Chapter 4.3, Problem 11PPS

To determine

To Write: An equation of line in slope intercept form that passes through point $\left(3,-2\right)$ and is parallel to line $y=x+4$

Answer to Problem 11PPS

An equation of line in slope- intercept form is $y=x-5$

Explanation of Solution

Given information:

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

Concept and Formula Used:

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

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

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

Calculation:

The required line is parallel to

  $y=x+4$

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

The slope of line $y=x+4$ is $1$ .

Let the slope of required line be ‘m’. As the required line is parallel to $y=x+4$

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

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

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

Conclusion:

An equation of line in slope- intercept form is $y=x-5$