Mathematics - Point Question with Solution | TestHub

Let the third vertex of $△ABC$ be $(a,b)$. Orthocentre $=H(0,0)$ Let $A(5,-1)$ and $B(-2,3)$ be other two vertices of $△ABC$. Now, (Slope of $AH)\times ($ Slope of $BC)=-1$ $$\begin{array}{rl}& \Rightarrow \left(\frac{-1-0}{5-0}\right)\left(\frac{b-3}{a+2}\right)=-1\\ & \Rightarrow b-3=5(a+2)\end{array}$$ Similarly, $($ Slope of $BH)\times ($ Slope of $AC)=-1$ $$\begin{array}{rl}& \Rightarrow -\left(\frac{3}{2}\right)\times \left(\frac{b+1}{a-5}\right)=-1\\ & \Rightarrow 3b+3=2a-10\\ & \Rightarrow 3b-2a+13=0\end{array}$$ On solving equation (1) and (2) we get $a=-4,b=-7$ Hence, third vertex is $(-4,-7)$.