Question

The value of K, So that the equation $-2x^2+xy+y^2-5x+y+K=0$ may represent a pair of Straight line is

Answer 1

Solution :-

Given equation $-2x^2+xy+y^2-5x+y+k=\mathrm{0...}\left(1\right)$

Condition for an equation to represent a pair of straight line is

$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0...\left(ii\right)$

comparing equation (i) & (ii)

$a=-2$ $b=-1$ $c=k$

$h=1/2$

$g=-5/2$

$f=1/2$

substituting we get

$(-2)(-1)\left(k\right)+2\left(\frac{1}{2}\right)\left(\frac{-5}{2}\right)\left(\frac{1}{2}\right)-(-2)(\frac{1}{2}{)}^2-(-1\left)\right(\frac{-5}{2}{)}^2-\frac{k}{4}=0$

$\Rightarrow 2k+\left(\frac{-5}{4}\right)+\frac{1}{2}+\frac{25}{4}-\frac{k}{4}=0$

$\Rightarrow \frac{7k}{4}+\left(\frac{-5+2+25}{4}\right)=0$

$\Rightarrow 7k+22=0$

$\Rightarrow k=\frac{22}{7}$