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:
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
Slope of PQ is
[tex]m_{PQ}=\dfrac{-10-(-2)}{5-12}=\dfrac{-8}{-7}=\dfrac{8}{7}[/tex]
Slope of RS is
[tex]m_{RS}=\dfrac{y-10}{4-(-4)}=\dfrac{y-10}{8}[/tex]
The product of slopes of two perpendicular lines is -1.
[tex]m_{PQ}\times m_{RS}=-1[/tex]
[tex]\dfrac{8}{7}\times \dfrac{y-10}{8}=-1[/tex]
[tex]\dfrac{y-10}{7}=-1[/tex]
[tex]y-10=-7[/tex]
Add 10 on both sides.
[tex]y=-7+10[/tex]
[tex]y=3[/tex]
Therefore, the value of y is 3.
