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**Type – IV B: ****To Find Value of Constant Using Auxiliary Equation of the Pair of lines**

**ALGORITHM:**

- Write the auxiliary equation of joint equation.
- Find the slope of given line m.
- Substitute value of m in auxillary equation.
- Simplify and get the value of constant.

**Example – 12:**

- Find the value of ‘k’ if one of the line given by 6x
^{2}+ kxy + y^{2}= 0 is 2x + y = 0. **Solution:**

Given joint equation of lines is 6x^{2} + kxy + y^{2} = 0

It is in the form ax^{2} + 2hxy + by^{2} = 0.

a = 6, 2h = k and b= 1

The auxiliary equation of given line is of the form

bm^{2} + 2hm + a = 0

∴ (1)m^{2} + km + 6 = 0

∴ m^{2} + 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 – 13:**

- Find λ, if 2x + 5y = 0 coincides with one of the lines x
^{2}– λxy + 5y^{2}= 0. **Solution:**

Given joint equation of lines is x^{2} – λxy + 5y^{2} = 0

It is in the form ax^{2} + 2hxy + by^{2} = 0.

a = 1, 2h = – λ and b= 5

The auxiliary equation of given line is of the form

bm^{2} + 2hm + a = 0

∴ (5)m^{2} – λm + 1 = 0

∴ 5m^{2} – 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 – 14:**

- Find k, if the one of the lines given by 2x
^{2}– xy + ky^{2}= 0 is x – 3y = 0 . **Solution:**

Given joint equation of lines is 2x^{2} – xy + ky^{2} = 0

It is in the form ax^{2} + 2hxy + by^{2} = 0.

a = 2, 2h = – 1 and b= k

The auxiliary equation of given line is of the form

bm^{2} + 2hm + a = 0

∴ (k)m^{2} – 1m + 2 = 0

∴ km^{2} – 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 – 15:**

- Find k, if the one of the lines given by kx
^{2}+ 3xy – y^{2}= 0 is 2x + y = 0 **Solution:**

Given joint equation of lines is kx^{2} + 3xy – y^{2} = 0

It is in the form ax^{2} + 2hxy + by^{2} = 0.

a = k, 2h = 3 and b = – 1

The auxiliary equation of given line is of the form

bm^{2} + 2hm + a = 0

∴ (-1)m^{2} + 3m + k = 0

∴ m^{2} – 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 – 16:**

- Find k, if the one of the lines given by 6x
^{2}– 14xy + 14ky^{2}= 0 coincides with y = 2x **Solution:**

Given joint equation of lines is 6x^{2} – 14xy + 14ky^{2} = 0

It is in the form ax^{2} + 2hxy + by^{2} = 0.

a = 6, 2h = – 14 and b = 14k

The auxiliary equation of given line is of the form

bm^{2} + 2hm + a = 0

∴ (14k)m^{2} – 14 m + 6 = 0

∴ 14km^{2} – 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 – 17:

- Find k, if the one of the lines given by kx
^{2}– 5xy – 6y^{2}= 0 is 4x + 3y = 0 **Solution:**

Given joint equation of lines is kx^{2} – 5xy – 6y^{2} = 0

It is in the form ax^{2} + 2hxy + by^{2} = 0.

a = k, 2h = – 5 and b = – 6

The auxiliary equation of given line is of the form

bm^{2} + 2hm + a = 0

∴ (- 6)m^{2} – 5m + k = 0

∴ 6m^{2} + 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 – 18:

- Find k, if the one of the lines given b 3x
^{2}+ kxy + 2y^{2}= 0 is 2x + y = 0 **Solution:**

Given joint equation of lines is 3x^{2} + kxy + 2y^{2} = 0

It is in the form ax^{2} + 2hxy + by^{2} = 0.

a = 3, 2h = k and b = 2

The auxiliary equation of given line is of the form

bm^{2} + 2hm + a = 0

∴ 2m^{2} + 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 – 19:**

- Find the value of ‘k’ if one of the line given by 4x
^{2}+ kxy – y^{2}= 0 is 2x + y = 0. **Solution:**

Given joint equation of lines is 4x^{2} + kxy – y^{2} = 0

It is in the form ax^{2} + 2hxy + by^{2} = 0.

a = 4, 2h = k and b = -1

The auxiliary equation of given line is of the form

bm^{2} + 2hm + a = 0

∴ (-1)m^{2} + k m + 4 = 0

∴ m^{2} – 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 – 20:**

- Find the value of ‘a’ if the line given by ax
^{2}+ xy – 3y^{2}= 0 is perpendicular to 3x – 5y – 1 = 0. **Solution:**

Given joint equation of lines is ax^{2} + xy – 3y^{2} = 0

It is in the form Ax^{2} + 2Hxy + By^{2} = 0.

A = a, 2H = 1 and B = – 3

The auxiliary equation of given line is of the form

Bm^{2} + 2Hm + A = 0

∴ (-3)m^{2} + 1 m + a = 0

∴ 3m^{2} – 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 – 21:**

- Find the value of ‘k’ if the line given by 2x
^{2}– 5xy + ky^{2}= 0 is perpendicular to x – 2y = 8.

**Solution:**

Given joint equation of lines is 2x^{2} – 5xy + ky^{2}= 0

It is in the form ax^{2} + 2hxy + by^{2} = 0.

a = 2, 2h = – 5 and b = k

The auxiliary equation of given line is of the form

bm^{2} + 2hm + a = 0

∴ km^{2} – 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 – 22:**

- Find the value of ‘k’ if the line given b 3x
^{2}– kxy + 5y^{2}= 0 is perpendicular to 5x + 3y = 0.

**Solution:**

Given joint equation of lines is 3x^{2} – kxy + 5y^{2}= 0

It is in the form ax^{2} + 2hxy + by^{2} = 0.

a = 3, 2h = – k and b = 5

The auxiliary equation of given line is of the form

bm^{2} + 2hm + a = 0

∴ 5m^{2} – 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

Science > Mathematics > Pair of Straight Lines > You are Here |

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