Science > Mathematics > Pair of Straight Lines > Use of Auxiliary Equation of Pair of Lines
In this article, we shall study the use of the auxiliary equation of par of lines to find the required condition.
Algorithm:
- Write the auxiliary equation of the joint equation.
- Find the slope of given line m.
- Substitute value of m in auxiliary equation.
- Simplify and get the value of constant.
Example 01:
- Find the value of ‘k’ if one of the line given by 6x2 + kxy + y2 = 0 is 2x + y = 0.
- Solution:
Given joint equation of lines is 6x2 + kxy + y2 = 0
It is in the form ax2 + 2hxy + by2 = 0.
a = 6, 2h = k and b= 1
The auxiliary equation of given line is of the form
bm2 + 2hm + a = 0
∴ (1)m2 + km + 6 = 0
∴ m2 + km + 6 = 0 ……… (1)
One of the line is 2x + y = 0. Its slope is – 2/1 = -2
Now – 2 must be one of the roots of the auxiliary equation (1),
Substituting m = – 2 in equation (1)
(-2)2 + k(-2) + 6 = 0
∴ 4 – 2k + 6 = 0
∴ – 2k = – 10
∴ k = 5
Example 02:
- Find λ, if 2x + 5y = 0 coincides with one of the lines x2 – λxy + 5y2 = 0.
- Solution:
Given joint equation of lines is x2 – λxy + 5y2 = 0
It is in the form ax2 + 2hxy + by2 = 0.
a = 1, 2h = – λ and b= 5
The auxiliary equation of given line is of the form
bm2 + 2hm + a = 0
∴ (5)m2 – λm + 1 = 0
∴ 5m2 – λm + 1 = 0 ……… (1)
One of the line is 2x + 5y = 0. Its slope is – 2/5
Now – 2/5 must be one of the roots of the auxiliary equation (1),
Substituting m = – 2/5 in equation (1)
5(-2/5)2 – λ(-2/5) + 1 = 0
∴ 5(4/25) – λ(-2/5) + 1 = 0
∴ (4/5) – λ(-2/5) + 1 = 0
Multiplying both sides of equation by 5
∴ 4 + 2λ + 5 = 0
∴ 2λ = – 9
∴ λ = – 9/2
Example 03:
- Find k, if the one of the lines given by 2x2 – xy + ky2 = 0 is x – 3y = 0 .
- Solution:
Given joint equation of lines is 2x2 – xy + ky2 = 0
It is in the form ax2 + 2hxy + by2 = 0.
a = 2, 2h = – 1 and b= k
The auxiliary equation of given line is of the form
bm2 + 2hm + a = 0
∴ (k)m2 – 1m + 2 = 0
∴ km2 – m + 2 = 0 ……… (1)
One of the line is x – 3y = 0. Its slope is – 1/-3 = 1/3
Now 1/3 must be one of the roots of the auxiliary equation (1),
Substituting m = 1/3 in equation (1)
k(1/3)2 – (1/3) + 2 = 0
∴ K(1/9) – 1/3 + 2 = 0
Multiplying both sides of equation by 9
∴ k – 3 + 18 = 0
∴ k = -15
Example 04:
- Find k, if the one of the lines given by kx2 + 3xy – y2 = 0 is 2x + y = 0
- Solution:
Given joint equation of lines is kx2 + 3xy – y2 = 0
It is in the form ax2 + 2hxy + by2 = 0.
a = k, 2h = 3 and b = – 1
The auxiliary equation of given line is of the form
bm2 + 2hm + a = 0
∴ (-1)m2 + 3m + k = 0
∴ m2 – 3m – k = 0 ……… (1)
One of the line is 2x + y = 0. Its slope is – 2/1 = -2
Now -2 must be one of the roots of the auxiliary equation (1),
Substituting m = -2 in equation (1)
(-2)2 – 3(-2) – k = 0
∴ 4 + 6 – k = 0
∴ k = 10
Example 05:
- Find k, if the one of the lines given by 6x2 – 14xy + 14ky2 = 0 coincides with y = 2x
- Solution:
Given joint equation of lines is 6x2 – 14xy + 14ky2 = 0
It is in the form ax2 + 2hxy + by2 = 0.
a = 6, 2h = – 14 and b = 14k
The auxiliary equation of given line is of the form
bm2 + 2hm + a = 0
∴ (14k)m2 – 14 m + 6 = 0
∴ 14km2 – 14 m + 6 = 0 ……… (1)
One of the line is y = 2x. Its slope is 2
Now 2 must be one of the roots of the auxiliary equation (1),
Substituting m = 2 in equation (1)
14k(2)2 – 14 (2) + 6 = 0
∴ 56 k – 28 + 6 = 0
∴ 56 k = 22
∴ k = 22/56 = 11/28
Example 06:
- Find k, if the one of the lines given by kx2 – 5xy – 6y2 = 0 is 4x + 3y = 0
- Solution:
Given joint equation of lines is kx2 – 5xy – 6y2 = 0
It is in the form ax2 + 2hxy + by2 = 0.
a = k, 2h = – 5 and b = – 6
The auxiliary equation of given line is of the form
bm2 + 2hm + a = 0
∴ (- 6)m2 – 5m + k = 0
∴ 6m2 + 5 m – k = 0……… (1)
One of the line is 4x + 3y = 0. Its slope is – 4/3
Now – 4/3 must be one of the roots of the auxiliary equation (1),
Substituting m = – 4/3 in equation (1)
6(- 4/3)2 + 5 (- 4/3) – k = 0
∴ 6(16/9) + 5 (- 4/3) – k = 0
Multiplying both sides of equation by 3
∴ 32 – 20 – 3k = 0
– 3k = – 12
∴ k = 4
Example 07:
- Find k, if the one of the lines given b 3x2 + kxy + 2y2 = 0 is 2x + y = 0
- Solution:
Given joint equation of lines is 3x2 + kxy + 2y2 = 0
It is in the form ax2 + 2hxy + by2 = 0.
a = 3, 2h = k and b = 2
The auxiliary equation of given line is of the form
bm2 + 2hm + a = 0
∴ 2m2 + k m + 3 = 0 ……… (1)
One of the line is 2x + y = 0. Its slope is – 2/1 = -2
Now – 2 must be one of the roots of the auxiliary equation (1),
Substituting m = – 2 in equation (1)
2(-2)2 + k (-2) + 3 = 0
∴ 8 – 2k + 3 = 0
∴ – 2k = – 11
∴ k = 11/2
Example 08:
- Find the value of ‘k’ if one of the line given by 4x2 + kxy – y2 = 0 is 2x + y = 0.
- Solution:
Given joint equation of lines is 4x2 + kxy – y2 = 0
It is in the form ax2 + 2hxy + by2 = 0.
a = 4, 2h = k and b = -1
The auxiliary equation of given line is of the form
bm2 + 2hm + a = 0
∴ (-1)m2 + k m + 4 = 0
∴ m2 – k m – 4 = 0 ……… (1)
One of the line is 2x + y = 0. Its slope is – 2/1 = -2
Now – 2 must be one of the roots of the auxiliary equation (1),
Substituting m = – 2 in equation (1)
(-2)2 – k (-2) – 4 = 0
∴ 4 + 2k – 4 = 0
∴ 2k = 0
∴ k = 0
Example 09:
- Find the value of ‘a’ if the line given by ax2 + xy – 3y2 = 0 is perpendicular to 3x – 5y – 1 = 0.
- Solution:
Given joint equation of lines is ax2 + xy – 3y2 = 0
It is in the form Ax2 + 2Hxy + By2 = 0.
A = a, 2H = 1 and B = – 3
The auxiliary equation of given line is of the form
Bm2 + 2Hm + A = 0
∴ (-3)m2 + 1 m + a = 0
∴ 3m2 – m – a = 0 ……… (1)
Given line is 3x – 5y – 1 = 0. slope of this line is – 3/-5 = 3/5
One of the line is perpendicular to 3x – 5y – 1 = 0. Hence its slope = – 5/3
Now – 5/3 must be one of the roots of the auxiliary equation (1),
Substituting m = – 5/3 in equation (1)
∴ 3(- 5/3)2 – (- 5/3) – a = 0
∴ 3(25/9) + (5/3) – a = 0
Multiplying both sides of equation by 3
∴ 25 + 5 – 3a = 0
∴ – 3a = – 30
∴ a = 10
Example 10:
- Find the value of ‘k’ if the line given by 2x2 – 5xy + ky2= 0 is perpendicular to x – 2y = 8.
- Solution:
Given joint equation of lines is 2x2 – 5xy + ky2= 0
It is in the form ax2 + 2hxy + by2 = 0.
a = 2, 2h = – 5 and b = k
The auxiliary equation of given line is of the form
bm2 + 2hm + a = 0
∴ km2 – 5m + 2 = 0 ……… (1)
Given line is x – 2y = 8. slope of this line is – 1/-2 = 1/2
One of the line is perpendicular to x – 2y = 8. Hence its slope = – 2
Now – 2 must be one of the roots of the auxiliary equation (1),
Substituting m = – 2 in equation (1)
∴ k(-2)2 – 5(-2) + 2 = 0
∴ 4 k + 10 + 2 = 0
∴ 4k = – 12
∴ k = – 3
Example 11:
- Find the value of ‘k’ if the line given b 3x2 – kxy + 5y2= 0 is perpendicular to 5x + 3y = 0.
- Solution:
Given joint equation of lines is 3x2 – kxy + 5y2= 0
It is in the form ax2 + 2hxy + by2 = 0.
a = 3, 2h = – k and b = 5
The auxiliary equation of given line is of the form
bm2 + 2hm + a = 0
∴ 5m2 – km + 3 = 0 ……… (1)
Given line is 5x + 3y = 0. slope of this line is – 5/3
One of the line is perpendicular to 5x + 3y = 0. Hence its slope = 3/5
Now 3/5 must be one of the roots of the auxiliary equation (1),
Substituting m = 3/5 in equation (1)
∴ 5(3/5)2 – k(3/5) + 3 = 0
∴ 5(9/25) – k(3/5) + 3 = 0
Multiplying both sides of equation by 5
∴ 9 – 3k + 15 = 0
∴ – 3k = – 24
∴ k = 8