# Joint Equation of Pair of Lines Perpendicular To Another Pair of Lines

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### Type – VI: To Find Joint Equation of Pair of Lines Perpendicular To Lines Represented by Given Joint Equation.

#### Joint Equation of Pair of Lines:

• 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:

1. Write given joint equation.
2. Compare with the standard equation.
3. Find values of a, h and b.
4. let m1 and  m2 be the slopes of the lines represented by the given joint equation.
5. Find values of m1 + m2 and m1.m2
6. Required lines are perpendicular to given lines. Hence slopes of required lines are – 1/m1 and – 1/m2.
7. Write equations of required lines in the form u = 0 and v = 0.
8. Find u.v = 0
9. SImplify and write the joint equation of line in standard form

#### Example – 1:

• 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/m

Equation of first required line is

y =  – 1/mx

∴   m1y = – x

∴   x + m1y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/mx

∴   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 – 2:

• 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/m

Equation of first required line is

y =  – 1/mx

∴   m1y = – x

∴   x + m1y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/mx

∴   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  – 3:

• 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/m

Equation of first required line is

y =  – 1/mx

∴   m1y = – x

∴   x + m1y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/mx

∴   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 – 4:

• Find the joint equation of pair of lines through origin and perpendicular to the line pairb 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/m

Equation of first required line is

y =  – 1/mx

∴   m1y = – x

∴   x + m1y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/mx

∴   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 – 5:

• 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/m

Equation of first required line is

y =  – 1/mx

∴   m1y = – x

∴   x + m1y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/mx

∴   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 – 6:

• 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/m

Equation of first required line is

y =  – 1/mx

∴   m1y = – x

∴   x + m1y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/mx

∴   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 – 7:

• 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/m

Equation of first required line is

y =  – 1/mx

∴   m1y = – x

∴   x + m1y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/mx

∴   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 – 8:

• 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/m

Equation of first required line is

y =  – 1/mx

∴   m1y = – x

∴   x + m1y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/mx

∴   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.

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