The value of k such that the lines $6x^2+17xy+12y^2=0andk{x}^2+5xy+3y^2=0$ may have one line in common is

$\begin{array}{l}GivenEquation6x^2+17xy+12y^2=0\\ \Rightarrow 6x^2+9xy+8xy+12y^2=0\\ \Rightarrow 3x(2x+3y)+4y(2x+3y)=0\\ \Rightarrow (2x+3y)(3x+4y)=0\\ \therefore 2x+3y=0\Rightarrow y=\frac{-2x}{3}--\left(1\right)\\ 3x+4y=0\Rightarrow y=\frac{-3x}{4}--\left(2\right)\\ If\left(1\right)iscommonlinethenput\left(1\right)ink{x}^2+5xy+3y^2=0\\ \Rightarrow k{x}^2-\frac{10x^2}{3}+3\left(\frac{4x^2}{9}\right)=0\\ \Rightarrow x^2(k-\frac{10}{3}+\frac{4}{3})=0\\ \therefore x^2\ne 0\Rightarrow k-\frac{10}{3}+\frac{4}{3}=0\\ \Rightarrow k=2\end{array}$