Two sides of a triangle are $y=m_{1}xandy=m_{2}x;m_1,m_2$ are the roots of the equation $x^2+ax-1=0.$ For all values of ‘a’, the orthocenter of the triangle lies at 

Given lines $y=m_{1}xandy=m_{2}x$

Given  m₁, m₂ are roots $x^2+ax-1=0$

$\begin{array}{l}{m}_1+m_2=-a{m}_1{m}_2=-1\\ \therefore Thelinesare\perp r\\ \therefore orthocentre=P.Iof\perp rlines\\ \therefore P.Iofy=m_{1}x,y=m_{2}xis\left(0,0\right)\\ \therefore Orthocentre=\left(0,0\right)\end{array}$