Mathematics - Point Question with Solution | TestHub

${\mathrm{M}}_{\mathrm{AB}}=\frac{4}{-5}\Rightarrow {\mathrm{M}}_{\mathrm{DP}}=\frac{5}{4}$ Equation of $\mathrm{PC}$ is $\mathrm{y}-1=\frac{5}{4}(\mathrm{x}-1)\dots (1)$ ${\mathrm{M}}_{\mathrm{AP}}=\frac{2}{-2}=-1\Rightarrow {\mathrm{M}}_{\mathrm{BC}}=+1$ Equation of $\mathrm{BC}$ is $\mathrm{y}-3=(\mathrm{x}+2)\dots (2)$ On solving (1) and (2) $\begin{array}{rl}& x+4=\frac{5}{4}(x-1)\Rightarrow 4x+16=5x-5\Rightarrow \alpha =21\\ & \Rightarrow \beta =y=x+5=26\\ & \alpha +\beta =47\end{array}$ Equation of $\perp$ bisector of AP $y-0=(x-2)\dots (3)$ Equation of $\perp$ bisector of $\mathrm{AB}$ $y-1=\frac{5}{4}\left(x-\frac{1}{2}\right)\dots (4)$ $\begin{array}{rl}& \text{ On solving (3) \& (4) }\\ & (\mathrm{x}-3)4=5\mathrm{x}-\frac{5}{2}\\ & \mathrm{x}=\frac{-19}{2}=\mathrm{h}\\ & \mathrm{y}=\frac{-23}{2}=\mathrm{k}\\ & \Rightarrow 2(\mathrm{h}+\mathrm{k})=-42\end{array}$