Mathematics - Straight Lines Question with Solution | TestHub

Coordinates of A and B are $(-3,4)$ and $\left(-\frac{3}{5},\frac{8}{5}\right)$ if orthocentre $\mathrm{p}(\mathrm{h},\mathrm{k})$

Coordinates of A and B are $(-3,4)$ and $\left(-\frac{3}{5},\frac{8}{5}\right)$ if orthocentre $\mathrm{p}(\mathrm{h},\mathrm{k})$

Then, (slope of PA) × (slope of BC) $=-1$

$\frac{\mathrm{k}-4}{\mathrm{h}+3}\times 4=-1$

$\Rightarrow 4k-16=-h-3$

$\Rightarrow h+4k=13$ ....(i)

and slope of $\mathrm{PB}\times$ slope of $\mathrm{AC}=-1$

$\begin{array}{ll}\Rightarrow & \frac{\mathrm{k}-\frac{8}{5}}{\mathrm{h}+\frac{3}{5}}\times -\frac{2}{3}=-1\\ \Rightarrow & \frac{5\mathrm{k}-8}{5\mathrm{h}+3}\times \frac{2}{3}=1\\ \Rightarrow & \begin{array}{l}10k-16=15h+9\\ 15h-10k+25=0\\ 3h-2k+5=0\dots \text{ (ii) }\end{array}\end{array}$

Solivng Eqs. (i) and (ii), we get $\mathrm{h}=\frac{3}{7},\mathrm{k}=\frac{22}{7}$

Hence, orthocentre lies in I quadrant.