Separate Equations of Lines (Factorization Method)

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

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)