WORD PROBLEMS ON EQUATION OF PERPENDICULAR BISECTOR

A perpendicular bisector can be defined as a line segment which intersects another line perpendicularly and divides it into two equal parts. 

To find equation of perpendicular bisector which connects following points (x₁, y₁) and (x₂, y₂).

Example 1 :

Two post offices are located at P(3, 8) and Q(7, 2) on a council map. What is the equation of the line which should from the boundary between the two regions being serviced by the post offices ?

Solution :

Step 1 :

Midpoint of PQ  =  [(x₁ + x₂)/2, (y₁ + y₂)/2]

P(3, 8)------>(x₁, y₁)

Q(7, 2)------>(x₂, y₂)

Midpoint of PQ  =  [(3 + 7)/2, (8 + 2)/2]

=  [(10/2), (10/2)]

Midpoint of PQ  =  (5, 5)

Step 2 :

Slope of the line joining the points P and Q.

m  =  (y₂–y₁)/(x₂–x₁)

m  =  (2-8)/(7 – 3)

m  =  -6/4

m  =  -3/2

Step 3 :

Slope of perpendicular bisector  =  -1/(-3/2)

=  2/3

Step 4 :

Equation of perpendicular bisector :

y–y₁  =  m (x–x₁)

y-5  =  2/3 (x–5)

3y-15  =  2x–10

2x–3y+5  =  0

So, the equation of perpendicular bisector is

2x–3y+5  =  0

Example 2  :

Recall that the perpendicular bisector of a chord of a circle passes through The center of the circle. 

A circle passes through points P(5, 7), Q(7, 1) and R(- 1, 5). Find the Perpendicular bisectors of PQ and QR and solve them simultaneously to find The center of the circle.

Solution :

P(5, 7), Q(7, 1) and R(- 1, 5)

First, we find the equation of the perpendicular bisector of PQ and QR.

Step 1 :

Midpoint of PQ  =  [(x₁+x₂)/2, (y₁+y₂)/2]

P(5, 7)------>(x₁, y₁)

Q(7, 1)------>(x₂, y₂)

Midpoint of PQ  =  [(5+7)/2, (7+1)/2]

Midpoint of PQ  =  (6, 4)

Step 2 :

Slope of PQ :

m  =  (y₂–y₁)/(x₂–x₁)

m  =  (1-7)/(7–5)

m  =  -6/2

m  =  -3

Step 3 :

Slope of perpendicular line  =  -1/(-3)  ==>  1/3

Equation of perpendicular bisector PQ :

y–y₁  =  m (x–x₁)

y-4  =  1/3 (x-6 )

3y-12  =  x–6

3y  =  x – 6 + 12

3y  =  x + 6

y  =  (1/3)x + 2-------(1)

Step 1 :

Midpoint of QR :

Q(7, 1)------>(x₁, y₁)

R(- 1, 5)------>(x₂, y₂)

Midpoint of QR  =  [(7-1)/2, (1+5)/2]

Midpoint of QR  =  (3, 3)

Step 2 :

Slope of QR :

m  =  (y₂ – y₁)/(x₂ – x₁)

m  =  (5 - 1)/(- 1 – 7)

m  =  - 4/8

m  =  - 1/2

Step 3 :

Slope of perpendicular(-1/m)  =  2

Step 4 :

y–y₁  =  m (x–x₁)

y-3  =  2 (x-3 )

y-3  =  2x–6

y  =  2x - 3

y  =  2x – 3 -------(2)

Now, To find the center of the circle,

By equating (1) and (2), we get

(1)  =  (2)

1/3x + 2  =  2x – 3

x + 6  =  6x – 9

x – 6x  =  - 9 – 6

- 5x  =  - 15

x  =  3

By applying x value in (2), we get

y  =  2x – 3

y  =  2(3) – 3

y  =  3

So, x  =  3 and y  =  3

Therefore, the center of the circle is (3, 3). 

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