If 3x^2+ xy-y^2-3x+6y+k=0 represents a pair of lines, then k=

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Accepted Answer

Correct answer: B. (b) : 3x^2+ xy-y^2-3x+6y+k=0 represents a pair of lines. Comparing this equation with ax^2+2 hxy+by^2+2 gx+2 fy+c=0 a=3 ,h=1/2 ,b=-1 ,g=-3/2 ,f=3 ,c=k Now, abc+2 fgh-af^2-bg^2-ch^2=0 (Condition for pair of straight lines) 3 (-1 )k+2 3 (- 32) ( 12 )-3 (3 )^2 - (-1 )(- 32)^2-k ( 12 )^2=0 ⇒-3k- 92-27+ 94- k4=0 ⇒3k+ k4= -18-108+94 ⇒13k=-117 ⇒k=-9. ⇒ 13k=-117 ⇒k=-9.

$If 3 x^{2} + xy - y^{2} - 3 x + 6 y + k = 0 represents a pair of lines, then k =$

Solution

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