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**Type – IV A: ****To Find Condition That Given Line is Represented by Homogeneous Equation**

**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 condition.

**Example – 1:**

- If the joint equation of lines is ax
^{2}+ 2hxy + by^{2}= 0. Find the condition that the line 5x + 3y = 0 is one of them. **Solution:**

Given joint equation of lines is ax^{2} + 2hxy + by^{2} = 0.

Its auxiliary equation is

bm^{2} + 2hm + a = 0 ……… (1)

One of the line is 5x + 3y = 0. Its slope is – 5/3

Now – 5/3 must be one of the roots of the auxiliary equation (1),

Substituting m = – 5/3 in equation (1)

Multiplying both sides of the equation by 9

25 b -15 h + 9a = 0

9a + 25 b -15 h = 0

This is the required condition.

**Example – 2:**

- If the joint equation of lines is ax
^{2}+ 2hxy + by^{2}= 0. Find the condition that the line 3x – y = 0 is one of them. **Solution:**

Given joint equation of lines is ax^{2} + 2hxy + by^{2} = 0.

Its auxiliary equation is

bm^{2} + 2hm + a = 0 ……… (1)

One of the line is 3x – y = 0. Its slope is – 3/-1 = 3

Now 3 must be one of the roots of the auxiliary equation (1),

b(3)^{2} + 2h(3) + a = 0

∴ 9b + 6h + a = 0

∴ a + 9b + 6h = 0

This is the required condition.

**Example – 3:**

- If the joint equation of lines is ax
^{2}+ 2hxy + by^{2}= 0. the line 4x – 3y = 0 coincides with one of them. Show that 16 b + 24 h + 9a = 0. **Solution:**

Given joint equation of lines is ax^{2} + 2hxy + by^{2} = 0.

Its auxiliary equation is

bm^{2} + 2hm + a = 0 ……… (1)

One of the line is 4x – 3y = 0. Its slope is – 4/-3 = 4/3

Now 3 must be one of the roots of the auxiliary equation (1),

b(4/3)^{2} + 2h(4/3) + a = 0

∴ b(16/9) + 2h(4/3) + a = 0

Multiplying both sides of the equation by 9

16b + 24 h + 9a = 0, Proved as required.

**Example – 4:**

- Find the condition that one of the line represented by ax
^{2}+ 2hxy + by^{2}= 0 is y = mx **Solution:**

Given joint equation of lines is ax^{2} + 2hxy + by^{2} = 0.

Its auxiliary equation is

bm^{2} + 2hm + a = 0 ……… (1)

One of the line is y = mx . Its slope is m

Now m must be one of the roots of the auxiliary equation (1),

b(m)^{2} + 2h(m) + a = 0

∴ bm^{2} + 2hm + a = 0

This is the required condition..

**Example – 5:**

- Find the condition that one of the line represented by ax
^{2}+ 2hxy + by^{2}= 0 is px + qy = 0. **Solution:**

Given joint equation of lines is ax^{2} + 2hxy + by^{2} = 0.

Its auxiliary equation is

bm^{2} + 2hm + a = 0 ……… (1)

One of the line is px + qy = 0. Its slope is – p/q

Now – p/q must be one of the roots of the auxiliary equation (1),

This is the required condition

**Example – 6:**

- Find the condition that one of the line represented by ax
^{2}2hxy + by^{2}= 0. is 3x – 2y = 0 **Solution:**

Given joint equation of lines is ax^{2} + 2hxy + by^{2} = 0.

Its auxiliary equation is

bm^{2} + 2hm + a = 0 ……… (1)

One of the line is 3x – 2y = 0. Its slope is – 3/-2 = 3/2

Now 3/2 must be one of the roots of the auxiliary equation (1),

b(3/2)^{2} + 2h(3/2) + a = 0

∴ b(9/4) + 2h(3/2) + a = 0

Multiplying both sides of the equation by 4

9b + 12 h + 4a = 0

This is the required condition

** Example – 7:**

- Find the condition that the line 4x + 5y = 0 coincides with one of the line represented by ax
^{2}+ 2hxy + by^{2}= 0.**OR**If the line 4x + 5y = 0 coincides with one of the line epresented by ax^{2}+ 2hxy + by^{2}= 0, prove that 25a – 40h + 16 b = 0. **Solution:**

Given joint equation of lines is ax^{2} + 2hxy + by^{2} = 0.

Its auxiliary equation is

bm^{2} + 2hm + a = 0 ……… (1)

One of the line is 4x + 5y = 0. Its slope is – 4/5

Now – 4/5 must be one of the roots of the auxiliary equation (1),

b(-4/5)^{2} + 2h(-4/5) + a = 0

∴ b(16/25) + 2h(-4/5) + a = 0

Multiplying both sides of the equation by 25

16b – 40 h + 25a = 0

25a – 40h + 16 b = 0. Proved as required

**Example – 8:**

- If the joint equation of lines is ax
^{2}+ 2hxy + by^{2}= 0. Find the condition that the line 3x + 4y = 7 is perpendicular to one of them. **Solution:**

Given joint equation of lines is ax^{2} + 2hxy + by^{2} = 0.

Its auxiliary equation is

bm^{2} + 2hm + a = 0 ……… (1)

Given line is 3x + 4y = 7. Its slope is – 3/4

As one of the lines is perpendicular to given line

Hence slope of one of the line represented by joint equation is 4/3

Now 4/3 must be one of the roots of the auxiliary equation (1),

b(4/3)^{2} + 2h(4/3) + a = 0

∴ b(16/9) + 2h(4/3) + a = 0

Multiplying both sides of the equation by 9

16b + 24h + 9a = 0

9a + 16b + 24h = 0

This is the required condition

**Example – 9:**

- If the joint equation of lines is ax
^{2}+ 2hxy + by^{2}= 0. Find the condition that the line x + 2y = 0 is perpendicular to one of them.

**Solution:**

Given joint equation of lines is ax^{2} + 2hxy + by^{2} = 0.

Its auxiliary equation is

bm^{2} + 2hm + a = 0 ……… (1)

Given line is x + 2y = 0. Its slope is – 1/2

As one of the lines is perpendicular to given line

Hence slope of one of the line represented by joint equation is 2

Now 2 must be one of the roots of the auxiliary equation (1),

b(2)^{2} + 2h(2) + a = 0

4b + 4h + a = 0

a + 4b + 4h = 0

This is the required condition

**Example – 10:**

- If the joint equation of lines is ax
^{2}+ 2hxy + by^{2}= 0. Find the condition that the line 3x + y = 0 is perpendicular to one of them.

**Solution:**

Given joint equation of lines is ax^{2} + 2hxy + by^{2} = 0.

Its auxiliary equation is

bm^{2} + 2hm + a = 0 ……… (1)

Given line is 3x + y = 0. Its slope is – 3/1 = -3

As one of the lines is perpendicular to given line

Hence slope of one of the line represented by joint equation is 1/3

Now 1/3 must be one of the roots of the auxiliary equation (1),

b(1/3)^{2} + 2h(1/3) + a = 0

b(1/9) + 2h(1/3) + a = 0

Multiplying both sides by 9

b + 6h + 9a = 0

9a + b + 6h = 0

This is the required condition

**Example – 11:**

- If the joint equation of lines is ax
^{2}+ 2hxy + by^{2}= 0. Find the condition that the line px + qy = 0 is perpendicular to one of them. **Solution:**

Given joint equation of lines is ax^{2} + 2hxy + by^{2} = 0.

Its auxiliary equation is

bm^{2} + 2hm + a = 0 ……… (1)

Given line is px + qy = 0. Its slope is – p/q

As one of the lines is perpendicular to given line

Hence slope of one of the line represented by joint equation is q/p

Now q/p must be one of the roots of the auxiliary equation (1),

b(q/p)^{2} + 2h(q/p) + a = 0

b(q^{2}/p^{2}) + 2h(q/p) + a = 0

Multiplying both sides by p^{2}

bq^{2} + 2pqh + ap^{2} = 0

ap^{2} + bq^{2} + 2pqh = 0

This is the required condition

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