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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 m
_{1}and m_{2}are the slopes of the two lines represented by joint equation ax^{2}+ 2hxy + by^{2}= 0. Then m_{1}. m_{2}= a/b ; m_{1}+ m_{2}= -2h/b

**ALGORITHM:**

- Write given joint equation.
- Compare with the standard equation.
- Find values of a, h and b.
- let m
_{1}and m_{2}be the slopes of the lines represented by the given joint equation. - Find values of m
_{1}+ m_{2}and m_{1}.m_{2} - Required lines are perpendicular to given lines. Hence slopes of required lines are – 1/m
_{1}and – 1/m_{2}. - Write equations of required lines in the form u = 0 and v = 0.
- Find u.v = 0
- 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 5x
^{2}– 8xy + 3y^{2}= 0. **Solution:**

Given joint equation is 5x^{2} – 8xy + 3y^{2} = 0.

Comparing with ax^{2} + 2hxy + by^{2} = 0

a = 5, 2h = -8, and b = 3.

Now let m_{1} and m_{2} be the slopes of the lines represented by given joint equation.

m_{1} + m_{2} = -2h/b = -(-8)/3 = 8/3

and m_{1}. m_{2} = a/b = 5/3

Now since the required lines are perpendicular to the lines represented by given joint equation,

their slopes are – 1/m_{1} and – 1/m_{2 }

Equation of first required line is

y = – 1/m_{1 }x

∴ m_{1}y = – x

∴ x + m_{1}y = 0 ……. (1)

and similarly equation second required line is

y = – 1/m_{2 }x

∴ m_{2}y = – x

∴ x + m_{2}y = 0 …. (2)

combined equation is

(x + m_{1}y)( x + m_{2}y) = 0

∴ x ( x + m_{2}y) + m_{1}y ( x + m_{2}y) = 0

∴ x^{2} + m_{2}xy + m_{1}xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (8/3)xy + (5/3)y^{2} = 0

Multiplying both sides by 3

3x^{2} + 8xy + 5y^{2} = 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 x
^{2}– xy + 2y^{2}= 0 **Solution:**

Given joint equation is x^{2} – xy + 2y^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

a = 1, 2h = -1 and b = 2.

Now let m_{1} and m_{2} be the slopes of the lines represented by given joint equation.

m_{1} + m_{2} = -2h/b = – (-1)/2 = 1/2

and m_{1}. m_{2} = a/b = 1/2

Now since the required lines are perpendicular to the lines represented by given joint equation,

their slopes are – 1/m_{1} and – 1/m_{2 }

Equation of first required line is

y = – 1/m_{1 }x

∴ m_{1}y = – x

∴ x + m_{1}y = 0 ……. (1)

and similarly equation second required line is

y = – 1/m_{2 }x

∴ m_{2}y = – x

∴ x + m_{2}y = 0 …. (2)

combined equation is

(x + m_{1}y)( x + m_{2}y) = 0

∴ x ( x + m_{2}y) + m_{1}y ( x + m_{2}y) = 0

∴ x^{2} + m_{2}xy + m_{1}xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (1/2)xy + (1/2)y^{2} = 0

Multiplying both sides by 2

2x^{2} + xy + y^{2} = 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 ax
^{2}+ 2hxy + by^{2}= 0 **Solution:**

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

Now let m_{1} and m_{2} be the slopes of the lines represented by given joint equation.

m_{1} + m_{2} = -2h/b and m_{1}. m_{2} = a/b

Now since the required lines are perpendicular to the lines represented by given joint equation,

their slopes are – 1/m_{1} and – 1/m_{2 }

Equation of first required line is

y = – 1/m_{1 }x

∴ m_{1}y = – x

∴ x + m_{1}y = 0 ……. (1)

and similarly equation second required line is

y = – 1/m_{2 }x

∴ m_{2}y = – x

∴ x + m_{2}y = 0 …. (2)

combined equation is

(x + m_{1}y)( x + m_{2}y) = 0

∴ x ( x + m_{2}y) + m_{1}y ( x + m_{2}y) = 0

∴ x^{2} + m_{2}xy + m_{1}xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (-2h/b)xy + (a/b)y^{2} = 0

Multiplying both sides by b

bx^{2} – 2hxy + ay^{2} = 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 5x
^{2}– 8xy + 3y^{2}= 0.

Here a = 5, 2h = -8 and b = 3

Hence answer is 3x^{2} + 8xy + 5y^{2} = 0

**Example – 4:**

- Find the joint equation of pair of lines through origin and perpendicular to the line pairb 2x
^{2}– 8xy + 3y^{2}= 0. **Solution:**

Given joint equation is 2x^{2} – 8xy + 3y^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

a = 2, 2h = – 8 and b = 3.

Now let m_{1} and m_{2} be the slopes of the lines represented by given joint equation.

m_{1} + m_{2} = -2h/b = – (-8)/3 = 8/3

and m_{1}. m_{2} = a/b = 2/3

Now since the required lines are perpendicular to the lines represented by given joint equation,

their slopes are – 1/m_{1} and – 1/m_{2 }

Equation of first required line is

y = – 1/m_{1 }x

∴ m_{1}y = – x

∴ x + m_{1}y = 0 ……. (1)

and similarly equation second required line is

y = – 1/m_{2 }x

∴ m_{2}y = – x

∴ x + m_{2}y = 0 …. (2)

combined equation is

(x + m_{1}y)( x + m_{2}y) = 0

∴ x ( x + m_{2}y) + m_{1}y ( x + m_{2}y) = 0

∴ x^{2} + m_{2}xy + m_{1}xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (8/3)xy + (2/3)y^{2} = 0

Multiplying both sides by 3

3x^{2} + 8xy + 2y^{2} = 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 5x
^{2}+ 2xy – 3y^{2}= 0. **Solution:**

Given joint equation is 2x^{2} – 8xy + 3y^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

a = 5, 2h = 2 and b = -3.

Now let m_{1} and m_{2} be the slopes of the lines represented by given joint equation.

m_{1} + m_{2} = -2h/b = – (2)/-3 = 2/3

and m_{1}. m_{2} = 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/m_{1} and – 1/m_{2 }

Equation of first required line is

y = – 1/m_{1 }x

∴ m_{1}y = – x

∴ x + m_{1}y = 0 ……. (1)

and similarly equation second required line is

y = – 1/m_{2 }x

∴ m_{2}y = – x

∴ x + m_{2}y = 0 …. (2)

combined equation is

(x + m_{1}y)( x + m_{2}y) = 0

∴ x ( x + m_{2}y) + m_{1}y ( x + m_{2}y) = 0

∴ x^{2} + m_{2}xy + m_{1}xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (2/3)xy + (-5/3)y^{2} = 0

Multiplying both sides by 3

3x^{2} + 2xy – 5y^{2} = 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 x
^{2}+ 4xy – 5y^{2}= 0. **Solution:**

Given joint equation is 2x^{2} – 8xy + 3y^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

a = 1, 2h = 4 and b = -5.

Now let m_{1} and m_{2} be the slopes of the lines represented by given joint equation.

m_{1} + m_{2} = -2h/b = – (4)/-5 = 4/5

and m_{1}. m_{2} = 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/m_{1} and – 1/m_{2 }

Equation of first required line is

y = – 1/m_{1 }x

∴ m_{1}y = – x

∴ x + m_{1}y = 0 ……. (1)

and similarly equation second required line is

y = – 1/m_{2 }x

∴ m_{2}y = – x

∴ x + m_{2}y = 0 …. (2)

combined equation is

(x + m_{1}y)( x + m_{2}y) = 0

∴ x ( x + m_{2}y) + m_{1}y ( x + m_{2}y) = 0

∴ x^{2} + m_{2}xy + m_{1}xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (4/5)xy + (-1/5)y^{2} = 0

Multiplying both sides by 5

5x^{2} + 4xy – y^{2} = 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 2x
^{2}– 3xy – 9y^{2}= 0. **Solution:**

Given joint equation is 2x^{2} – 8xy + 3y^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

a = 2, 2h = -3 and b = – 9.

Now let m_{1} and m_{2} be the slopes of the lines represented by given joint equation.

m_{1} + m_{2} = -2h/b = – (-3)/-9 = -1/3

and m_{1}. m_{2} = 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/m_{1} and – 1/m_{2 }

Equation of first required line is

y = – 1/m_{1 }x

∴ m_{1}y = – x

∴ x + m_{1}y = 0 ……. (1)

and similarly equation second required line is

y = – 1/m_{2 }x

∴ m_{2}y = – x

∴ x + m_{2}y = 0 …. (2)

combined equation is

(x + m_{1}y)( x + m_{2}y) = 0

∴ x ( x + m_{2}y) + m_{1}y ( x + m_{2}y) = 0

∴ x^{2} + m_{2}xy + m_{1}xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (-1/3)xy + (-2/9)y^{2} = 0

Multiplying both sides by 9

9x^{2} – 3xy – 2y^{2} = 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 x
^{2}+ xy – y^{2}= 0. **Solution:**

Given joint equation is 2x^{2} – 8xy + 3y^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

a = 1, 2h = 1 and b = -1.

Now let m_{1} and m_{2} be the slopes of the lines represented by given joint equation.

m_{1} + m_{2} = -2h/b = – 1/-1 = 1

and m_{1}. m_{2} = a/b = 1/-1 = -1

Now since the required lines are perpendicular to the lines represented by given joint equation,

their slopes are – 1/m_{1} and – 1/m_{2 }

Equation of first required line is

y = – 1/m_{1 }x

∴ m_{1}y = – x

∴ x + m_{1}y = 0 ……. (1)

and similarly equation second required line is

y = – 1/m_{2 }x

∴ m_{2}y = – x

∴ x + m_{2}y = 0 …. (2)

combined equation is

(x + m_{1}y)( x + m_{2}y) = 0

∴ x ( x + m_{2}y) + m_{1}y ( x + m_{2}y) = 0

∴ x^{2} + m_{2}xy + m_{1}xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + (1)xy + (- 1)y^{2} = 0

x^{2} + xy – y^{2} = 0

This is the required joint equation of pair of lines.

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

Physics |
Chemistry |
Biology |
Mathematics |