Question

give an example of a line that is parallel and a line that is perpendicular to each given line. 15. 9x + 6y=-6 16. y = -5 m= parallel: parallel: m= perpendicular: perpendicular: find the x - and y - intercepts of each line, then graph the line. 17. y=-5x - 3 x - int=( , ) y - int=( , )

Explanation:

Step1: Rewrite 9x + 6y=-6 in slope - intercept form

Solve 9x + 6y=-6 for y. 6y=-9x - 6, so y =-\frac{3}{2}x - 1. The slope m of the line 9x + 6y=-6 is -\frac{3}{2}.

Step2: Find parallel and perpendicular lines for 9x + 6y=-6

A parallel line has the same slope. So a parallel line could be y =-\frac{3}{2}x+1. A perpendicular line has a slope that is the negative - reciprocal. The negative - reciprocal of -\frac{3}{2} is \frac{2}{3}, so a perpendicular line could be y=\frac{2}{3}x + 1.

Step3: Analyze y=-5

The line y=-5 is a horizontal line with slope m = 0. A parallel line is also a horizontal line, for example y = 0. A perpendicular line is a vertical line, for example x = 0.

Step4: Find x - and y - intercepts of y=-5x - 3

For the x - intercept, set y = 0. Then 0=-5x - 3, 5x=-3, x=-\frac{3}{5}. So the x - intercept is (-\frac{3}{5},0).
For the y - intercept, set x = 0. Then y=-3. So the y - intercept is (0,-3).

Answer:

For 9x + 6y=-6:
m =-\frac{3}{2}, Parallel: y =-\frac{3}{2}x + 1, m=\frac{2}{3}, Perpendicular: y=\frac{2}{3}x + 1
For y=-5:
Parallel: y = 0, Perpendicular: x = 0
For y=-5x - 3:
x - int=(-\frac{3}{5},0), y - int=(0,-3)