The base $BC$ of a triangle $ABC$ is bisected at the point $(p, q)$ and the equations to the sides $AB$ and $AC$ are respectively $px+qy= 1$ and $qx + py = 1.$ Then the equation to the median through $A$ is

The triangle $PQR$ is inscribed in the circle ${x^2} + {y^2} = 25$. If $Q$ and $R$ have co-ordinates $(3,4)$ and $(-4, 3)$ respectively, then $\angle QPR$ is equal to

Let $PS$ be the median of the triangle with vertices $P(2,2) , Q(6,-1) $ and $R(7,3) $. The equation of the line passing through $(1,-1) $ and parallel to $PS $ is :

A square of side a lies above the $x$ -axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha ,(0 < \alpha < \frac{\pi }{4})$ with the positive direction of $x$-axis. The equation of its diagonal not passing through the origin is

Consider a triangle $\mathrm{ABC}$ having the vertices $\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)$ and $\mathrm{C}(\gamma, \delta)$ and angles $\angle \mathrm{ABC}=\frac{\pi}{6}$ and $\angle \mathrm{BAC}=\frac{2 \pi}{3}$. If the points $\mathrm{B}$ and $\mathrm{C}$ lie on the line $\mathrm{y}=\mathrm{x}+4$, then $\alpha^2+\gamma^2$ is equal to………………..