Question

A straight line passed through P(-4,8) and Q(0-2). Find the equation of the perpendicular

bisector of the line PQ​

Answer 1

y = 2/5x + 11/5

Explanation:

midpoint:

`((x1 + x2)/(2), (y1 + y2)/(2) )`

P(-4 , 8)

Q(0 , -2)

[(-4+0)/2 , (8-2)/2]

=(-4/2 , 6/2)

=(-2 , 3)

gradient:

`m = (y1 - y2)/(x1 - x2)`

m = (8+2)/(-4-0)

= 10/(-4)

= -5/2

= 2/5 (opp. of reciprocal)

equation:

y = mx + c

y = 2/5x + c

3 = 2/5(-2) + c (using coordinate of midpoint)

c = 3 + (-4/5)

= 11/5

y = 2/5x + 11/5

(Correct me if i am wrong)