HOW TO FIND EQUATION OF PERPENDICULAR BISECTOR 

Problem 1 :

Find the equation of the perpendicular bisector of P(9, 5) and Q(-1, 3).

Solution :

Midpoint of the line segment PQ :

Midpoint = (x₁ + x₂)/2, (y₁ + y₂)/2

= (9 + (-1))/2, (5 + 3)/2

= 8/2, 8/2

= (4, 4)

Slope of the line joining the points P and Q :

m = (y₂ - y₁) / (x₂ - x₁)

m = (3 - 5)/(-1 - 9) 

m = -2/-10

m = 1/5

Slope of the perpendicular line = -1/(1/5) = -5

The perpendicular bisector is also a line which is perpendicular to the line joining the points P and Q and it will pass through the point (4, 4).

(y - y₁) = m(x - x₁)

y - 4 = -5(x - 4)

y - 4 = -5x + 20

y = -5x + 24

Problem 2 :

Given that the triangle QRS have vertices (-3, 1), (-4, -2) and (7, 0).Find the equation of the perpendicular bisector of RS.

Solution :

Let the vertices be Q(-3, 1), R(-4, -2) and S(7, 0)

Midpoint of the line joining the points RS :

midpoint = (x₁ + x₂)/2, (y₁ + y₂)/2

= (-4 + 7)/2, (-2 + 0)/2

= 3/2, -2/2

= (3/2, -1)

Slope of the line RS :

m = (0 + 2)/(7 + 4)

m = 2/11

Slope of the perpendicular bisector = -11/2

Equation of perpendicular bisector :

(y + 1) = (-11/2)(x - (3/2))

2(y + 1) = -11(x - (3/2))

2y + 2 = -11x + 33/2

2y = -11x + (33/2) - 2

2y = -11x + 29/2

y = (-11/2)x + (29/4)