The perpendicular bisector of x+y+2=0 and x-y-1=0 of sides AB and AC of a $\Delta ABC$intersects them at $(-1,-1)and(2,1)$respectively. If the mid point of side BC is P, then distance of this point from the orthocenter of triangle is

$\begin{array}{l}GivenEquationx+y+2=0---\left(1\right)\\ x-y-1=0---\left(2\right)\\ andQ(-1,-1)R(2,1)\\ solving\left(1\right)\&\left(2\right)\\ xy1\\ \begin{array}{cccc}1& 2& 1& 1\\ -1& -1& 1& -1\end{array}\\ \Rightarrow \frac{x}{-1+2}=\frac{y}{2+1}=\frac{1}{-1-1}\\ \Rightarrow \frac{x}{1}=\frac{y}{3}=\frac{1}{-2}\\ A(x,y)=(\frac{-1}{2},\frac{-3}{2})\\ Sismidpo\mathrm{int}ofAB\\ B=2QA=(-2+\frac{1}{2},-2+\frac{3}{2})=(\frac{-3}{2},\frac{-1}{2})\end{array}$

$\begin{array}{l}Rismidpo\mathrm{int}of\overline{AC}\\ C=2R-QA\\ =(4+\frac{1}{2},2+\frac{3}{2})=(\frac{9}{2},\frac{7}{2})\\ P=midpo\mathrm{int}ofBC\\ =(\frac{\frac{-3}{2}+\frac{9}{2}}{2},\frac{-\frac{1}{2}+\frac{7}{2}}{2})=(\frac{3}{2},\frac{3}{2})\\ \left(1\right)\&\left(2\right)are\perp r\Rightarrow \angle A={90}^0\\ \therefore orthocentre=A(\frac{-1}{2},\frac{-3}{2})\\ AP=\sqrt{{(\frac{3}{2}+\frac{1}{2})}^2+{(\frac{3}{2}+\frac{3}{2})}^2}\\ =\sqrt{4+9}=\sqrt{13}\end{array}$