FIND EQUATION OF PERPENDICULAR BISECTOR WITH TWO POINTS

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).

Find the equation of the perpendicular bisector of AB for :

Example 1 :

A(3, - 3) and B(1, - 1)

Solution :

Step 1 :

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

A(3, - 3)------>(x₁, y₁)

B(1, - 1)------>(x₂, y₂)

Midpoint of AB  =  [(3+1)/2, (-3-1)/2]

Midpoint of AB  =  (2, - 2)

Step 2 :

Slope of the line joining the point A and B.

m  =  (y₂–y₁)/(x₂–x₁)

m  =  (-1+3)/(1–3)

m   =  - 2/2

m  =  - 1

Step 3 :

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

=  1

Step 4 :

Equation of the perpendicular bisector of AB :

y–y₁  =  m(x–x₁)

y+2  =  1(x–2)

y+2  =  x–2

x–y  =  4

So, equation of perpendicular bisector is x-y  =  4.

Example 2 :

A(1, 3) and B(- 3, 5)

Solution :

Step 1 :

A(1, 3)------>(x₁, y₁)

B(- 3, 5)------>(x₂, y₂)

Midpoint of AB  =  [(1-3)/2, (3+5)/2]

Midpoint of AB  =  (- 1, 4)

Step 2 :

Slope of the line joining the point A and B.

m  =  (y₂ – y₁)/(x₂ – x₁)

m  =  (5 - 3)/(- 3 – 1)

m  =  - 2/4

m  =  - 1/2

Step 3 :

Slope of perpendicular line  =  -1/(-1/2)

=  2

Step 4 :

Equation of the perpendicular bisector of AB :

y–y₁  =  m (x–x₁)

y-4  =  2 (x+1 )

y-4  =  2x+2

2x–y+2+4  =  0

2x–y+6  =  0

So, equation of perpendicular bisector is 2x–y+6  =  0

Example 3 :

A(3, 1) and B(- 3, 6)

Solution :

Step 1 :

A(3, 1)------>(x₁, y₁)

B(- 3, 6)------>(x₂, y₂)

Midpoint of AB  =  [(3 - 3 )/2, (1 + 6)/2]

=  [(0/2), (7/2)]

Midpoint of AB  =  (0, 7/2)

Step 2 :

Slope of the line joining the point A and B.

m  =  (y₂ – y₁)/(x₂ – x₁)

m  =  (6 - 1)/(- 3 – 3)

m  =  - 5/6

m  =  - 5/6

Step 3 :

Slope of perpendicular line  =  -1/(-5/6)

=  6/5

Step 4 :

Equation of the perpendicular bisector of AB :

y–y₁  =  m (x–x₁)

y–7/2  =  6/5 (x–0)

(2y–7)/2  =  (6/5) (x)

5(2y–7)  =  12x

10y–35  =  12x

12x–10y+35  =  0

12x–10y  =  -35

So, equation of perpendicular bisector is 12x–10y  =  -35.

Example 4 :

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

Solution :

Step 1 :

A(4, - 2)------>(x₁, y₁)

B(4, 4)------>(x₂, y₂)

Midpoint of AB  =  [(4+4)/2, (-2+4)/2]

Midpoint of AB  =  (4, 1)

Step 2 :

Slope of the line joining the point A and B.

m  =  (y₂–y₁)/(x₂–x₁)

m  =  (4+2)/(4–4)

m  =  6/0

m  =  undefined

Step 3 :

Slope of perpendicular line  =  -1/undefined

=  0

Step 4 :

y–y₁  =  m (x–x₁)

y - 1  =  0 (x–4)

y - 1  =  0

y  =  1

So, the required equation is y  =  1.

Example 5 :

A is the point (-2,-6) and B is (4,-4). Find the equation of the perpendicular bisector of the line AB

Solution :

Step 1 :

A(-2, -6)------>(x₁, y₁)

B(4, -2)------>(x₂, y₂)

Midpoint of AB  =  [(-2+ 4)/2, (-6-2)/2]

= (2/2, -8/2)

= (1, -4)

Midpoint of AB  =  (1, -4)

Step 2 :

Slope of the line joining the point A and B.

m  =  (y₂–y₁)/(x₂–x₁)

m  =  (-2-6)/(4+2)

m  =  -8/6

m  =  -4/3

Step 3 :

Slope of perpendicular line  =  -1/(-4/3)

=  3/4

Step 4 :

y–y₁  =  m (x–x₁)

y - (-4) =  (3/4)(x – 1)

y + 1  =  (3/4)(x - 1)

4(y + 1) = 3(x - 1)

4y + 4 = 3x - 3

3x - 4y - 3 - 4 = 0

3x - 4y - 7 = 0

So, the required equation is 3x - 4y = 7

Example 6 :

A triangle DEF has vertices D(-1,-1), E(3,8) and F(11,3). Find the equation of the perpendicular bisector of the line DF.

Solution :

Midpoint of D and F, D(-1, -1) and F(11, 3)

= (-1 + 11)/2, (-1 + 3)/2

= 10/2, 2/2

= (5, 1)

Slope of the line DF :

= (3 - (-1))/(11 - (-1))

= (3 + 1) / (11 + 1)

= 4/12

= 1/3

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

= -3

Equation of the perpendicular bisector :

y – y₁ = m (x – x₁)

y – 1 = -3 (x – 5)

y = -3x + 15 + 1

y = -3x + 16

So, the equation of perpendicular bisector is y = -3x + 16.

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