• # question_answer Two vertices of a triangle are $(5,-1)$ and $(-\,2,3)$ if orthocenter of the triangle is origin, then the coordinate of third vertex is A) $(4,7)$              B) $(3,7)$ C) $(-4,-7)$            D) $(4,-\,7)$

[c]  Slope of $AH=\frac{k}{h}$ Slope of$BC=\frac{4}{-\,7}$ Slope of $AH\times$slope of $BC=-1$ $\therefore$      $\frac{k}{h}\times \frac{4}{-\,7}=-1$ $\frac{k}{h}=\frac{7}{4}$                                                        ? (i)
 Slope of $AC=\frac{k+1}{h-5}$ Slope of $BH=\frac{-\,3}{2}$ Slope of $AC\times$Slope of $BH=-\,1$ $\therefore$      $\left( \frac{k+1}{h-5} \right)\left( \frac{-\,3}{2} \right)=-1$ $\Rightarrow$   $\frac{k+1}{h-5}=\frac{2}{3}$                         ? (ii) On solving Eqs. (i) and (ii), we get $h=-\,4,k=-\,7$ $\therefore$Third vertex of triangle is $(-\,4,-7).$