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), $(1)(-1)(k)+2(1)(3)(-2)-(1){(1)}^2-(-1)$
${(3)}^2-(k){(-2)}^2=0$
$\Rightarrow$ $-k-12-1+9-4k=0$
$\Rightarrow$ $-5k=4\Rightarrow k=-4/5$