Science > Mathematics > Pair of Straight Lines > Joint Equation of Pair of Lines Perpendicular To Another Pair of Lines
In this article, we shall study to find a joint equation of lines perpendicular to another pair of lines.
Note: If m1 and m2 are the slopes of the two lines represented by joint equation ax2 + 2hxy + by2 = 0. Then m1. m2 = a/b ; m1 + m2 = -2h/b
Algorithm:
- Write given joint equation.
- Compare with the standard equation.
- Find values of a, h and b.
- let m1 and m2 be the slopes of the lines represented by the given joint equation.
- Find values of m1 + m2 and m1.m2
- Required lines are perpendicular to given lines. Hence slopes of the required lines are – 1/m1 and – 1/m2.
- Write equations of required lines in the form u = 0 and v = 0.
- Find u.v = 0
- Simplify and write the joint equation of the line in standard form
Example 01:
- Find the joint equation of pair of lines through the origin and perpendicular to the line pair 5x2 – 8xy + 3y2 = 0.
- Solution:
Given joint equation is 5x2 – 8xy + 3y2 = 0.
Comparing with ax2 + 2hxy + by2 = 0
a = 5, 2h = -8, and b = 3.
Now let m1 and m2 be the slopes of the lines represented by given joint equation.
m1 + m2 = -2h/b = -(-8)/3 = 8/3
and m1. m2 = a/b = 5/3
Now since the required lines are perpendicular to the lines represented by given joint equation,
their slopes are – 1/m1 and – 1/m2
Equation of first required line is
y = – 1/m1 x
∴ m1y = – x
∴ x + m1y = 0 ……. (1)
and similarly equation second required line is
y = – 1/m2 x
∴ m2y = – x
∴ x + m2y = 0 …. (2)
combined equation is
(x + m1y)( x + m2y) = 0
∴ x ( x + m2y) + m1y ( x + m2y) = 0
∴ x2 + m2xy + m1xy + m1m2y2 = 0
∴ x2 + (m1 + m2)xy + m1m2y2 = 0
∴ x2 + (8/3)xy + (5/3)y2 = 0
Multiplying both sides by 3
3x2 + 8xy + 5y2 = 0
This is the required joint equation of pair of lines.
Example 02:
- Find the joint equation of pair of lines through origin and perpendicular to the line pair x2 – xy + 2y2 = 0
- Solution:
Given joint equation is x2 – xy + 2y2 = 0
Comparing with ax2 + 2hxy + by2 = 0
a = 1, 2h = -1 and b = 2.
Now let m1 and m2 be the slopes of the lines represented by given joint equation.
m1 + m2 = -2h/b = – (-1)/2 = 1/2
and m1. m2 = a/b = 1/2
Now since the required lines are perpendicular to the lines represented by given joint equation,
their slopes are – 1/m1 and – 1/m2
Equation of first required line is
y = – 1/m1 x
∴ m1y = – x
∴ x + m1y = 0 ……. (1)
and similarly equation second required line is
y = – 1/m2 x
∴ m2y = – x
∴ x + m2y = 0 …. (2)
combined equation is
(x + m1y)( x + m2y) = 0
∴ x ( x + m2y) + m1y ( x + m2y) = 0
∴ x2 + m2xy + m1xy + m1m2y2 = 0
∴ x2 + (m1 + m2)xy + m1m2y2 = 0
∴ x2 + (1/2)xy + (1/2)y2 = 0
Multiplying both sides by 2
2x2 + xy + y2 = 0
This is the required joint equation of pair of lines.
Example 03:
- Find the joint equation of pair of lines through origin and perpendicular to the line pair ax2 + 2hxy + by2 = 0
- Solution:
Given joint equation is ax2 + 2hxy + by2 = 0
Now let m1 and m2 be the slopes of the lines represented by given joint equation.
m1 + m2 = -2h/b and m1. m2 = a/b
Now since the required lines are perpendicular to the lines represented by given joint equation,
their slopes are – 1/m1 and – 1/m2
Equation of first required line is
y = – 1/m1 x
∴ m1y = – x
∴ x + m1y = 0 ……. (1)
and similarly equation second required line is
y = – 1/m2 x
∴ m2y = – x
∴ x + m2y = 0 …. (2)
combined equation is
(x + m1y)( x + m2y) = 0
∴ x ( x + m2y) + m1y ( x + m2y) = 0
∴ x2 + m2xy + m1xy + m1m2y2 = 0
∴ x2 + (m1 + m2)xy + m1m2y2 = 0
∴ x2 + (-2h/b)xy + (a/b)y2 = 0
Multiplying both sides by b
bx2 – 2hxy + ay2 = 0
This is the required joint equation of pair of lines.
Note:
- This result can be directly used in competitive exams
- To find the combined equation of the pair of lines through origin and perpendicular to the line pair 5x2 – 8xy + 3y2 = 0.
Here a = 5, 2h = -8 and b = 3
Hence answer is 3x2 + 8xy + 5y2 = 0
Example 04:
- Find the joint equation of pair of lines through origin and perpendicular to the line pair 2x2 – 8xy + 3y2 = 0.
- Solution:
Given joint equation is 2x2 – 8xy + 3y2 = 0
Comparing with ax2 + 2hxy + by2 = 0
a = 2, 2h = – 8 and b = 3.
Now let m1 and m2 be the slopes of the lines represented by given joint equation.
m1 + m2 = -2h/b = – (-8)/3 = 8/3
and m1. m2 = a/b = 2/3
Now since the required lines are perpendicular to the lines represented by given joint equation,
their slopes are – 1/m1 and – 1/m2
Equation of first required line is
y = – 1/m1 x
∴ m1y = – x
∴ x + m1y = 0 ……. (1)
and similarly equation second required line is
y = – 1/m2 x
∴ m2y = – x
∴ x + m2y = 0 …. (2)
combined equation is
(x + m1y)( x + m2y) = 0
∴ x ( x + m2y) + m1y ( x + m2y) = 0
∴ x2 + m2xy + m1xy + m1m2y2 = 0
∴ x2 + (m1 + m2)xy + m1m2y2 = 0
∴ x2 + (8/3)xy + (2/3)y2 = 0
Multiplying both sides by 3
3x2 + 8xy + 2y2 = 0
This is the required joint equation of pair of lines.
Example 05:
- Find the joint equation of pair of lines through origin and perpendicular to the line pair 5x2 + 2xy – 3y2 = 0.
- Solution:
Given joint equation is 2x2 – 8xy + 3y2 = 0
Comparing with ax2 + 2hxy + by2 = 0
a = 5, 2h = 2 and b = -3.
Now let m1 and m2 be the slopes of the lines represented by given joint equation.
m1 + m2 = -2h/b = – (2)/-3 = 2/3
and m1. m2 = a/b = 5/-3 = – 5/3
Now since the required lines are perpendicular to the lines represented by given joint equation,
their slopes are – 1/m1 and – 1/m2
Equation of first required line is
y = – 1/m1 x
∴ m1y = – x
∴ x + m1y = 0 ……. (1)
and similarly equation second required line is
y = – 1/m2 x
∴ m2y = – x
∴ x + m2y = 0 …. (2)
combined equation is
(x + m1y)( x + m2y) = 0
∴ x ( x + m2y) + m1y ( x + m2y) = 0
∴ x2 + m2xy + m1xy + m1m2y2 = 0
∴ x2 + (m1 + m2)xy + m1m2y2 = 0
∴ x2 + (2/3)xy + (-5/3)y2 = 0
Multiplying both sides by 3
3x2 + 2xy – 5y2 = 0
This is the required joint equation of pair of lines.
Example 06:
- Find the joint equation of pair of lines through origin and perpendicular to the line pair x2 + 4xy – 5y2 = 0.
- Solution:
Given joint equation is 2x2 – 8xy + 3y2 = 0
Comparing with ax2 + 2hxy + by2 = 0
a = 1, 2h = 4 and b = -5.
Now let m1 and m2 be the slopes of the lines represented by given joint equation.
m1 + m2 = -2h/b = – (4)/-5 = 4/5
and m1. m2 = a/b = 1/-5 = – 1/5
Now since the required lines are perpendicular to the lines represented by given joint equation,
their slopes are – 1/m1 and – 1/m2
Equation of first required line is
y = – 1/m1 x
∴ m1y = – x
∴ x + m1y = 0 ……. (1)
and similarly equation second required line is
y = – 1/m2 x
∴ m2y = – x
∴ x + m2y = 0 …. (2)
combined equation is
(x + m1y)( x + m2y) = 0
∴ x ( x + m2y) + m1y ( x + m2y) = 0
∴ x2 + m2xy + m1xy + m1m2y2 = 0
∴ x2 + (m1 + m2)xy + m1m2y2 = 0
∴ x2 + (4/5)xy + (-1/5)y2 = 0
Multiplying both sides by 5
5x2 + 4xy – y2 = 0
This is the required joint equation.
Example 07:
- Find the joint equation of pair of lines through origin and perpendicular to the line pair 2x2 – 3xy – 9y2 = 0.
- Solution:
Given joint equation is 2x2 – 8xy + 3y2 = 0
Comparing with ax2 + 2hxy + by2 = 0
a = 2, 2h = -3 and b = – 9.
Now let m1 and m2 be the slopes of the lines represented by given joint equation.
m1 + m2 = -2h/b = – (-3)/-9 = -1/3
and m1. m2 = a/b = 2/-9 = – 2/9
Now since the required lines are perpendicular to the lines represented by given joint equation,
their slopes are – 1/m1 and – 1/m2
Equation of first required line is
y = – 1/m1 x
∴ m1y = – x
∴ x + m1y = 0 ……. (1)
and similarly equation second required line is
y = – 1/m2 x
∴ m2y = – x
∴ x + m2y = 0 …. (2)
combined equation is
(x + m1y)( x + m2y) = 0
∴ x ( x + m2y) + m1y ( x + m2y) = 0
∴ x2 + m2xy + m1xy + m1m2y2 = 0
∴ x2 + (m1 + m2)xy + m1m2y2 = 0
∴ x2 + (-1/3)xy + (-2/9)y2 = 0
Multiplying both sides by 9
9x2 – 3xy – 2y2 = 0
This is the required joint equation of pair of lines.
Example 08:
- Find the joint equation of pair of lines through origin and perpendicular to the line pair x2 + xy – y2 = 0.
- Solution:
Given joint equation is 2x2 – 8xy + 3y2 = 0
Comparing with ax2 + 2hxy + by2 = 0
a = 1, 2h = 1 and b = -1.
Now let m1 and m2 be the slopes of the lines represented by given joint equation.
m1 + m2 = -2h/b = – 1/-1 = 1
and m1. m2 = a/b = 1/-1 = -1
Now since the required lines are perpendicular to the lines represented by given joint equation,
their slopes are – 1/m1 and – 1/m2
Equation of first required line is
y = – 1/m1 x
∴ m1y = – x
∴ x + m1y = 0 ……. (1)
and similarly equation second required line is
y = – 1/m2 x
∴ m2y = – x
∴ x + m2y = 0 …. (2)
combined equation is
(x + m1y)( x + m2y) = 0
∴ x ( x + m2y) + m1y ( x + m2y) = 0
∴ x2 + m2xy + m1xy + m1m2y2 = 0
∴ x2 + (m1 + m2)xy + m1m2y2 = 0
∴ x2 + (1)xy + (- 1)y2 = 0
x2 + xy – y2 = 0
This is the required joint equation of pair of lines.