Chapter 4.3, Problem 11PPS

To determine

To find: The equation of the straight line in slope-intercept form which passes through the point $\left(3,-2\right)$ and is parallel to the graph of line $y=x+4$ .

Answer to Problem 11PPS

Equation of line is $y=x-5$ .

Explanation of Solution

Given information: The given line is $y=x+4$ . And, another line is parallel to this line which passes through the point $\left(3,-2\right)$ .

Concept used: If a line is parallel to another non vertical line then there slope are defined and are also equal.

Calculation: Since, parallel non-vertical lines have equal slope. Therefore, line parallel to line $y=x+4$ , will have slope $\left(m=1\right)$ .

Now, equation of line using point-slope form parallel to line $y=x+4$ passing through the point $\left(3,-2\right)$ is evaluated as,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\mathrm{......}\left(\text{Using point-slope form}\right)\\ y+2=1\left(x-3\right)\mathrm{......}\left(\begin{array}{l}\text{Here, }m=1\text{ and}\\ \left(x_1,y_1\right)\to \left(3,-2\right)\end{array}\right)\\ y+2=x-3\\ y=x-\mathrm{5......}\left(\text{Slope-intercept form}\right)\end{array}$