Mathematics - Straight Lines Question with Solution | TestHub

Coordinates of A and B are $(-3,4)$ and $\left(-\frac{3}{5},\frac{8}{5}\right)$; let orthocentre be $\mathrm{P}(h,k)$. Then, (slope of PA) × (slope of BC) $=-1:\frac{k-4}{h+3}\times 4=-1\Rightarrow 4k-16=-h-3\Rightarrow h+4k=13$ and (slope of PB) × (slope of AC) $=-1:\frac{k-\frac{8}{5}}{h+\frac{3}{5}}\times \frac{-2}{3}=-1\Rightarrow \frac{5k-8}{5h+3}\times \frac{2}{3}=1\Rightarrow 10k-16=15h+9\Rightarrow$ $3h-2k+5=0\dots$ (ii). Solving Eqs. (i) and (ii), we get $h=\frac{3}{7},k=\frac{22}{7}$. Hence, orthocentre lies in I quadrant.