Question

Find the value of k so that the following equations may represent pairs of straight lines $12x^2-10xy+{2y}^2+11x-5y+k=0$

Answer 1

We have,

Equation of straight line

$12x^2-10xy+2y^2+11x-5y+k=0$

On comparing that,

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

So,

$a=12$

$h=-5$

$b=2$

$g=\frac{11}{2}$

$f=\frac{-5}{2}$

$c=k$

Then,

We know that,

$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

So,

$12\times -5\times k+2\times \frac{-5}{2}\times \frac{11}{2}\times (-5)-12\times {\left(\frac{-5}{2}\right)}^2-2\times {\left(\frac{11}{2}\right)}^2-k{(-5)}^2=0$

$\Rightarrow -60k+\frac{275}{2}-75-\frac{121}{2}-25k=0$

$\Rightarrow -75k+\frac{154}{2}-75=0$

$\Rightarrow -75k+2=0$

$\Rightarrow -75k=-2$

$\Rightarrow k=\frac{2}{75}$

Hence, this is the answer.