Find the slope of a line,which passes through the origin and the mid-point of the line segment joining the points $P(0, -4)$ and $B(8, 0)$.

If the coordinates of the vertices of $\Delta OAB$ are $(0, 0)$,$(\cos \alpha, \sin \alpha)$,and $(-\sin \alpha, \cos \alpha)$ respectively,then $OA^2 + OB^2 = $

The points which trisect the line segment joining the points $(0, 0)$ and $(9, 12)$ are

Consider three points $P = (-\sin(\beta - \alpha), -\cos \beta)$,$Q = (\cos(\beta - \alpha), \sin \beta)$,and $R = (\cos(\beta - \alpha + \theta), \sin(\beta - \theta))$,where $0 < \alpha, \beta, \theta < \frac{\pi}{4}$. Then:

The inclination of the straight line passing through the point $(-3, 6)$ and the midpoint of the line joining the points $(4, -5)$ and $(-2, 9)$ is

The line $2x + y - 3 = 0$ divides the line segment joining the points $A(1, 2)$ and $B(-2, 1)$ in the ratio $a : b$ at the point $C$. If the point $C$ divides the line segment joining the points $P\left(\frac{b}{3a}, -3\right)$ and $Q\left(-3, -\frac{b}{3a}\right)$ in the ratio $p : q$,then $\frac{p}{q} + \frac{q}{p} =$