If the pair of lines ${\mathrm{x}}^2+2\mathrm{xy}-35{\mathrm{y}}^2-4\mathrm{x}+44\mathrm{y}-12=0$ and $5\mathrm{x}+\mathrm{\lambda y}-8=0$are concurrent, then $\mathrm{\lambda }$ is equal to

$$\begin{gathered} a=1,2h=2,b=-35,2g=-4,2f=44, \\ Point of intersection=\left(\frac{hf-bg}{ab-h^2},\frac{gh-af}{ab-h^2}\right) \\ hf-bg=1\left(22\right)-(-35)(-2)=22-70=-48 \\ gh-af=(-2)\left(1\right)-1\left(22\right)=-24 \\ ab-h^2=1(-35)-1=-36 \\ Point of intersection=\left(\frac{hf-bg}{ab-h^2},\frac{gh-af}{ab-h^2}\right)=\left(\frac{-48}{-36},\frac{-24}{\lambda }\right)=(\frac{4}{3},\frac{2}{3}) \\ substitute (\frac{4}{3},\frac{2}{3}) in 5\mathrm{x}+\mathrm{\lambda y}-8=0 \\ 5\left(\frac{4}{3}\right)+\lambda \left(\frac{2}{3}\right)-8=0\Rightarrow 20+2\lambda -24=0 \\ \therefore \lambda =2 \end{gathered}$$