Mathematics - Circle Question with Solution | TestHub

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

Then, (slope of PA) × (slope of BC) = -1 $\frac{\mathrm{k}-4}{\mathrm{h}+3}\times 4=-1$

$$\begin{array}{llc}\Rightarrow & 4k-16=-h-3\Rightarrow & h+4k=13\end{array}$$

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

$$\begin{array}{llcc}\Rightarrow & \frac{\mathrm{k}-\frac{8}{5}}{\mathrm{h}+\frac{3}{5}}\times -\frac{2}{3}=-1\Rightarrow & \frac{5\mathrm{k}-8}{5\mathrm{h}+3}\times \frac{2}{3}=1\Rightarrow & \begin{array}{l}10\mathrm{k}-16=15\mathrm{th}+915\mathrm{th}-10\mathrm{k}+25=03\mathrm{h}-2\mathrm{k}+5=0\dots \text{ (ii) }\end{array}\end{array}$$

Solivng Eqs. (i) and (ii), we get $\mathrm{h}=\frac{3}{7},\mathrm{k}=\frac{22}{7}$ Hence, orthocentre lies in I quadrant.