Question

A line has a slope of - 3/5. Which ordered pairs could be points on a line that is perpendicular to this line? Check all that apply.
(–8, 8) and (2, 2)
(–5, –1) and (0, 2)
(–3, 6) and (6, –9)
(–2, 1) and (3, –2)
(0, 2) and (5, 5)

Answer 1

It's (-8,8) and (2,2)

     (-2,1) and (3,-2)

Answer 2

Answer:

(–3, 6) and (6, –9) and  (0, 2) and (5, 5)

Step-by-step explanation:

Given : Slope = $\frac{-3}{5}$.

To find: Which ordered pairs could be points on a line that is perpendicular to this line.

Solution : We have given that  Slope = $\frac{-3}{5}$.

Slope = $\frac{y_{2}-y_{1} }{x_{2}- x_{1}}$.

For (–8, 8) and (2, 2)

Slope = $\frac{2- 8 }{2 + 8}$.

Slope = $\frac{-6 }{10}$.

Slope = $\frac{-3 }{5}$.

For (–5, –1) and (0, 2).

Slope = $\frac{2+1 }{0 + 5}$.

Slope = $\frac{3 }{5}$.

For, (–3, 6) and (6, –9)

Slope = $\frac{-9- 6}{6 + 3}$.

Slope = $\frac{-15 }{9}$.

Slope = $\frac{-5}{3}$.

For (–2, 1) and (3, –2)

Slope = $\frac{-2 -1 }{3 +2}$.

Slope  = $\frac{-3 }{5}$.

For (0, 2) and (5, 5)

Slope = $\frac{5 -2 }{5 -0}$.

Slope =  $\frac{3 }{5}$.

Therefore,  (–3, 6) and (6, –9) and  (0, 2) and (5, 5)