The point of intersection of the straight lines represented by $6x^2+xy-40y^2-35x-83y+11=0is$ 

$\begin{array}{l}GivenEquation6x^2+xy-40y^2-35x-83y+11=0\\ Comparingwitha{x}^2+2hxy+b{y}^2+2gx+2fy+c=0\\ a=6,h=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,b=-40,g=\raisebox{1ex}{$-35$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,f=\raisebox{1ex}{$-83$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,c=11\\ P.I=(\frac{hf-bg}{ab-h^2},\frac{gh-af}{ab-h^2})\\ =(\frac{\frac{1}{2}\left(\frac{-83}{2}\right)+40\left(\frac{-35}{2}\right)}{6(-40)-\frac{1}{4}},\frac{\frac{-35}{2}\left(\frac{1}{2}\right)-6\left(\frac{-83}{2}\right)}{6(-40)-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.})\\ =(\frac{-83-2800}{-960-1},\frac{-35+996}{-960-1})\\ =(\frac{-2883}{-961},\frac{961}{-961})\Rightarrow (3,-1)\end{array}$