# Condition That Line is Represented by Homogeneous Equation

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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 ax2 + 2hxy  + by2 = 0. Find the condition that the line 5x + 3y = 0 is one of them.
• Solution:

Given joint equation of lines is ax2 + 2hxy  + by2 = 0.

Its auxiliary equation is

bm2 + 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  ax2 + 2hxy  + by2 = 0. Find the condition that the line 3x – y = 0 is one of them.
• Solution:

Given joint equation of lines is ax2 + 2hxy  + by2 = 0.

Its auxiliary equation is

bm2 + 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 ax2 + 2hxy  + by2 = 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 ax2 + 2hxy  + by2 = 0.

Its auxiliary equation is

bm2 + 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 ax2 + 2hxy  + by2 = 0 is y = mx
• Solution:

Given joint equation of lines is ax2 + 2hxy  + by2 = 0.

Its auxiliary equation is

bm2 + 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

∴  bm2 + 2hm + a = 0

This is the required condition..

#### Example – 5:

• Find the condition that one of the line represented by ax2 + 2hxy  + by2 = 0 is px + qy = 0.
• Solution:

Given joint equation of lines is ax2 + 2hxy  + by2 = 0.

Its auxiliary equation is

bm2 + 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 ax2 2hxy  + by2 = 0. is 3x – 2y = 0
• Solution:

Given joint equation of lines is ax2 + 2hxy  + by2 = 0.

Its auxiliary equation is

bm2 + 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 ax2 + 2hxy  + by2 = 0. OR  If the line 4x + 5y = 0 coincides with one of the line epresented by ax2 + 2hxy  + by2 = 0, prove that 25a – 40h + 16 b = 0.
• Solution:

Given joint equation of lines is ax2 + 2hxy  + by2 = 0.

Its auxiliary equation is

bm2 + 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 ax2 + 2hxy  + by2 = 0. Find the condition that the line 3x + 4y = 7 is perpendicular to one of them.
• Solution:

Given joint equation of lines is ax2 + 2hxy  + by2 = 0.

Its auxiliary equation is

bm2 + 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 ax2 + 2hxy  + by2 = 0. Find the condition that the line x + 2y = 0 is perpendicular to one of them.

Solution:

Given joint equation of lines is ax2 + 2hxy  + by2 = 0.

Its auxiliary equation is

bm2 + 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 ax2 + 2hxy  + by2 = 0. Find the condition that the line 3x + y = 0 is perpendicular to one of them.

Solution:

Given joint equation of lines is ax2 + 2hxy  + by2 = 0.

Its auxiliary equation is

bm2 + 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 ax2 + 2hxy  + by2 = 0. Find the condition that the line px + qy = 0 is perpendicular to one of them.
• Solution:

Given joint equation of lines is ax2 + 2hxy  + by2 = 0.

Its auxiliary equation is

bm2 + 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(q2/p2) + 2h(q/p) + a = 0

Multiplying both sides by p2

bq2 + 2pqh + ap2 = 0

ap2 + bq2 + 2pqh = 0

This is the required condition

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