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