Question

Find the value of k so that the following equations may represent pairs of straight lines :
$12x^2-kxy+2y^2+11x-5y+2=0.$

Answer 1

The general equation of second degree $a{x}^2+2hxy+{by}^2+2gx+2fy+c=0$

Represents a pair of straight lines if $\Delta =abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

For $12x^2-kxy+2y^2+11x-5y+2=0$

$a=12,b=2,h=\frac{-k}{2},g=\frac{11}{2},f=\frac{-5}{2},c=2\Delta =(12\times 2\times 2)+(2\times \frac{-5}{2}\times \frac{11}{2}\times \frac{-k}{2})-(12\times \frac{-5}{2}\times \frac{-5}{2})-(2\times \frac{11}{2}\times \frac{11}{2})-(2\times \frac{-k}{2}\times \frac{-k}{2})=048+\frac{55k}{4}-75-\frac{121}{2}-\frac{k^2}{2}=0\frac{k^2}{2}-\frac{55k}{4}+\frac{121}{2}+28=02k^2-55k+350=0$

Factorising the equation

$2k^2-20k-35k+350=02k(k-10)-35(k-10)=0(2k-35)(k-10)=0k=10,\frac{35}{2}\Rightarrow k=10,17\frac{1}{2}$