# 10th GRADE MATH SOLUTION EXERCISE 5.5 part3

This page 10th grade math solution exercise 5.5 part3 is going to provide you solution for every problems that you find in the exercise no 5.5

## 10th grade math solution exercise 5.5 part3

(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

= [(x + x)/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

(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

In the page10th grade math solution exercise 5.5 part3 we are going to see the solution of next problem

(12) Find the equation of the straight line which passes through the point of intersection of the straight lines 5x - 6y = 1 and 3x + 2y + 5 = 0 and is perpendicular to the straight
line 3x - 5y + 11 = 0.

Solution:

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

5x - 6y = 1  --(1)

3x + 2y = -5 ----(2)

5x - 6y = 1

(2) x 3 =>   9x + 6y = -15

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

14 x = -14

x = -1

substitute x = -1 in the first equation

5 (-1) -6 y = 1

-5  - 6y = 1

- 6 y = 1 + 5

- 6y = 6

y = -1

Point of intersection of those two lines is (-1,-1)

Slope of the perpendicular line 3x - 5y + 11 = 0

m = -3/(-5)

= 3/5

slope of the required line = -1/m

= -1/(3/5)

= -5/3

Equation of the line :

(y - y₁) = m (x - x₁)

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

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

3 y + 3 = - 5 x - 5

5 x + 3 y - 5 + 3 = 0

5 x + 3y - 2 = 0

(13) Find the equation of the straight line joining the point of intersection of the lines 3x – y + 9 = 0 and x + 2y = 4 and the point intersection of the lines 2x + y – 4 = 0 and x – 2y + 3 = 0

Solution:

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

3x – y + 9 = 0 ---- (1)

x + 2y - 4 = 0 ---- (2)

(1) - (2)

3x – y + 9 = 0

(2) x 3   3x + 6y - 12 = 0

(-)  (-)  (+)

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

- 7y + 21 = 0

7y = 21

y = 3

Substitute y = 3 in the first equation

3 x - 3 + 9 = 0

3 x + 6 = 0

3 x = - 6

x = -6/3

x = -2

the point of intersection is (-2,3)

2x + y – 4 = 0 ---- (3)

x – 2y + 3 = 0 ---- (4)

(1) - (2)

2x + y - 4 = 0

(2) x 2   2x - 4y + 6 = 0

(-)    (+)  (-)

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

5 y - 10 = 0

5y = 10

y = 2

Substitute y = 2 in the third equation

2 x + y - 4 = 0

2 x + 2 - 4 = 0

2 x - 2 = 0

2 x = 2

x = 1

the point of intersection is (1,2)

Equation of the line:

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

(-2,3) (1,2)

x₁ = -2, y₁ = 3, x₂ = 1 , y₂ = 2

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

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

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

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

3 y - 9 = - x - 2

x + 3 y - 9 + 2 = 0

x + 3y - 7 = 0