If the equation `3x^(2)+xy-y^(2)-3x+6y+k=0` represents a pair of straight lines, then the value of k, is

Options

A. 0
B. 9
C. 1
D. -9

Answer

D

Explanation

Given equation is `3x^2+xy-y^2-3x+6y+k=0`
Here , `a=b=-1, h =(1)/(2),g=-(3)/(2),f=-(3)/(2),f=3,c=k`
Given equation represents a pair of straight line , if `abc+2fgh-af^2-bh^2-ch^2=0`
`therefore 3(-1)(k)+2xx3xx(-(3)/(2))xx(1)/(2)-3(3)^2+1((-3)/(2))^2-k((1)/(2))^2=0`
`rArr -3k-(9)/(2)-27+(9)/(4)-(k)/(4)=0rArr (-13k)/(4)-(117)/(4)rArr k=-9`