How to split a pair of line equation of ax2+2hxy+by2+2gx+2fy+c=0 in the short method?
Let ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represent a pair of straight line :
Take a example : 6x2 +17xy +12y2 +22x + 31y +20 = 0
1) here first factorise the terms containing second degree .
6x2 +17xy +12y2 = 6x2 +8xy +9xy +12y2 = (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
Take a example : 6x2 +17xy +12y2 +22x + 31y +20 = 0
1) here first factorise the terms containing second degree .
6x2 +17xy +12y2 = 6x2 +8xy +9xy +12y2 = (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