Question

Given P(12,-2), Q(5,-10), R(-4,10) and S(4,y), find the value of y so that line PQ is perpendicular to line RS

Answer 1

Answer:

The value of y is 3.

Step-by-step explanation:

Given points: P(12,-2), Q(5,-10), R(-4,10) and S(4,y).

We need to find the value of y so that line PQ is perpendicular to line RS.

Slope formula:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

Slope of PQ is

$m_{PQ}=\dfrac{-10-(-2)}{5-12}=\dfrac{-8}{-7}=\dfrac{8}{7}$

Slope of RS is

$m_{RS}=\dfrac{y-10}{4-(-4)}=\dfrac{y-10}{8}$

The product of slopes of two perpendicular lines is -1.

$m_{PQ}\times m_{RS}=-1$

$\dfrac{8}{7}\times \dfrac{y-10}{8}=-1$

$\dfrac{y-10}{7}=-1$

$y-10=-7$

Add 10 on both sides.

$y=-7+10$

$y=3$

Therefore, the value of y is 3.