How to split a pair of line equation of ax^{2}+2hxy+by^{2}+2gx+2fy+c=0 in the short method?

^{2}+ 2hxy + by

^{2}+ 2gx + 2fy + c = 0 represent a pair of straight line :

Take a example : 6x

^{2}+17xy +12y

^{2}+22x + 31y +20 = 0

1) here first factorise the terms containing second degree .

6x

^{2}+17xy +12y

^{2 }= 6x

^{2}+8xy +9xy +12y

^{2}= (2x +3y)(3x +4y)

2) Now add constant p and q to these factors .

(2x +3y+p)(3x +4y+q) (1)

Compare the coefficient of x ,y and constant in product of (1)

So 3p + 2q = 22

4p +3q = 31

And pq = 20

On solving above equation, we have p = 4 and q =5

So the lines are (2x +3y+4) = 0 and (3x +4y+5) = 0

**
**