therefore, the slope of required line is equal to slope of $y=\frac{1}{3}x+4$ i.e. $m=\frac{1}{3}$

The required line passes through $\left(-5,2\right)$ and having slope $\frac{1}{3}$ . Using point slope form of equation of line

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(-5,2\right)\&m=\frac{1}{3}\\ y-2=\frac{1}{3}\left(x-\left(-5\right)\right)\\ y-2=\frac{1}{3}x+\frac{5}{3}\left(\text{Distributive property}\right)\\ \text{Add 2 on both sides}\\ y-2+2=\frac{1}{3}x+\frac{5}{3}+2\\ y=\frac{1}{3}x+\left(\frac{5+6}{3}\right)\\ y=\frac{1}{3}x+\frac{11}{3}\end{array}$

Conclusion:

An equation of line in slope- intercept form is $y=\frac{1}{3}x+\frac{11}{3}$