In this article, we shall study the use of the auxiliary equation of par of lines to find the required condition.

Algorithm:

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

Example 01:

• Find the value of ‘k’ if one of the line given by 6x² + kxy + y² = 0 is 2x + y = 0.
• Solution:

Given joint equation of lines is 6x² + kxy + y² = 0

It is in the form ax² + 2hxy  + by² = 0.

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

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   (1)m² + km + 6 = 0

∴   m² + 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 02:

• Find λ, if  2x + 5y = 0 coincides with one of the lines x² – λxy +  5y² = 0.
• Solution:

Given joint equation of lines is x² – λxy +  5y² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = 1, 2h = – λ and b= 5

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   (5)m² – λm + 1 = 0

∴   5m² – 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 03:

• Find k, if the one of the lines given by 2x² – xy  + ky² = 0 is x – 3y = 0 .
• Solution:

Given joint equation of lines is 2x² – xy  + ky² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = 2, 2h = – 1 and b= k

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   (k)m² – 1m + 2 = 0

∴   km² –  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 04:

• Find k, if the one of the lines given by kx² + 3xy  – y² = 0 is 2x + y = 0
• Solution:

Given joint equation of lines is kx² + 3xy  – y² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = k, 2h = 3 and b = – 1

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   (-1)m² + 3m + k = 0

∴   m² –  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 05:

• Find k, if the one of the lines given by 6x² – 14xy  + 14ky² = 0 coincides with y = 2x
• Solution:

Given joint equation of lines is 6x² – 14xy  + 14ky² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = 6, 2h = – 14 and b = 14k

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   (14k)m² – 14 m + 6 = 0

∴  14km² – 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 06:

• Find k, if the one of the lines given by kx² – 5xy  – 6y² = 0 is 4x + 3y = 0
• Solution:

Given joint equation of lines is kx² – 5xy  – 6y² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = k, 2h = – 5 and b = – 6

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   (- 6)m² – 5m + k = 0

∴  6m² + 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 07:

• Find k, if the one of the lines given b 3x² + kxy  + 2y² = 0 is 2x + y = 0
• Solution:

Given joint equation of lines is 3x² + kxy  + 2y² = 0

It is in the form ax² + 2hxy  + by² = 0.

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

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴  2m² + 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 08:

• Find the value of ‘k’ if one of the line given by 4x² + kxy – y² = 0 is 2x + y = 0.
• Solution:

Given joint equation of lines is 4x² + kxy – y² = 0

It is in the form ax² + 2hxy  + by² = 0.

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

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴  (-1)m² + k m + 4 = 0

∴  m² – 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 09:

• Find the value of ‘a’ if the line given by ax² + xy – 3y² = 0 is perpendicular to 3x – 5y – 1 = 0.
• Solution:

Given joint equation of lines is ax² + xy – 3y² = 0

It is in the form Ax² + 2Hxy  + By² = 0.

A = a, 2H = 1 and B = – 3

The auxiliary equation of given line is of the form

Bm² + 2Hm + A = 0

∴  (-3)m² + 1 m + a = 0

∴  3m² – 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 10:

• Find the value of ‘k’ if the line given by 2x² – 5xy + ky²= 0 is perpendicular to x – 2y = 8.
• Solution:

Given joint equation of lines is 2x² – 5xy + ky²= 0

It is in the form ax² + 2hxy  + by² = 0.

a = 2, 2h = – 5 and b = k

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴ km² – 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 11:

• Find the value of ‘k’ if the line given b 3x² – kxy + 5y²= 0 is perpendicular to 5x + 3y = 0.
• Solution:

Given joint equation of lines is 3x² – kxy + 5y²= 0

It is in the form ax² + 2hxy  + by² = 0.

a = 3, 2h = – k and b = 5

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴  5m² – 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 > Use of Auxiliary Equation of Pair of Lines

The post Use of Auxiliary Equation of Pair of Lines appeared first on The Fact Factor.

In this article, we shall study to find a joint equation of lines perpendicular to another pair of lines.

Note: 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 the 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 the line in standard form

Example 01:

• 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/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   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 02:

• 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/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   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 03:

• 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/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   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 04:

• Find the joint equation of pair of lines through origin and perpendicular to the line pair 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/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   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 05:

• 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/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   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 06:

• 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/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   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 07:

• 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/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   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 08:

• 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/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   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.

Science > Mathematics > Pair of Straight Lines > Joint Equation of Pair of Lines Perpendicular To Another Pair of Lines

The post Joint Equation of Pair of Lines Perpendicular To Another Pair of Lines appeared first on The Fact Factor.

In this article, we shall study to predict the nature of lines using the joint equations of lines.

Notes:

If ax² + 2hxy + by²= 0 is a joint equation of lines then the lines represented by joint equation are

• Real if and only if h² – ab ≥ 0
• Real and distinct if and only if h² – ab > 0
• Real and coincident if and only if h² – ab = 0
• Imaginary and can’t be drawn if and only if h² – ab < 0

Algorithm:

1. Write the joint equation of lines in standard form.
2. Compare with standard ax² + 2hxy + by² = 0. Find values of a, h and b.
3. Find the value of the quantity h² – ab
4. Decide the nature of the line using the notes given above.

Example 01:

• Determine the nature of lines represented by the joint equation x² + 2xy + y² = 0
• Solution:

Given joint equation is x² + 2xy + y² = 0

Comparing with ax² + 2hxy + by² = 0

a = 1, 2h = 2, h = 1, b = 1

Now, h² – ab = (1)² – (1)(1) = 1 – 1 = 0

Here h² – ab = 0, hence the lines are real and coincident.

Example 02:

• Determine the nature of lines represented by the joint equation x² + 2xy – y² = 0
• Solution:

Given joint equation is x² + 2xy + y² = 0

Comparing with ax² + 2hxy + by² = 0

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

Now, h² – ab = (1)² – (1)(-1) = 1 + 1 = 2

Here h² – ab > 0, hence the lines are real and distinct.

Example 03:

• Determine the nature of lines represented by the joint equation 4x² + 4xy + y² = 0
• Solution:

Given joint equation is 4x² + 4xy + y² = 0

Comparing with ax² + 2hxy + by² = 0

a = 4, 2h = 4, h = 2, b =  1

Now, h² – ab = (2)² – (4)(1) = 4 – 4 = 0

Here h² – ab = 0, hence the lines are real and coincident.

Example 04:

• Determine the nature of lines represented by the joint equation x² – y² = 0
• Solution:

Given joint equation is x² + 0xy – y² = 0

Comparing with ax² + 2hxy + by² = 0

a = 1, 2h = 0, h = 0, b = – 1

Now, h² – ab = (0)² – (1)(-1) = 0 + 1 = 1

Here h² – ab > 0, hence the lines are real and distinct.

Example 05:

• Determine the nature of lines represented by the joint equation x² + 7xy + 2y² = 0

Solution:

Given joint equation is x² + 7xy + 2y² = 0

Comparing with ax²+ 2hxy + by² = 0

a = 1, 2h = 7, h = 7/2, b =  2

Now, h² – ab = (7/2)² – (1)(2) = 41/4 – 2 = 41/4

Here h² – ab > 0, hence the lines are real and distinct.

Example 06:

• Determine the nature of lines represented by the joint equation xy  = 0
• Solution:

Given joint equation is xy  = 0

Comparing with ax² + 2hxy + by² = 0

a = 0, 2h = 1, h = 1/2, b =  0

Now, h² – ab = (1/2)² – (0)(0) =1/4

Here h² – ab > 0, hence the lines are real and distinct.

Example 07:

Example – 7:  

• Determine the nature of lines represented by the joint equation x² + 2xy + 2y² = 0
• Solution:

Given joint equation is x² + 2xy + 2y² = 0

Comparing with ax² + 2hxy + by² = 0

a = 1, 2h = 2, h = 1, b =  2

Now, h² – ab = (1)² – (1)(2) = 1 – 2 = – 1

Here h² – ab < 0, hence the lines are imaginary and can’t be drawn.

Example 08:

• Determine the nature of lines represented by the joint equation px² + 2qxy – py² = 0.
• Solution:

Given joint equation is px² + 2qxy – py² = 0

Comparing with ax² + 2hxy + by² = 0

a = p, 2h = 2q, h = q, b = – p

Now, h² – ab = (q)² – (p)(-p) = q² + p²

as p and q are real numbers, there squares are always positive. Thus sum of their squares is always positive. Here h² – ab > 0, hence the lines are real and distinct.

Example 09:

• Determine the nature of lines represented by the joint equation px² – qy² = 0
• Solution:

Given joint equation is px² + 0xy – qy² = 0

Comparing with ax² + 2hxy + by²= 0

a = p, 2h = 0, h = 0, b = – q

Now, h² – ab = (0)² – (p)(-q) = 0 + pq = pq

• If p and q are both positive, then the product pq is positive, then h² – ab > 0, hence the lines are real and distinct.
• If p and q are both negative, then the product pq is positive, then h² – ab > 0, hence the lines are real and distinct.
• If p and q have opposite signs then the product pq is negative,  then h²– ab < 0, hence the lines are imaginary and can’t be drawn.
• If p = q = 0, then pq = 0. Here h² – ab = 0, hence the lines are real and coincident.

Nature of Lines is Given. To Find the Value of Constant

Algorithm:

1. Write the joint equation of lines in standard form.
2. Compare with standard ax² + 2hxy + by² = 0. Find values of a, h and b.
3. Find the value of the quantity h² -ab.
4. Use the conditions depending upon the nature of the line use notes given at the top and find the value of the constant.

Example 10:

• If the lines represented by equation x² + 2hxy + 4y² = 0  are real and coincident, find h.
• Solution:

Given joint equation is x² + 2hxy + 4y² = 0

Comparing with Ax² + 2Hxy + By² = 0

A = 1, 2H = 2h, H = h, B =  4

Now, lines are real and coincident

H² – AB = 0

h² – (1)(4) = 0

h² – 4 = 0

h² = 4

h = ± 2

Example 11:

• If the lines represented by equation px² + 6xy + 9y² = 0  are real and distinct, find p.
• Solution:

Given joint equation is px2 + 6xy + 9y2  = 0

Comparing with ax² + 2hxy + by² = 0

a = p, 2h = 6, h = 3, b =  9

Now, lines are real and distinct

h² – ab > 0

32 – (p)(9) > 0

9 – 9p > 0

-9p > -9

∴ p < 1

∴ p ∈ (- ∞, 1)

Example 12:

• If the lines represented by equation px² + 4xy + 4y² = 0  are real and distinct, find p.
• Solution:

Given joint equation is px² + 4xy + 4y² = 0

Comparing with ax² + 2hxy + by² = 0

a = p, 2h = 4, h = 2, b =  4

Now, lines are real and distinct

h² – ab > 0

2² – (p)(4) > 0

4 – 4p > 0

-4p > -4

∴ p < 1

∴ p ∈ (- ∞, 1)

Example 13:

• If the lines represented by equation 3x² + 2hxy + 3y² = 0  are real , find h.
• Solution:

Given joint equation is 3x² + 2hxy + 3y² = 0

Comparing with Ax² + 2Hxy + By² = 0

A = 3, 2H = 2h, H = h, B =  3

Now, lines are real and coincident

H² – AB  = 0

h² – (3)(3) =  0

h² – 9  = 0

∴ h² =  9

∴  h  = ± 3

Science > Mathematics > Pair of Straight Lines > Nature of Lines Represented by Joint Equation

The post Nature of Lines Represented by Joint Equation appeared first on The Fact Factor.

In this article, we shall study to find separate equations of lines from a combined or joint equation of pair of lines by auxiliary equation method.

Algorithm:

1. Check if lines exist. use the same method used in the case to find the nature of lines. Proceed further if lines exist.
2. Divide both sides of the equation by x².
3. Simplify the equation and substitute y/x = m in it.
4. Find two roots m₁ and m₂ of quadratic equation in m
5. Find separate equations of lines by y = m₁ x and y = m₂x
6. Note that this method is applicable to any problem.

Example 01:

• Obtain the separate equations of the lines represented by  11x² + 8xy + y² = 0
• Solution:

The given joint equation is  11x² + 8xy + y² = 0

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in equation (1) we get

11 + 8m + m² = 0

m² + 8m + 11 = 0

This is a quadratic equation in m and has two roots say m₁ and m₂

which gives slopes of the two lines represented by the joint equation.

a = 1, b = 8, c = 11

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (-4 + √5) x

∴  y = – (4 – √5) x

 ∴  (4 – √5) x  + y = 0

The equation of the second line is

y = m₂x  i.e.  i.e.

∴  y = (-4 – √5) x

∴  y = – (4 + √5) x

 ∴  (4 + √5) x  + y = 0

Ans: The separate equations of the lines are (4 – √5) x  + y = 0 and (4 + √5) x  + y = 0

Example 02:

• Obtain separate equations of lines represented by joint equation x² – 4xy + y² = 0.
• Solution:

The given joint equation is  x² – 4xy + y² = 0

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

1 – 4m + m² = 0

∴  m² – 4m + 1= 0

This is a quadratic equation in m and has two roots say m₁ and m₁

which gives slopes of the two lines represented by the joint equation.

a = 1, b = -4, c = 1

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (2 + √3) x

 ∴  (2 + √3) x  – y = 0

The equation of the second line is

y = m₂x

∴  y = (2 – √3) x

 ∴  (2 – √3) x  – y = 0

Ans: The separate equations of the lines are  (2 + √3) x  – y = 0 and (2 – √3) x  – y = 0

Example 03:

• Obtain separate equations of lines represented by joint equation 22x² – 10xy + y² = 0.

Solution:

The given joint equation is 22x² – 10xy + y² = 0.

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

22 – 10m + m² = 0

∴  m² – 10m + 22= 0

This is a quadratic equation in m and has two roots say m₁ and m₁

which gives slopes of the two lines represented by the joint equation.

a = 1, b = – 10, c = 22

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (5 + √3) x

 ∴  (5 + √3) x  – y = 0

The equation of the second line is

y = m₂x

∴  y = (5 – √3) x

 ∴  (5 – √3) x  – y = 0

Ans: The separate equations of the lines are  (5 + √3) x  – y = 0 and (5 – √3) x  – y = 0

Example 04:

• Obtain separate equations of lines represented by joint equation x² + 2xy – y² = 0.
• Solution:

The given joint equation is x² + 2xy – y² = 0

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

1 + 2m – m² = 0

∴  m² – 2m  – 1= 0

This is a quadratic equation in m and has two roots say m₁ and m₁

which gives slopes of the two lines represented by the joint equation.

a = 1, b = – 2, c = -1

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (1 + √2) x

 ∴  (1 + √2) x  – y = 0

The equation of the second line is

y = m₂x

∴  y = (1 – √2) x

 ∴  (1 – √2) x  – y = 0

Ans: The separate equations of the lines are  (1 + √2) x  – y = 0 and (1 – √2) x  – y = 0

Example 05:

• Obtain separate equations of lines represented by joint equation 2x² + 2xy – y² = 0.

Solution:

The given joint equation is  2x² + 2xy – y² = 0

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

2 + 2m – m² = 0

∴  m² – 2m – 2= 0

This is a quadratic equation in m and has two roots say m₁ and m₁

which gives slopes of the two lines represented by the joint equation.

a = 1, b = – 2, c = -2

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (1 + √3) x

 ∴  (1 + √3) x  – y = 0

The equation of the second line is

y = m₂x

∴  y = (1 – √3) x

 ∴  (1 – √3) x  – y = 0

Ans: The separate equations of the lines are  (1 + √3) x  – y = 0 and (1 – √3) x  – y = 0

Example 06:

• Obtain separate equations of lines represented by joint equation x² + 2(tanα) xy – y² = 0.
• Solution:

The given joint equation is x² + 2(tanα) xy – y² = 0

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

1 + 2tanα m – m² = 0

∴  m² – 2tanα m – 1= 0

This is a quadratic equation in m and has two roots say m₁ and m₁

which gives slopes of the two lines represented by the joint equation.

a = 1, b = – 2tanα, c = -1

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (tanα + secα) x

 ∴  (tanα + secα) x  – y = 0

The equation of the second line is

y = m₂x

∴  y = (tanα – secα) x

 ∴  (tanα – secα) x  – y = 0

Ans: The separate equations of the lines are  (tanα + secα) x  – y = 0 and  (tanα – secα) x  – y = 0

Example 07:

• Obtain separate equations of lines represented by joint equation x² + 2(cosecα) xy + y² = 0.
• Solution:

The given joint equation is x² + 2(cosecα) xy + y² = 0.

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

1 + 2cosecα m + m² = 0

∴  m² + 2cosecα m + 1= 0

This is a quadratic equation in m and has two roots say m₁ and m₁

which gives slopes of the two lines represented by the joint equation.

a = 1, b = 2cosecα, c = 1

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (- cosec α + cot α) x

∴  y = –  (cosec α – cot α) x

 ∴  (cosec α – cot α) x  – y = 0

The equation of the second line is

y = m₂x

∴  y = (- cosec α – cot α) x

∴  y = – (cosec α + cot α) x

 ∴  (cosec α + cot α) x  – y = 0

Ans: The separate equations of the lines are  (cosec α – cot α) x  – y = 0 and  (cosec α + cot α) x  – y = 0

Science > Mathematics > Pair of Straight Lines > Separate Equations of Lines (Auxillary Equation Method)

The post Separate Equations of Lines (Auxiliary Equation Method) appeared first on The Fact Factor.

In this article, we shall study to obtain separate equations of lines from a combined or joint equation of pair of lines by factorization method.

Algorithm:

1. Check if lines exist. use the same method used in the case to find the nature of lines (do this orally). Proceed further if lines exist.
2. If factors of the equation can be found directly and easily then factorize the joint equation
3. Write each factor = 0. Thus you will get two separate equations.
4. Note that this method is useful only if the joint equation can be factorized.

Example 01:

• Obtain the separate equations of the lines represented by  3x² – 4xy + y² = 0.
• Solution:

The given joint equation is  3x² – 4xy + y² = 0

∴  3x² – 3xy – xy + y² = 0

∴  3x(x – y) – y(x – y)= 0

∴  (x – y)(3x – y) = 0

∴  (x – y)  = 0  and  (3x – y) = 0

Ans: Separate equations of the lines are x – y = 0 and  3x – y = 0

Example 02:

• Obtain separate equations of the lines represented by joint equation y² – 5x² = 0.

Solution:

Given joint equation is y² – 5x² = 0.

5x² – y² = 0.

∴  (√5x)² – y² = 0

∴  (√5x + y) (√5x – y) = 0.

Ans: The separate equations of the lines are √5x + y = 0 and 5x – y = 0

Example 03:

• Obtain separate equations of lines represented by joint equation x² – 5xy + 6y² = 0.
• Solution:

Given joint equation is x² – 5xy + 6y² = 0.

∴  x² – 3xy – 2xy + 6y² = 0

∴  x(x – 3y) – 2y(x – 3y) = 0

∴  (x – 3y)(x – 2y) = 0

Ans: The separate equations of the lines are x – 3y = 0 and x – 2y = 0

Example 04:

• Obtain separate equations of lines represented by joint equation 3x²  + 10xy + 8y²  = 0.
• Solution:

Given joint equation is 3x² + 10xy + 8y² = 0.

∴  3x² + 6xy + 4xy + 8y² = 0

∴  3x(x + 2y) + 4y(x + 2y) = 0

∴  (x + 2y)(3x + 4y) = 0

Ans: The separate equations of the lines are x + 2y = 0 and 3x + 4y = 0

Example 05:

• Obtain separate equations of lines represented by joint equation 3x² + 10xy + 8y² = 0.
• Solution:

The given joint equation is  x² – 2xy + y² = 0

x² – xy – xy + y² = 0

∴  x(x – y) – y(x – y)= 0

∴  (x – y) (x – y)= 0

∴  (x – y)²= 0

∴  (x – y)  = 0

∴   x = y

Ans: These are coincident lines x = y

Example 06:

• Obtain separate equations of lines represented by joint equation 3x² + 5xy – 2y²= 0.
• Solution:

The given joint equation is  3x² + 5xy – 2y² = 0

∴  3x²+  6xy – xy – 2y² = 0

∴  3x(x + 2y) – y(x + 2y)= 0

∴  (x + 2y) (3x – y)= 0

Ans: The separate equations of the lines are x + 2y = 0 and 3x – y = 0

Example 07:

• Obtain separate equations of lines represented by joint equation 10x² + xy – 3y² = 0.
• Solution:

Given joint equation is 10x² + xy – 3y² = 0.

∴10x² + 6xy – 5xy – 3y² = 0

∴ 2x(5x + 3y) – y(5x + 3y) = 0

∴  (5x + 3y)(2x – y) = 0

Ans: The separate equations of the lines are 5x + 3y = 0 and 2x – y = 0

Example 08:

• Obtain separate equations of lines represented by joint equation x² – 4y² = 0.
• Solution:

Given joint equation is x² – 4y² = 0.

∴ x² – (2y)² = 0

∴ (x + 2y)(x – 2y) = 0

Ans: The separate equations of the lines are x + 2y = 0 and  x – 2y = 0

Example 09:

• Obtain separate equations of lines represented by joint equation 5x² -3y² = 0.
• Solution:

Given joint equation is 5x² -3y² = 0.

∴ (√5x)² – (√3y)² = 0

∴ (√5x + √3y)(x – √3y) = 0

Ans: The separate equations of the lines are √5x + √3y = 0 and  √5x – √3y = 0

Example 10:

• Obtain separate equations of lines represented by joint equation 6x² – 5xy + y² = 0.
• Solution:

Given joint equation is 6x² – 5xy + y² = 0.

∴ 6x² – 3xy – 2xy + y² = 0

∴ 3x(2x – y) – y(2x – y) = 0

∴ (2x – y)(3x – y) = 0

Ans: The separate equations of the lines are 2x – y = 0 and 3x – y = 0

Example 11:

• Obtain separate equations of lines represented by joint equation 3x² – y² = 0.
• Solution:

Given joint equation is  3x² – y² = 0.

∴   (√3x)² – y² = 0

∴  (√3x + y)(√3x – y) = 0

Ans: The separate equations are √3x + y = 0 and √3x – y = 0

Example 12:

• Obtain separate equations of lines represented by joint equation 3x² + 2xy + 7y² = 0.
• Solution:

Given joint equation is 3x² + 2xy + 7y² = 0.

Comparing with ax² + 2hxy + by² = 0

a = 3, 2h = 2, h = 1, b =  7

Now, h² – ab = (1)² – (3)(7) = 1 – 21 = – 20

Here h² – ab < 0, hence the lines are imaginary and can’t be drawn. Thus lines do not exist.

Example 13:

• Obtain separate equations of lines represented by joint equation 3x² – 7xy + 4y² = 0.
• Solution:

Given joint equation is 3x² – 7xy + 4y² = 0.

∴ 3x² – 3xy – 4xy + 4y² = 0

∴ 3x(x – y) – 4y(x – y) = 0

∴ (3x – 4y)(x – y) = 0

AQns: The separate equations of the lines are 3x – 4y = 0 and x – y = 0

Example 14:

• Obtain separate equations of lines represented by joint equation 3y² + 7xy = 0.
• Solution:

Given joint equation is 3y² + 7xy = 0.

∴  y(3y + 7x) = 0

Ans: The separate equations of the lines are 7x + 3y = 0 and y = 0

Example 15:

• Obtain separate equations of lines represented by joint equation 5x² – 9y² = 0.
• Solution:

Given joint equation is 5x² -9y² = 0.

∴   (√5x)² – (3y)² = 0

∴   (√5x + 3y)(√5x – 3y) = 0

Ans: The separate equations of the lines are √5x + 3y = 0 and  √5x – 3y = 0

Example 16:

• Obtain separate equations of lines represented by joint equation x² – 4xy = 0.
• Solution:

Given joint equation is x² – 4xy = 0.

∴ x(x – 4y) = 0

Ans: The separate equations of the lines are x – 4y = 0 and x = 0

Example 17:

• Obtain separate equations of lines represented by joint equation 3x² – 10xy – 8y² = 0.
• Solution:

Given joint equation is 3x² – 10xy – 8y² = 0.

∴ 3x² – 12xy + 2xy – 8y² = 0

∴ 3x(x – 4y) + 2y(x – 4y) = 0

∴ (3x + 2y)(x – 4y) = 0

Ans: The separate equations of the lines are 3x + 2y = 0 and x – 4y = 0

Example 18:

• Obtain separate equations of lines represented by joint equation 3x² – 2xy – 3y² = 0.
• Solution:

Given joint equation is 3x² – 2xy – y² = 0.

∴ 3x² – 3xy + xy – y² = 0

∴ 3x(x – y) + y(x – y) = 0

∴ (3x + y)(x – y) = 0

Ans: The separate equations of the lines are 3x + y = 0 and x – y = 0

Example 19:

• Obtain separate equations of lines represented by joint equation 6x² – 5xy – 6y² = 0.
• Solution:

Given joint equation is 6x² – 5xy – 6y² = 0.

∴ 6x² – 9xy+ 4xy – 6y² = 0

∴ 3x(2x – 3y) + 2y(2x – 3y) = 0

∴ (3x + 2y)(2x – 3y) = 0

Ans: The separate equations of the lines are 3x + 2y = 0 and 2x – 3y = 0

Example 20:

• Obtain separate equations of lines represented by joint equation 4x² – y² + 2x + y = 0. Also find point of intersection..
• Solution:

The given joint equation is  4x² – y²+ 2x + y = 0

∴  (2x)² – y²+ (2x + y) = 0

∴  (2x + y)(2x – y) + (2x + y) = 0

∴  (2x + y)(2x – y + 1] = 0

∴  Separate equations of the lines are 2x + y = 0 and 2x – y + 1 = 0

Let,   2x + y = 0    ……….. (1)

        2x – y = -1    ……….. (2)

Adding equations (1) and (2)

4x = -1 i.e x = -1/4

Substituting in equation (1) we get

2(-1/4) + y = 0

∴  -1/2 + y = 0

∴  y = 1/2

Hence the point of intersection is (-1/4, 1/2)

Ans: The separate equations of the lines are 2x + y = 0 and 2x – y + 1 = 0

and their point of intersection is (-1/4, -1/2)

Science > Mathematics > Pair of Straight Lines > Separate Equations of Lines (Factorization Method)

The post Separate Equations of Lines (Factorization Method) appeared first on The Fact Factor.

In this article, we shall study to find a combined or joint equation of pair of lines when a point on the line and equations of two other lines are given.

You Should Know It:

1. Slope of ax + by + c = 0 is – a/b (b ≠ 0) i.e. Slope = – coefficient of x / coefficient of y
2. When two lines are parallel their slopes are equal.
3. When two lines are perpendicular to each other then the product of their slopes is -1. Thus, If the slope of one line is m, then the slope of a line perpendicular to it is – 1/m.

Algorithm:

1. Find the slopes of the given line using the formula given in point 1 of notes.
2. Find the slopes of required lines forming a pair using relations given in points 2 and 3 of notes.
3. Use slope point form  y – y₁ =  m (x – x₁ ) to find equations of given lines. If Lines are passing through origin use y = mx form.
4. Write equations of lines in the form u = 0 and  v = 0.
5. Find u.v = 0.
6. Simplify the L.H.S. of the joint equation.

Example 01:

• Find the joint equation of a pair of lines through the origin such that one is parallel to x + 2y = 5 and the other is perpendicular to 2x – y + 3 = 0.
• Solution:

Let l₁ and l₂  be the two lines whose joint equation is to be found.

The equation of first given line is x + 2y = 5, 

Hence the slope of this line is – 1/2

As required first line (l₁) is parallel to this line.

The slope of l₁ is  – 1/2

The equation of the line (l₁) passing through the origin is

y = -1/2 x

∴  2y = – x

∴  x + 2y = 0   ….  (1)

Equation of second given line is 2x – y + 3 = 0. 

Hence the slope of this line is -2/-1 = 2

As required first line (l₂) is perpendicular to this line 

The slope of l₂ is – 1/2

The equation of line (2) passing through origin is

y =-1/2  x

∴  2y = -x

∴  x + 2y = 0   …….  (2)

From (1) and (2) the required combined equation is

(x + 2y) (x + 2y) = 0

∴   x² + 4xy + 4y² = 0

This is the required combined equation for the pair of lines.

Example 02:

• Find the joint equation of a pair of lines through the origin such that one is parallel to 2x + y – 5 = 0 and other is perpendicular to 3x – 4y + 7 = 0.
• Solution:

Let l₁ and l₂  be the two lines whose joint equation is to be found.

Equation of first given line is 2x + y – 5 = 0. 

Hence slope of this line is -2/1 = -2

As required first line ( l₁) is parallel to this line.

Hence slope of l₁ is – 2

The equation of line (1) passing through origin is

∴   y = – 2 x

  ∴   2x + y = 0  ….  (1)

Equation of second given line is 3x – 4y + 7 = 0. 

Hence slope of this line is – 3/-4 = 3/4

As required first line (l₂) is perpendicular to this line.

Hence slope of l₂ is -4/3

The equation of line (l₂) passing through origin is

y = – 4/3 x

∴   3y = – 4x

∴   4x + 3y = 0  …….  (2)

From (1) and (2) the required combined equation is

(2x + y) (4x + 3y) = 0

∴   8x² + 6xy + 4xy + 3y² = 0

∴   8x² + 10xy + 3y² = 0

This is the required combined equation for the pair of lines.

Example 03:

• Find the joint equation of a pair of lines through the origin such that one is parallel to 3x – y = 7 and the other is perpendicular to 2x + y = 8.
• Solution:

Let l₁ and l₂  be the two lines whose joint equation is to be found.

Equation of first given line is 3x + y – 7 = 0. 

Hence slope of this line is – 3/1 = – 3

As required first line ( l₁) is parallel to this line.

Hence slope of l₁ is – 3

The equation of line ( l₁) passing through origin is

y = – 3 x

 ∴   3x + y = 0   ….  (1)

Equation of second given line is 2x + y – 8 = 0. 

Hence slope of this line is -2/1 = – 2

As required first line (l₂) is perpendicular to this line.

Hence slope of l₂ is – 2

The equation of line (2) passing through origin is

y = -2x

∴   2x + y = 0   ….  (2)

From (1) and (2) the required combined equation is

(3x + y) (2x + y) = 0

∴   6x² + 6xy + 2xy + y² = 0

∴   6x² + 8xy + y² = 0

This is the required combined equation for the pair of lines.

Example 04:

• Find the joint equation of pair of lines through the origin, one of which is parallel to and another is perpendicular to the line 2x + 3y – 2 = 0.
• Solution:

Let l₁ and l₂  be the two lines whose joint equation is to be found.

The equation of the given line is 2x + 3y – 2 = 0.

Slope of the given line = m = -2/3

Let  l₁ be parallel to given line slope of l₁  is -2/3

The equation of line (1) passing through origin is

y =  – 2/3 x

∴   3 y = – 2 x

∴   2x + 3y = 0  …… (1)

Let  l₂ be perpendicular to given line slope of l₂  is 3/2

The equation of line (l₂) passing through origin is

y =  3/2 x

∴   2 y =  3 x

∴    3x – 2y = 0  …. (2)

From equations (1) and (2) the combined equation is

(2x + 3y) (3x – 2y) = 0

∴    2x(3x – 2y) + 3y(3x – 2y) = 0

∴    6x² – 4xy + 9xy – 6y² = 0

∴    6x² + 5xy – 6y² =  0

This is the required combined equation for the pair of lines.

Example 05:

• Find the joint equation of pair of lines through the origin, one of which is parallel to and another is perpendicular to the line 5x + 3y = 7.
• Solution:

Let l₁ and l₂  be the two lines whose joint equation is to be found.

The equation of the given line is  5x + 3y – 7 = 0.

Slope of the given line = m = – 5/3

Let  l₁ be parallel to given line slope of l₁  is -5/3

The equation of line (l₁) passing through origin is

y =  – 5/3 x

∴    3 y = – 5 x

∴    5x + 3y = 0 …………… (1)

Let  l₂ be perpendicular to given line slope of l₂  is 3/5

The equation of line (l₂) passing through origin is

y =  3/5 x

∴    5 y =  3 x

∴    3x – 5y = 0  …. (2)

From equations (1) and (2) the combined equation is

(5x + 3y) (3x – 5y) = 0

∴    5x(3x – 5y) + 3y(3x – 5y) = 0

∴    15x² – 25xy + 9xy – 15y² = 0

∴    15x² – 16xy – 15y² =  0

This is the required combined equation.

Example 06:

Example 35:   

• Find the joint equation of a pair of lines through the origin such that one is parallel to and the other is perpendicular to x + 2y + 1857 = 0.
• Solution:

Let l₁ and l₂  be the two lines whose joint equation is to be found.

The equation of the given line is x + 2y + 1857 = 0.

The slope of the given line = m = =1/2

Let l₁ be parallel to given line slope of l₁  is – 1/2

The equation of line (l₁) passing through origin is

y =  – 1/2 x

∴    2y = – x

∴    x + 2y = 0 …… (1)

Let  l₂ be perpendicular to given line slope of l₂  is 2

The equation of line (l₂) passing through origin is

y = 2 x

∴    2x – y = 0 …. (2)

From equations (1) and (2) the combined equation is

(x + 2y) (2x – y) = 0

∴    x(2x – y) + 2y(2x – y) = 0

∴    2x² – xy + 4xy – 2y² = 0

∴    2x² + 3xy – 2y² =  0

This is the required combined equation.

Example 07:

• Find the joint equation of pair of lines through the origin, and perpendicular to the lines x + 2y = 19 and 3x + y = 18.
• Solution:

Let l₁ and l₂  be the two lines whose joint equation is to be found.

The equation of the first given line is x + 2y – 19=0.

Slope of the first given line = m = – 1/2

Let  l₁ be perpendicular to firstgiven line.

Hence  slope of l₁  is  2

The equation of line (l₁) passing through origin is

y = 2x

∴    2x – y = 0  …… (1)

The equation of the second given line is 3x + y – 18 = 0

Slope of the second given line = m =  -3/1 = -3

Let l₂  be perpendicular to second given line.

Hence slope of l₂  is 1/3

The equation of line ( l₂) passing through origin is

y = 1/3 x

∴   3 y =  x

∴    x – 3y = 0  …. (2)

From equations (1) and (2) the combined equation is

(2x – y) (x – 3y) = 0

∴    2x(x – 3y) – y(x – 3y) = 0

∴    2x² – 6xy – xy + 3y² = 0

∴    2x² – 7xy + 3y² =  0

This is the required combined equation.

Example 08:

• Find the joint equation of a pair of lines through the point (3, 4) and such that one is parallel to 2x + 3y + 7 = 0 and the other is perpendicular to 3x – 5y = 8.
• Solution:

Let l₁ and l₂  be the two lines whose joint equation is to be found.

Equation of first given line is 2x + 3y + 7 = 0

Hence slope of this line is – 2/3

As required first line (l₁) is parallel to this line

Slope of l₁ is  = m = -2/3

The equation of line (l₁) passing through (3, 4) ≡ (x₁, y₁) is

(y – y₁)  = m(x – x₁)

∴    (y – 4)  = -2/3 (x – 3)

∴    3y -12 = – 2x + 6

∴    2x + 3y – 18 = 0 …….  (1)

Equation of second given line is 3x – 5y – 8 = 0.

Hence slope of this line is – 3/-5 = 3/5

As required first line (l₂) is perpendicular to this line

Slope of l₂ is = m = -5/3

The equation of line (l₂) passing through  (3, 4) ≡ (x₁, y₁) is

(y – y₁)  = m(x – x₁)

∴    (y – 4)  = – 5/3 (x – 3)

∴    3y -12 = – 5x + 15

∴    5x + 3y – 27 = 0   ….  (2)

From (1) and (2) the required combined equation is

(2x + 3y – 18) (5x + 3y – 27) = 0

∴     2x(5x + 3y – 27) +3y(5x + 3y – 27) -18(5x + 3y – 27) = 0

∴    10x² + 6xy – 54x + 15xy + 9y² – 81y – 90x -54 y + 486= 0

∴     10x² + 21xy + 9y² – 144x – 135y + 486= 0

This is the required combined equation.

Example 09:

• Find the joint equation of a pair of lines through the point (- 1, 2) and perpendicular to lines  x + 2y + 3 = 0 and 3x – 4y – 5 = 0.
• Solution:

Let l₁ and l₂  be the two lines whose joint equation is to be found.

Equation of first given line is  x + 2y + 3 = 0

Hence slope of this line is – 1/2

As required first line (l₁) is perpendicular to this line

Slope of l₁ is  = m = 2

The equation of line (l₁) passing through (-1, 2) ≡ (x₁, y₁) is

(y – y₁)  = m(x – x₁)

∴     (y – 2)  = 2(x + 1)

∴     y -2 =  2x + 2

∴     2x – y + 4 = 0  …….  (1)

Equation of second given line is 3x – 4y – 5 = 0.

Hence slope of this line is – 3/-4 = 3/4

As required first line (l₂) is perpendicular to this line

Slope of l₂ is = m = -4/3

The equation of line (l₂) passing through (-1, 2) ≡ (x₁, y₁) is

∴     (y – y₁)  = m(x – x₁)

∴     (y – 2)  = – 4/3 (x + 1)

∴     3y – 6 = – 4x – 4

∴     4x + 3y – 2 = 0   …….  (2)

From (1) and (2) the required combined equation is

( 2x – y + 4) (4x + 3y – 2) = 0

∴     2x(4x + 3y – 2) – y(4x + 3y – 2) + 4(4x + 3y – 2) = 0

∴     8x² + 6xy – 4x – 4xy – 3y² + 2y + 16x + 12 y – 8 = 0

∴      8x² + 2xy – 3y² + 12x + 14y – 8= 0

This is the required combined equation.

Example 10:

• Find the joint equation of a pair of lines through the point (3, 2) one of which is parallel to x – 2y = 2 and the other is perpendicular to y = 3.
• Solution:

Let l₁ and l₂  be the two lines whose joint equation is to be found.

Equation of first given line is  x – 2y – 2 = 0

Hence slope of this line is -1 /-2 = 1/2

As required first line (l₁) is parallel to this line

Slope of l₁ is  = m = 1/2

The equation of line (l₁) passing through (3, 2) ≡ (x₁, y₁) is

(y – y₁)  = m(x – x₁)

∴     (y – 2)  = (x – 3)

∴     2y – 4 = x – 3

∴     x – 2y+ 1 = 0   ….  (1)

Equation of second given line is y = 3.

The equation of line (l₂) passing through (3, 2) and perpendicular to y = 3  is

x = 3

∴     x – 3 = 0  …………….  (2)

From (1) and (2) the required combined equation is

( x – 3) (x – 2y+ 1) = 0

∴     x(x – 2y+ 1) – 3(x – 2y+ 1)  = 0

∴     x² – 2xy + x – 3x + 6y – 3 = 0

∴     x² – 2xy – 2x + 6y – 3= 0

This is the required combined equation.

Example 11:

• Find the joint equation of a pair of lines through the point (1, 2) and perpendicular to both the lines 3x + 2y – 5 = 0 and 2x – 5y + 1 = 0.
• Solution:

Let l₁ and l₂  be the two lines whose joint equation is to be found.

Equation of first given line is 3x + 2y – 5 = 0

Hence slope of this line is -3/2

As required first line (l₁) is perpendicular to this line

Slope of l₁ is  = 2/3 = m

The equation of line (l₁) passing through (1, 2) ≡ (x₁, y₁) is

(y – y₁)  = m(x – x₁)

∴    (y – 2)  = 2/3(x – 1)

∴    3y – 6 =  2x – 2

∴    2x – 3y + 4 = 0  ….  (1)

Equation of second given line is 2x – 5y + 1 = 0.

Hence slope of this line is – 2/-5 = 2/5

As required first line (l₂) is perpendicular to this line

Slope of l₂ is = -5/2 = m

The equation of line (l₂) passing through (3, 4) = (x₁, y₁) is

(y – y₁)  = m(x – x₁)

∴    (y – 2)  = – 5/2(x – 1)

∴    2y -4 = – 5x + 5

∴    5x + 2y – 9 = 0  ….  (2)

From (1) and (2) the required combined equation is

(2x – 3y + 4) (5x + 2y – 9) = 0

∴    x(5x + 2y – 9) – 3y(5x + 2y – 9) +4(5x + 2y – 9) = 0

∴    10x² + 4xy – 18x – 15xy – 6y² + 27y + 20x + 8y – 36= 0

∴    10x² – 11xy – 6y² + 2x + 35y – 36= 0

This is the required combined equation for the pair of lines.

Example 12:

• Find the joint equation of a pair of lines through the point (2, 3) and perpendicular to both the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0.
• Solution:

Let l₁ and l₂  be the two lines whose joint equation is to be found.

Equation of first given line is 3x + 2y – 1 = 0

Hence slope of this line is – 3/2

As required first line (l₁) is perpendicular to this line

Slope of l₁ is  = 2/3 = m

The equation of line (l₁) passing through (1, 2) = (x₁, y₁) is

(y – y₁)  = m(x – x₁)

∴    (y – 3)  = 2/3 (x – 2)

∴    3y – 9 =  2x – 4

∴    2x – 3y + 5 = 0  …….  (1)

Equation of second given line is x – 3y + 2 = 0.

Hence slope of this line is – 1/-3 = 1/3

As required first line (l₂) is perpendicular to this line

Slope of l₂ is – 3 = m

The equation of line (l₂) passing through (3, 4) = (x₁, y₁) is

(y – y₁)  = m(x – x₁)

∴    (y – 3)  = -3 (x – 2)

∴    y – 3 = – 3x + 6

∴    3x + y – 9 = 0   ….  (2)

From (1) and (2) the required combined equation is

(2x – 3y + 5) (3x + y – 9) = 0

∴    2x(3x + y – 9) – 3y(3x + y – 9) +5(3x + y – 9) = 0

∴    6x² + 2xy – 18x – 9xy – 3y² + 27y + 15x + 5y – 45= 0

∴    6x² –  7xy – 3y² – 3x + 32y – 45= 0

This is the required combined equation.

Science > Mathematics > Pair of Straight Lines > Combined Equation of Lines When Point and Two Other Lines are Given

The post Combined Equation of Lines When Point on it and Two Other Lines are Given appeared first on The Fact Factor.

In this article, we shall study to find a combined or joint equation of pair of lines when lines are parallel to coordinate axes.

Algorithm:

• When line passes through (h, k) and is parallel to co-ordinate axes are given. Equations of lines are x = h and y = k
• Write equations of lines in the form u = 0 and  v = 0.
• Find u.v = 0.
• Simplify the L.H.S. of the joint equation.

Example 01:

• Find the joint equation of a pair of lines which passes through (1, 2) and parallel to co-ordinate axes.
• Solution:

Let l₁ and l₂  be the two lines.

Equation of l₁ passing through (1, 2) and parallel to y-axis is x = 1

∴  x – 1 = 0    …………. (1)

Equation of l₂ passing through (1, 2) and parallel to  x-axis is y = 2

∴  y – 2 = 0   …………. (2)

The required joint equation is

(x – 1)(y – 2) = 0

∴ xy – 2x – y + 2 = 0

This is the required combined equation.

Example 02:

• Find the joint equation of a pair of lines which passes through (-3, 4) and parallel to co-ordinate axes.
• Solution:

Let l₁ and l₂  be the two lines.

Equation of l₁ passing through (-3, 4) and parallel to y – axis is x = -3

∴  x + 3 = 0    …………. (1)

Equation of l₂ passing through (-3, 4) and parallel to  x – axis is y = 4

∴   y – 4 = 0   …………. (2)

The required joint equation is

(x + 3)(y – 4) = 0

∴ xy – 4x + 3y – 12 = 0

This is the required combined equation.

Example 03:

• Find the joint equation of a pair of lines which passes through (3, 2) and parallel to the lines x = 2 and y = 3.
• Solution:

Let l₁ and l₂  be the two lines.

Equation of l₁ passing through (3, 2) and parallel to x = 2 is x = 3

∴  x – 3 = 0    …………. (1)

Equation of l₂ passing through (3, 2) and parallel to y = 3 is y = 2

∴   y – 2 = 0   …………. (2)

The required joint equation is

(x – 3)(y – 2) = 0

∴  xy – 2x – 3y + 6 = 0

This is the required combined equation.

Example 04:

• Find the joint equation of a pair of lines which passes through (2, 3) and parallel to co-ordinate axes.
• Solution:

Let l₁ and l₂  be the two lines.

Equation of l₁ passing through (1, 2) and parallel to y – axis is x = 2

∴  x – 2 = 0    …………. (1)

Equation of l₂ passing through (1, 2) and parallel to x – axis is y = 3

∴  y – 3 = 0   …………. (2)

The required joint equation is

(x – 2)(y – 3) = 0

∴  xy – 3x – 2y + 6 = 0

This is the required combined equation.

Example 05:

• Find the joint equation of a pair of lines which are at a distance of 9 units from the y-axis and parallel to it.
• Solution:

Let l₁ and l₂  be the two lines.

Equation of l₁ which is at a distance of 9 from y – axis and parallel to it is x = – 9 i.e. x + 9 = 0  …. (1)

Equation of l₂ which is at a distance of 9 from y – axis and parallel to it is x =  9 i.e. x – 9 = 0  …… (1)

The required joint equation is

(x + 9)(x – 9) = 0

∴  x² – 81 = 0

This is the required combined equation.

Example 06:

• Find the joint equation of a pair of lines which are at a distance of 5 units from the x-axis and parallel to it.
• Solution:

Let l₁ and l₂  be the two lines.

Equation of l₁ which is at a distance of 5 from x – axis and parallel to it is y = 5i.e. y – 5  = 0   ……… (1)

Equation of l₂ which is at a distance of 5 from x – axis and parallel to it is y =  -5 i.e. y + 5 = 0   …… (2)

The required joint equation is

(y – 5)(y + 5) = 0

∴ y² – 25 = 0

This is the required combined equation.

Science > Mathematics > Pair of Straight Lines > Combined Equation of Lines When Lines are Parallel to Axes

The post Combined Equation of Lines When Lines are Parallel to Axes appeared first on The Fact Factor.

In this article, we shall study to find a combined or joint equation of pair of lines when the slopes of lines are given.

Algorithm:

• Locate slopes m1 and m2 of the two lines.
• Use y = mx form to find equations of the two lines.
• Write equations of lines in the form u = 0 and  v = 0. Find u.v = 0.
• Simplify the L.H.S. of the joint equation.

Example 01:

• Find the joint equation of a pair of lines through the origin having slopes  1 and 3.
• Solution:

Let l₁ and l₂  be the two lines. Slope of l₁ is 1 and that of l₂  is 3

Therefore the equation of line (l₁) passing through origin and having slope 1 is

y = 1 (x)

∴ x  –  y = 0 ……. (1)

Similarly the equation of line (l₂) passing through origin and having slope 3 is

y = 3 (x)

∴ 3 x – y = 0  …. (2)

From (1) and (2) the required combined equation is

(x – y) (3x – y) = 0

∴ x(3x – y) – y(3x – y) = 0

∴ 3x² – xy – 3xy + y² = 0

∴ 3x² – 4xy + y² = 0

This is the required cobined equation.

Example 02:

• Find the joint equation of a pair of lines through the origin having slopes  4 and -1.

Solution:

Let l₁ and l₂  be the two lines. Slope of l₁ is 4 and that of l₂  is – 1

Therefore the equation of line (l₁) passing through origin and having slope 4 is

 y = 4 (x)

∴   4x – y = 0  ……. (1)

Similarly the equation of line (l₂) passing through origin and having slope -1 is

y = -1 (x)

∴ x + y = 0  …. (2)

From (1) and (2) the required combined equation is

(4x – y) (x + y) = 0

∴ 4x(x + y) – y(x + y) = 0

∴ 4x² + 4xy – xy + y² = 0

∴ 4x² + 3xy + y² = 0

This is the required cobined equation.

Example 03:

• Find the joint equation of a pair of lines through the origin having slopes  2 and – 1/2.
• Solution:

Let l₁ and l₂  be the two lines. Slope of l₁ is 2 and that of l₂  is – 1/2

Therefore the equation of line ( l₁) passing through origin and having slope 2 is

y = 2 (x)

∴ 2x – y = 0 ……. (1)

Similarly the equation of line (l₂) passing through origin and having slope  is

y =  -1/2 (x)

∴ 2y = – x

∴  x + 2y = 0  ………. (2)

From (1) and (2) the required combined equation is

(2x – y) (x + 2y) = 0

∴ 2x(x + 2y) – y(x + 2y) = 0

∴ 2x² + 4xy – xy – 2y² = 0

∴ 2x² + 3xy – 2y² = 0

This is the required cobined equation.

Example 04:

• Find the joint equation of a pair of lines through the origin having slopes 1/3  and -1/2.

Solution:

Let l₁ and l₂  be the two lines. Slope of l₁ is 1/3 and that of l₂  is – 1/2

Therefore the equation of line (l₁) passing through origin and having slope  is

y = 1/3 (x)

∴  3y = x

∴  x – 3y = 0  ……. (1)

Similarly the equation of line (l₂) passing through origin and having slope  is

y =  (x)

∴  2y = -x

∴  x + 2y = 0  ……. (2)

From (1) and (2) the required combined equation is

(x – 3y) (x + 2y) = 0

∴ x(x + 2y) – 3y(x + 2y) = 0

∴ x² + 2xy – 3xy – 6y² = 0

∴ x² – xy = 6y² = 0

This is the required cobined equation.

Example 05:

• Find the joint equation of a pair of lines through the origin having slopes 1 + √3  and 1 – √3.
• Solution:

Let l₁ and l₂  be the two lines. Slope of l₁ is 1 + √3 and that of l₂  is 1 – √3

Therefore the equation of line (l₁) passing through origin and having slope  is

y = (1 + √3)x

∴   (1 + √3)x – y = 0 …. (1)

Similarly, the equation of the line (l₂) passing through the origin and having slope   is

y = ( 1 – √3)x

∴  ( 1 – √3)x – y = 0 …. (2)

From (1) and (2) the required combined equation is

This is the required combined equation.

Science > Mathematics > Pair of Straight Lines > Combined Equation of Lines When Their Slopes are Given

The post Combined Equation of Lines When Their Slopes are Given appeared first on The Fact Factor.

In this article, we shall study to find a combined equation of pair of lines when the inclinations of lines are given.

Algorithm:

1. If θ is the inclination of a line, then its slope is given by m = tan θ.
2. Find slopes m₁ and m₂ of the two lines.
3. Use y = mx form to find equations of the two lines.
4. Write equations of lines in the form u = 0 and  v = 0.
5. Find u.v = 0.
6. Simplify the L.H.S. of the joint equation.

Example 01:

• Find the combined equation of the lines passing through the origin and making an angle of 30° with the positive direction of the x-axis
• Solution:

 Slope of the first line = m₁ = tan 30° =  1/√3

The equation the first line is y = m₁ x

y = 1/√3 x

√3 y  = x

x – √3 y  = 0     …………. (1)

 Slope of the second line = m₂ = tan (-30°)= – tan 30°  =  -1/√3

The equation the second line is y = m₂ x

y =-  1/√3x

√3 y  = – x

x + √3 y  = 0     …………. (2)

Their joint equation is (x – √3 y) (x + √3 y) = 0

x² – 3y² = 0

This is the required combined equation.

Example 02 :

• Find the combined equation of the lines passing through the origin and making an angle of 45° with the positive direction of the x-axis
• Solution:

Slope of the first line = m₁ = tan 45° =  1

The equation the first line is y = m₁ x

y =  1.x

x – y  = 0     …………. (1)

 Slope of the second line = m₂ = tan (-45°)= – tan 45°  =  -1

The equation the second line is y = m₂ x

y = – 1 x

x +  y  = 0     …………. (2)

Their joint equation is (x – y) (x + y) = 0

x² – y² = 0

This is the required combined equation.

Example 03:

• Find the combined equation of the lines passing through the origin and making an angle of 60° with the positive direction of the x-axis.
• Solution:

 Slope of the first line = m₁ = tan 60° =  √3

The equation the first line is y = m₁ x

y = √3x

√3 x  – y = 0     …………. (1)

 Slope of the second line = m₂ = tan (-60°)= – tan 60°  =  -√3

The equation the second line is y = m₂ x

y =-  √3x

√3 x  + y = 0     …………. (2)

Their joint equation is (√3 x  – y) (√3 x + y) = 0

3x² – y² = 0

This is the required combined equation.

Example 04:

• Find the combined equation of the lines passing through the origin and making an angle of 60° with the y-axis
• Solution:

 Slope of the first line = m₁ = tan 30° =  1/√3

The equation the first line is y = m₁ x

y = 1/√3x

√3 y  = x

x – √3 y  = 0     …………. (1)

 Slope of the second line = m₂ = tan (150°)= tan(180° – 30°) = -tan 30°  =  -1/√3

The equation the second line is y = m₂ x

y =-  1/√3x

√3 y  = – x

x + √3 y  = 0     …………. (2)

Their joint equation is (x – √3 y) (x + √3 y) = 0

x² – 3y² = 0

This is the required combined equation.

Example 05:

• Find the joint equation of a pair of lines through the origin having and having inclination 30° and 150°.
• Solution:

 Slope of the first line = m₁ = tan 30° =  1/√3

The equation the first line is y = m₁ x

y = 1/√3x

√3 y  = x

x – √3 y  = 0     …………. (1)

 Slope of the second line = m₂ = tan (150°)= tan(180° – 30°) = -tan 30°  =  -1/√3

The equation the second line is y = m₂ x

y =-  1/√3x

√3 y  = – x

x + √3 y  = 0     …………. (2)

Their joint equation is (x – √3 y) (x + √3 y) = 0

x² – 3y² = 0

This is the required combined equation.

Example 06:

• Find the joint equation of a pair of lines through the origin having and having inclination 60° and 120°
• Solution:

 Slope of the first line = m₁ = tan 60° =  √3

The equation the first line is y = m₁ x

y = √3x

√3 x  – y = 0     …………. (1)

 Slope of the second line = m₂ = tan (120°)= tan (180° – 60°) = – tan 60°  =  -√3

The equation the second line is y = m₂ x

y =-  √3x

√3 x  + y = 0     …………. (2)

Their joint equation is (√3 x  – y) (√3 x + y) = 0

3x² – y² = 0

This is the required combined equation.

Example 07:

Find the joint equation of a pair of lines through the origin having and having inclination π/3 and 5π/3.

Solution:

 Slope of the first line = m₁ = tan π/3 =  √3

The equation the first line is y = m₁ x

y = √3x

√3 x  – y = 0     …………. (1)

 Slope of the second line = m₂ = tan (5π/3)= tan (2π – π/3) = – tan π/3  =  -√3

The equation the second line is y = m₂ x

y =-  √3x

√3 x  + y = 0     …………. (2)

Their joint equation is (√3 x  – y) (√3 x + y) = 0

3x² – y² = 0

This is the required combined equation.

Example 08:

• Find the joint equation of the straight lines through the origin which forms the equilateral triangle with the straight line x = 1.
• Solution:

 Let OA and OB be two lines such that triangle OAB is an equilateral triangle.

Equation of AB is x = 1 and AB, is parallel to the y-axis.

Now triangle OAB is an equilateral triangle,

by symmetry, m ∠ MOA  = m ∠MOB = 30°

Slope of OA  = m₁ = tan(-30°) =  – tan 30° = – 1/√3  and

Slope of OB = m₂ = tan( 30°)  =  1/√3

Equation of OA is passing through the origin, is

y =  m₁x. i.e.   y = – 1/√3  x

∴  √3 y = –  x

∴  x + √3 y = 0  …….. (1)

As OB is also passing through the origin, the form of the equation of OB is

y =  m₂x. i.e.   y = 1/√3  x

∴  √3 y =  x

∴  x – √3 y = 0  …….. (2)

Hence, the combined equation of the pair OA and OB is

(x + √3 y)(x – √3 y ) = 0

∴  x ² – 3y ² = 0.

This is the required combined equation.

Note: The joint equation of the straight lines through the origin which forms the equilateral triangle with any straight line x = k is always x ² – 3y ² = 0.

Example 09:

• Find the joint equation of the straight lines through the origin which forms the equilateral triangle with the straight line y  =  2.
• Solution:

Let OA and OB be two lines such that the triangle OAB is an equilateral triangle.

The equation of AB is y  =  2 and AB is parallel to the x-axis.

Now since triangle OAB is an equilateral triangle,

m ∠ XOA = 60° and m ∠ XOB = 120°

∴  Slope of OA =  tan 60° =  √3 and

∴  slope of OB = tan 120° =  – √3 

As OA is passing through the origin, the form of equation of OA is

 y = m₁x.  i.e. y = √3 x

∴  √3 x – y = 0  ………. (1)

As OB is also passing through the origin, the form of equation of OB is y = mx.

 y = m₂x.  i.e. y = – √3 x

∴  √3 x +  y = 0  ………. (2)

Hence their combined equation is

( √3 x – y)(  √3 x +  y) = 0

∴  3x² – y² = 0.

This is the required combined equation.

Note: The joint equation of the straight lines through the origin which forms the equilateral triangle with any straight line y = k is always  3x² – y² = 0.

Science > Mathematics > Pair of Straight Lines > Combined Equation of Lines When Inclinations are Given

The post Combined Equation of Lines When Their Inclinations are Given appeared first on The Fact Factor.

In this article, we shall study to find a combined equation of pair of lines when a separate equation of each line is given.

Algorithm:

1. Write equations of lines in the form u = 0 and  v = 0.
2. Find u.v = 0.
3. Simplify the L.H.S. of the joint equation.

Example 01:

• Obtain the joint equation of the co-ordinate axes.
• Solution:

Equation of the x-axis is  y =  0 .

Equation of the y-axis is  x =  0

The joint equation of the two co-ordinate axis is x y = 0

Example 02:

• Find the joint equation of the two lines whose separate equations are 3x – 2y = 0 and 4x + y = 0.
• Solution:

Equation of the first line is 3x -2y = 0

Equation of the second line is  4x + y= 0

Thus the joint equation of a line is given by

(3x – 2y)(4x + y) = 0

∴ 3x (4x + y)   – 2y (4x + y) = 0

∴ 12x² + 3xy – 8xy – 2y² = 0

∴ 12x² – 5xy – 2y² = 0

This is the required combined equation.

Example 03:

• Find the combined equation of the two lines whose separate equations are 3x + 4y = 0 and 2x = 3y
• Solution:

Equation of the first line is 3x + 4y = 0

Equation of the second line is  2x = 3y i.e 2x – 3y = 0

Thus the joint equation of a line is given by

(3x + 4y)(2x – 3y ) = 0

∴ 3x (2x – 3y)   + 4y (2x – 3y ) = 0

∴ 6x² – 9xy + 8xy – 12y² = 0

∴  6x² – xy – 12y² = 0

This is the required combined equation.

Example 04:

• Find the combined equation of the lines whose separate equations are 2x + y = 0 and 3x – 5y = 0.
• Solution:

Given equations of lines are 2x + y = 0 and 3x – 5y = 0

(2x + y )(3x – 5y) = 0

∴ 2x(3x – 5y) + y(3x – 5y) = 0

∴ 6x² – 10xy + 3xy  – 5y² = 0

∴  6x² – 7xy  – 5y² = 0

This is the required combined equation.

Example 05:

• Find the combined equation of the lines whose separate equations are 3x – 2y + 1 = 0 and 4x – 3y + 5 = 0
• Solution:

Given equations of lines are 3x – 2y + 1 = 0 and 4x – 3y + 5 = 0

(3x – 2y + 1)(4x – 3y + 5) = 0

3x(4x – 3y + 5) – 2y(4x – 3y + 5) + 1(4x – 3y + 5) =

∴     12x² – 9xy + 15x – 8xy + 6y² – 10y + 4x – 3y + 5 = 0

∴      12x² – 17 xy  + 6y² + 19x  – 13y + 5 = 0

This is the required combined equation.

Example 06:

• Find the combined equation of the lines whose separate equations are x + y = 3 and  2x + y – 1 = 0.
• Solution:

Given equations of lines are x + y – 3 = 0 and  2x + y – 1 = 0

(x + y – 3)(2x + y – 1) = 0

x(2x + y – 1) + y(2x + y – 1) – 3(2x + y – 1) = 0

∴   2x² + xy – x + 2xy + y² – y – 6x – 3y + 3 = 0

∴    2x² + 3xy  + y² – 7x  – 4y + 3 = 0

This is the required combined equation.

Example 07:

• Find the combined equation of the lines whose separate equations are 3x + 2y – 1 = 0 and x + 3y – 2 = 0.
• Solution:

Given equations of lines are 3x + 2y -1 = 0 and x + 3y -2 = 0.

(3x + 2y -1 )(x + 3y -2) = 0

3x(x + 3y -2) + 2y(x + 3y -2) – 1(x + 3y -2) = 0

∴   3x² + 9xy – 6x + 2xy + 6y² – 4y – x – 3y + 2 = 0

∴   3x² + 11xy  + 6y² – 7x  – 7y + 2 = 0

This is the required combined equation.

Example 08:

• Find the combined equation of the lines whose separate equations are 2x + y = 0 and 3x – 5y = 0.
• Solution:

Given equations of lines are 2x+y = 0 and 3x-5y = 0

(2x+y )(3x-5y) = 0

∴ 2x(3x-5y) + y(3x-5y) = 0

∴ 6x² – 10xy + 3xy  – 5y² = 0

∴  6x² – 7xy  – 5y² = 0

This is the required combined equation.

Example 09:

• Find the combined equation of the lines whose separate equations are x + 2y – 1 = 0 and 2x – 3y + 2 = 0.
• Solution:

Given equations of lines are x + 2y – 1 = 0 and 2x – 3y + 2 = 0.

(x + 2y – 1 )(2x – 3y + 2) = 0

∴  x(2x – 3y + 2) + 2y(2x – 3y + 2) – 1(2x – 3y + 2) = 0

∴ 2x² – 3xy + 2x + 4xy – 6y² + 4y – 2x + 3y – 2 = 0

∴  2x² + xy  – 6y² + 7y – 2 = 0

This is the required combined equation.

Science > Mathematics > Pair of Straight Lines > Combined Equation of Lines when Separate Equations are Given

The post Combined Equation of Lines when Separate Equations are Given appeared first on The Fact Factor.