a traingle has the lines y=m₁x and y=m_{2 ​}x for two of its sides,where m₁ and m₂ are the roots of the equation x²+ax-1=0,then find the orthocentre of triangle

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{y}={\mathrm{m}}_1\mathrm{x} \mathrm{and} \mathrm{y}={\mathrm{m}}_2\mathrm{x} \\ {\mathrm{m}}_1 \mathrm{and} {\mathrm{m}}_2 \mathrm{are} \mathrm{roots} \mathrm{of} {\mathrm{x}}^2+\mathrm{ax}-1=0 \\ \mathrm{Product} \mathrm{of} \mathrm{roots}={\mathrm{m}}_1{\mathrm{m}}_2=-1 \\ \mathrm{which} \mathrm{means} \mathrm{y}={\mathrm{m}}_1\mathrm{x} \mathrm{and} \mathrm{y}={\mathrm{m}}_2\mathrm{x} \mathrm{are} \mathrm{perpendicular} \mathrm{to} \mathrm{each} \mathrm{other} \\ \mathrm{Hence} \mathrm{orthocentre} \mathrm{will} \mathrm{be} \mathrm{at} \mathrm{their} \mathrm{point} \mathrm{of} \mathrm{intersection}. \\ \mathrm{And} \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of} \mathrm{y}={\mathrm{m}}_1\mathrm{x} \mathrm{and} \mathrm{y}={\mathrm{m}}_2\mathrm{x} \mathrm{is} \left(0,0\right) \\ \mathrm{Hence} \mathrm{orthocentre} \mathrm{is} \left(0,0\right) \end{gathered}$$

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