The line segment joining $A(2, -7)$ and $B(6, 5)$ is divided into $4$ equal parts by the points $P, Q$ and $R$ such that $AP = PQ = QR = RB$. The midpoint of $PR$ is

Three vertices of a parallelogram taken in order are $(-1, -6)$,$(2, -5)$ and $(7, 2)$. The fourth vertex is

If $P, Q, R$ are collinear points such that $P(7, 7)$,$Q(3, 4)$ and $PR = 10$,then what are the coordinates of $R$?

The point $A$ divides the join of the points $(-5, 1)$ and $(3, 5)$ in the ratio $k : 1$. The coordinates of the points $B$ and $C$ are $(1, 5)$ and $(7, -2)$ respectively. If the area of the triangle $ABC$ is $2$ square units,then $k =$

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 line joining $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta),$ where $a \neq b,$ is produced to the point $M(x, y)$ such that $AM : MB = b : a$. Then,$x \cos \frac{\alpha+\beta}{2} + y \sin \frac{\alpha+\beta}{2}$ is equal to