What should be the value of $m$ , so that the expression $x^{2} + y^{2} + m x y - 3 x - 3 y + 2$ can be resolved into two linear factors ?

2

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3

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Solution

The correct option is D $2$
We know that a second degree equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0 . . . . (1)$
represents a pair of straight lines if $a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0 . . . (2)$

Compare $(1)$ with $x^{2} + y^{2} + m x y - 3 x - 3 y + 2 = 0$ we get

$a = 1, b = 1, h = \frac{m}{2}, g = - \frac{3}{2}, f = - \frac{3}{2}, c = 2$

Put the values of $a, b, c, f, g a n d h$ in $(2)$ we get

$2 + \frac{9}{4} m - \frac{9}{4} - \frac{9}{4} - \frac{2 m^{2}}{4} = 0$

$\Rightarrow 8 + 9 m - 18 - 2 m^{2} = 0$

$\Rightarrow 2 m^{2} - 9 m + 10 = 0$

$\Rightarrow (2 m - 5) (m - 2) = 0$

$\Rightarrow m = \frac{5}{2}, 2$

But $m = 2$ satisfies the condition $(2)$

Therefore when $m = 2$ the expression $x^{2} + y^{2} + m x y - 3 x - 3 y + 2 = 0$ can be resolved into two linear factors.

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