Chapter 4, Problem 33SGR

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 $y=-\frac{1}{2}x+4$

Answer to Problem 33SGR

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 $y=-\frac{1}{2}x+4$

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

  $y=-\frac{1}{2}x+4$

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

The slope of line $y=-\frac{1}{2}x+4$ is $m=-\frac{1}{2}$ .

Let the slope of required line be ‘m’. As the required line is parallel to $y=-\frac{1}{2}x+4$

Therefore, the slope of required line is equal to slope of $y=-\frac{1}{2}x+4$ i.e. $m=-\frac{1}{2}$

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

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(-5,-2\right)\&m=-\frac{1}{2}\\ y-\left(-2\right)=-\frac{1}{2}\left(x-\left(-5\right)\right)\\ y+2=-\frac{1}{2}x-\frac{5}{2}\left(\text{Distributive property}\right)\\ \text{Subtract 2 from both sides}\\ y+2-2=-\frac{1}{2}x-\frac{5}{2}-2\\ y=-\frac{1}{2}x-\left(\frac{5+2}{2}\right)\\ y=-\frac{1}{2}x-\frac{7}{2}\end{array}$

Conclusion:

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