Question

Find the separate equation of the following pair of straight lines
2x² – xy – 3y² – 6x + 19y – 20 = 0

Answer 1

2x² – xy – 3y² – 6x + 19y – 20 = 0

Factorising 2x² – xy – 3y² we get

2x² – xy – 3y² = 2x² + 2xy – 3xy – 3y²

= 2x(x + y) – 3y(x + y) = (2x – 3y)(x + y)

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

Equating coefficient of x 2m + l = -6  .......(1)

Equating coefficient of y – 3m + l = 19   .......(2)

Constant term – 20 = lm

Solving (1) and (2) we get l = 4 and m = – 5 where lm = – 20.

So the separate equations are 2x – 3y + 4 = 0 and x + y – 5 = 0