EQUATION OF PERPENDICULAR BISECTOR

Example 1 :

Find the equation of perpendicular bisector of the line joining the points A(-4, 2) and B(6, -4) in general form.

Solution :

Let E be the mid point and CD be the perpendicular bisector of AB. 

Mid point of AB : 

= E((x₁ + x₂)/2, (y₁ + y₂)/2)

Substitute (x₁, y₁) = A(-4, 2) and (x₂, y₂) = B(6, -4).

= E((-4 + 6)/2, (2 - 4)/2)

= E(2/2, -2/2)

= E(1, -1)

Slope of AB : 

= (y₂ - y₁)/(x₂ - x₁)

Substitute (x₁, y₁) = A(-4, 2) and (x₂, y₂) = B(6, -4).

= (-4 - 2)/(6 + 4)

= -6/10

= -3/5

Slope of AD : 

= -1/slope of AB

= -1/(-3/5)

= 5/3

Equation of the perpendicular bisector CD :  

y = mx + b

Substitute m = 5/3.

y = (5/3)x + b ----(1)

Substitute (x, y) = E(1, -1). 

-1 = (5/3)(1) + b

-1 = 5/3 + b

-1 - 5/3 = b

-8/3 = b

(1)----> y = (5/3)x - 8/3

Multiply each side by 5.

3y = 5x - 8

-5x + 3y + 8 = 0

5x - 3y - 8 = 0

Example 2 :

Write an equation of the line in y = mx + b form that is the perpendicular bisector of the line segment having endpoints of (1, 2) and (2, 4)

Solution :

Let the given points be A(1, 2) and B(2, 4)

Mid point of AB : 

= ((x₁ + x₂)/2, (y₁ + y₂)/2)

Substitute (x₁, y₁) = A(1, 2) and (x₂, y₂) = B(2, 4).

= ((1 + 2)/2, (2 + 4)/2)

= (3/2, 6/2)

= (3/2, 3)

Slope of AB : 

= (y₂ - y₁)/(x₂ - x₁)

Substitute (x₁, y₁) = A(1, 2) and (x₂, y₂) = B(2, 4).

= (4 - 2)/(2 - 1)

= 2/1

= 2

Slope of perpendicular bisector : 

= -1/slope of AB

= -1/2

Equation of perpendicular bisector :

y - y₁ = m(x - x₁)

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

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

4(y - 3) = -1(2x - 3)

4y - 12 = -2x + 3

2x + 4y - 12 - 3 = 0

2x + 4y - 15 = 0

2x + 4y = 15

4y = -2x + 15

y = (-1/2)x + (15/4)

Example 3 :

The straight line p has the equation 3x − 4y + 8 = 0. The straight line q is parallel to p and passes through the point with coordinates (8, 5).

a Find the equation of q in the form y = mx + c. The straight line r is perpendicular to p and passes through the point with coordinates (−4, 6).

b Find the equation of r in the form ax + by + c = 0, where a, b and c are integers.

c Find the coordinates of the point where lines q and r intersect.

Solution :

a) The lines p and q are parallel to each other, then their slopes will be equal.

3x − 4y + 8 = 0

4y = 3x + 8

y = (3/4)x + (8/4)

y = (3/4)x + 2

Slope of the line q = 3/4

Line q is passing through the point (8, 5).

Equation of line q :

y - y1 = m(x - x1)

y - 5 = (3/4)(x - 8)

4(y - 5) = 3(x - 8)

4y - 20 = 3x - 24

3x - 4y - 24 + 20 = 0

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

b) Line r is perpendicular to p

Slope of the line r = -1/(3/4)

= -4/3

Line r passes through the point (-4, 6)

y - 6 = (-4/3)(x - (-4))

3(y - 6) = -4(x + 4)

3y - 18 = -4x - 16

4x + 3y - 18 + 16 = 0

4x + 3y = 2 -----(2)

c)  3x - 4y = 4 and 4x + 3y = 2

(1) x 3 ==> 9x - 12y = 12

(2) x 4 ==> 16x + 12y = 8

25x = 20

x = 20/25

x = 4/5

Applying x = 4/5 in (1), we get

3(4/5) - 4y = 4

(12/5) - 4y = 4

4y = (12/5) - 4

4y = -8/5

y = -2/5

So, the point of intersection is (4/5, -2/5)

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