# HOW TO FIND SEPARATE EQUATION FROM PAIR OF STRAIGHT LINES

If the given equation is in the form

ax2 + 2hxy + by2  =  0

by factoring we may get two linear factors and that will be the separate equations for the straight line.

If the given equation is in the form

ax2 + 2hxy + by2 + 2gx + 2fy + c =  0

then we have to take the first three terms and find factors.

• In the factors, we will have only x and y terms.
• Now we have to take constants with the factors.
• After adding the constants, the product of those factors will be equal to the original equation.
• Now we have to equate the coefficients of x and y terms and find the values of constants.

Let us see an example problem for better understanding.

Question 1 :

Find the separate equation of the following pair of straight lines

(i) 3x2 + 2xy − y2 = 0

Solution :

3x2 - xy  + 3xy − y2 = 0

x(3x - y) + y(3x - y)  =  0

(3x - y) (x + y)  =  0

(ii) 6(x − 1)2 + 5(x − 1)(y − 2) − 4(y − 2)2 = 0

Solution :

6(x − 1)2 + 5(x − 1)(y − 2) − 4(y − 2)2 = 0

6(x − 1)2 - 3(x − 1)(y − 2) + 8(x − 1)(y − 2) − 4(y − 2)2 = 0

3(x- 1) [2(x-  1) - (y -2)] + 4(y - 2) [2(x - 1) - (y - 2)]  =  0

3(x- 1) [2x - 2 - y + 2] + 4(y - 2) [2x - 2 - y + 2]  =  0

3(x- 1) (2x- y) + 4(y - 2) (2x - y)  =  0

(2x - y) (3x - 3 + 4y - 8)  =  0

(2x - y) (3x + 4y - 11)  =  0

(iii) 2x2 − xy − 3y2 − 6x + 19y − 20 = 0

Solution :

=  2x2 − xy − 3y2

=  2x2 + 2xy - 3xy − 3y2

=  2x (x + y) - 3y(x + y)

=  (2x - 3y) (x + y)

Separate equations of the straight line are

(2x - 3y + l) (x + y + m)

In order to find the values of l and m, we have to equate the coefficients of x and y.

2x2 − xy − 3y2 − 6x + 19y − 20  =  (2x - 3y + l) (x + y + m)

-6  =  2m + l ------------(1)

19  =  -3m + l ------------(2)

(1) - (2)

(2m + l) - (-3m + l)  =  -6 - 19

2m + l + 3m - l  =  -25

5m  =  -25   ==>  m  =  -5

By applying the value of m in the first equation, we get

2(-5) + l  =  -6

-10 + l  =  -6

l  =  -6 + 10  ==>  4

Hence the required equations are (2x - 3y + 4) (x + y - 5).

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