Example 1 :
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 + 8 = 0
Example 2 :
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 = -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
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