Chapter 4.3, Problem 14PPS

To determine

To Write: An equation of line in slope intercept form that passes through point $\left(-4,2\right)$ and is parallel to line $y=-\frac{1}{2}x+6$

Answer to Problem 14PPS

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

Explanation of Solution

Given information:

Line passes through point $\left(-4,2\right)$ and is parallel to line $y=-\frac{1}{2}x+6$

Concept and Formula Used:

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

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

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

Calculation:

The required line is parallel to

  $y=-\frac{1}{2}x+6$

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

The slope of line $y=-\frac{1}{2}x+6$ is $-\frac{1}{2}$ .

Let the slope of required line be ‘m’. As the required line is parallel to $y=-\frac{1}{2}x+6$

Therefore, the slope of required line is equal to slope of $y=-\frac{1}{2}x+6$ i.e. $m=-\frac{1}{2}$

The required line passes through $\left(-4,2\right)$ and having slope $-\frac{1}{2}$ . Using point slope form of equation of line

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(-4,2\right)\&m=-\frac{1}{2}\\ y-\left(2\right)=-\frac{1}{2}\left(x-\left(-4\right)\right)\\ y-2=-\frac{1}{2}x-2\left(\text{Distributive property}\right)\\ \text{Add 2 on both sides}\\ y-2+2=-\frac{1}{2}x-2+2\\ y=-\frac{1}{2}x\end{array}$

Conclusion:

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