Question

The value of $k$ for which the equation
$ {{x}^{2}}-4xy-{{y}^{2}}+6x+2y+k=0 $ represents a pair of straight lines is

• A. $ k=4 $
• B. $ k=-1 $
• C. $ k=\frac{-4}{5} $
• D. $ k=\frac{-22}{5} $

Answer 1

Answer: (C)

The given equation $ {{x}^{2}}-4xy-{{y}^{2}}+6x+2y+k-0 $ ..(i)
represent the point of straight line, if $ \Delta =0, $
where $ \Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 $ ..(ii)
Comparing Eq. (i) with the following equation
$ a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 $
We get, $ a=1,\,h=-2,\,b=-1,\,\,g=3,\,f=1,\,c=k $
From Eq. (ii), $ (1)\,(-1)\,(k)+2(1)\,(3)\,(-2)-(1)\,{{(1)}^{2}}-(-1) $
$ {{(3)}^{2}}-(k)\,{{(-2)}^{2}}=0 $
$ \Rightarrow $ $ -k-12-1+9-4k=0 $
$ \Rightarrow $ $ -5k=4\,\,\Rightarrow \,k=-4/5 $