**Find the Equation of Perpendicular Bisector :**

Here we are going to see how to find the equation of perpendicular bisector.

**Question 10 :**

Find the equation of the perpendicular bisector of the straight line segment joining the points (3,4) and (-1,2)

**Solution :**

midpoint of the line segment joining the points (3,4) and (-1,2)

x₁ = 3, y₁ = 4, x₂ = -1 , y₂ = 2

= [(x₁ + x₂)/2 , (y₁ + y₂)/2]

= (3 + (-1))/2 , (4 + 2)/2

= (2/2, 6/2)

= (1 , 3)

slope of the line joining the points (3,4) and (-1,2)

m = (y₂ - y₁)/(x₂ - x₁)

= (2 - 4)/(-1-3)

= -2/(-4)

= 1/2

Slope of the required line = -1/(1/2)

= -2

Equation of the required line :

(y - y₁)/(y₂ - y₁) = (x - x₁)/(x₂ - x₁)

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

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

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

- 4 y + 16 = - 2 x + 6

2 x -4 y + 16 - 6 = 0

2 x - 4 y + 10 = 0

dividing the whole equation by 2, we get

x - 2 y - 5 =0

**Question 11 :**

Find the equation of the straight line passing through the point of intersection of the lines 2x+y-3=0 and 5x + y - 6 = 0 and parallel to the line joining the points (1,2) and (2,1)

**Solution :**

To find the point of intersection of any two lines we need to solve them

2x + y - 3 = 0 ---- (1)

5x + y - 6 = 0 ---- (2)

(1) - (2)

2x + y - 3 = 0

5x + y - 6 = 0

(-) (-) (+)

-----------------

- 3x + 3 = 0

- 3 x = -3

x = 1

Substitute x = 1 in the first equation

2(1) + y - 3 = 0

-1 + y = 0

y = 1

the point of intersection is (1,1)

The required line is parallel to the line joining the points (1, 2) and (2, 1). So their slopes will be equal.

m = (y₂ - y₁)/(x₂ - x₁)

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

= -1/1 = -1

Slope of the required line is -1 and a point on the line is (1, 1)

Equation of the Line :

(y - y₁) = m (x - x₁)

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

y - 1 = - x + 1

x + y - 1 - 1 = 0

x + y - 2 = 0

After having gone through the stuff given above, we hope that the students would have understood, how to find the equation of perpendicular bisector.

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