To Find Separate Equations from Joint Equation

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Science > Mathematics > Pair of Straight LinesYou are Here

Type – III: To Find Separate Equations of Lines When Joint Equation is Given

The Algorithm of Method – I:

  1. Check if lines exist. use the same method used in the case to find nature of lines (do this orally).
  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.

The Algorithm of Method – II:

  1. Check if lines exist. use the same method used in the case to find nature of lines.
  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 – 1:

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

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  – 2:

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

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 – 3:

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

Solution:

Given joint equation is y2 – 5x2 = 0.

5x2 – y2 = 0.



∴  (5x)2 – y= 0

∴  (5x + y) (5x – y) = 0.

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

Example – 3:

  • 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 – 4:

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

Given joint equation is 3x2 + 10xy + 8y= 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 – 4:

  • 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  – 5:

  • 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 – 6:

  • 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 – 7:

  • 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 – 8:

  • 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 – 9:

  • 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 – 10:

  • 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 – 11:

  • 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 – 12:

 

  • 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 – 13:

  • 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 – 14:

  • 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 – 15:

  • 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 – 16:

  • 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 – 17:

  • 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 – 18:

  • 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 – 19:

  • 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 – 20:

  • 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 – 21:

  • 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 – 22:

  • 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 – 23:

  • 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 – 24:

  • 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

Example – 25:

  • 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 LinesYou are Here
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