The orthocentre of the triangle formed by the lines x + 3y – 10 = 0 and $6x^2+xy-y^2=0is$

$\begin{array}{l}GivenEquation6x^2+xy-y^2=0\\ Comparingwitha{x}^2+2hxy+b{y}^2=0\\ a=6,2h=1,b=-1\\ andx+3y-10=0withlx+my+n=0\\ l=1,m=3,n=-10\\ k=\frac{-n(a+b)}{a{m}^2-2hlm+b{l}^2}\end{array}$

$\begin{array}{l}=\frac{10(6-1)}{6\left(9\right)-1\left(1\right)\left(3\right)+(-1)1}\Rightarrow \frac{10\left(5\right)}{54-3-1}=\frac{50}{50}=1\\ \therefore Orthocentre=(kl,km)=(1\left(1\right),1\left(3\right))\Rightarrow (1,3)\end{array}$