FINDING THE EQUATION OF A PERPENDICULAR BISECTOR

Write the equation of the line which is perpendicular bisector to each pair of points.

Problem 1 :

A(-4, -2) and (8, 4)

(i) Midpoint

(ii) Slope of AB

(iii) Perpendicular slope

(iv) Equation of the line

Solution :

A(-4, -2) and (8, 4)

(i) Midpoint = (x₁ + x₂)/2, (y₁ + y₂)/2

(x₁, y₁) ==> (-4, -2) and (x₂, y₂) ==> (8, 4)

= (-4 + 8)/2, (-2 + 4)/2

= 4/2, 2/2

= (2, 1)

(ii) Slope of AB = (y₂ - y₁)/(x₂ - x₁)

= (4 - (-2))/ (8 - (-4))

= (4 + 2) / (8 + 4)

= 6/12

= 1/2

(iii) Perpendicular slope = -1/(1/2) ==> -2

(iv) Equation of the perpendicular bisector :

(y - y₁) = m(x - x₁)

(y - 2) = -2(x - 1)

y - 2 = -2x + 2

y = -2x + 2 + 2

y = -2x + 4

Problem 2 :

A(-9, 11) and (-15,19)

(i) Midpoint

(ii) Slope of AB

(iii) Perpendicular slope

(iv) Equation of the line

Solution :

A(-9, 11) and (-15,19)

(i) Midpoint = (x₁ + x₂)/2, (y₁ + y₂)/2

(x₁, y₁) ==> (-9, 11) and (x₂, y₂) ==> (-15, 19)

= (-9 - 15)/2, (11 + 19)/2

= -24/2, 30/2

= (-12, 15)

(ii) Slope of AB = (y₂ - y₁)/(x₂ - x₁)

= (19 - 11)/ (-15 + 9)

= -8/6

= -4/3

(iii) Perpendicular slope = -1/(-4/3) ==> 3/4

(iv) Equation of the perpendicular bisector :

(y - y₁) = m(x - x₁)

(y - 15) = (3/4)(x - (-12))

4(y - 15) = 3(x + 12)

4y - 60 = 3x + 36

4y = 3x + 36 + 60

4y = 3x + 96

y = (3/4)x + 24

Problem 3 :

A(11, -5) and (1, -10)

(i) Midpoint

(ii) Slope of AB

(iii) Perpendicular slope

(iv) Equation of the line

Solution :

A(11, -5) and (1, -10)

(i) Midpoint = (x₁ + x₂)/2, (y₁ + y₂)/2

(x₁, y₁) ==> (11, -5) and (x₂, y₂) ==> (1, -10)

= (11+1)/2, (-5-10)/2

= 12/2, -15/2

= (6, -15/2)

(ii) Slope of AB = (y₂ - y₁)/(x₂ - x₁)

= (-10+5)/ (1-11)

= -5 / -10

= 1/2

(iii) Perpendicular slope = -1/(1/2) ==> -2

(iv) Equation of the perpendicular bisector :

(y - y₁) = m(x - x₁)

y + (15/2) = -2(x - 6)

(2y + 15) = -4(x - 6)

2y + 15 = -4x + 24

2y = -4x + 24 - 15

2y = -4x + 9

y = -2x + (9/2)

Problem 4 :

A(14, 18) and (-6, 10)

(i) Midpoint

(ii) Slope of AB

(iii) Perpendicular slope

(iv) Equation of the line

Solution :

A(14, -18) and (-6, 10)

(i) Midpoint = (x₁ + x₂)/2, (y₁ + y₂)/2

(x₁, y₁) ==> (14, -18) and (x₂, y₂) ==> (-6, 10)

= (14 - 6)/2, (-18 + 10)/2

= 8/2, -8/2

= (4, -4)

(ii) Slope of AB = (y₂ - y₁)/(x₂ - x₁)

= (10+18)/ (-6-14)

= 28 / -20

= -7/5

(iii) Perpendicular slope = -1/(-7/5) ==> 5/7

(iv) Equation of the perpendicular bisector :

(y - y₁) = m(x - x₁)

(y - (-4)) = (5/7)(x - 4)

7(y + 4) = 5(x - 4)

7y + 28 = 5x - 20

5x - 7y - 20 - 28 = 0

5x - 7y - 48 = 0

Problem 5 :

A triangle has vertices K(-2,-3), L(5,-7) and M(6,1).

(a) Find the equation of the perpendicular bisector of the line KM.

(b) Show that the point L lies on this line.

(c) What kind of triangle is triangle KLM?

Solution :

a)  Vertices of the triangle K(-2,-3), L(5,-7) and M(6,1).

Midpoint of KM = (-2 + 6)/2, (-3 + 1)/2

= 4/2, -2/2

= (2, -1)

Let P be the midpoint of line segment KM.

Finding the slope of the line KM :

K(-2,-3) and M(6,1).

m = (1 - (-3)) / (6 - (-2))

= (1 + 3) / (6 + 2)

= 4/8

= 1/2

Slope of the perpendicular bisector KM = -1/(1/2) ==> -2

Equation of the line LP :

(y - y₁) = m(x - x₁)

Equation of the perpendicular bisector KM :

(y - (-1)) = -2(x - 2)

y + 1 = -2(x - 2)

y + 1 = -2x + 4

y = -2x + 4 - 1

y = -2x + 3

b) Check if the point L is on the above line :

L(5, -7)

-7 = -2(5) + 3

-7 = -10 + 3

-7 = -7

Since the point L satisfies the equation, we decide it is on the line.

c) The triangle KLM must be right triangle since it has altitude and