$A$ ray of light coming from the point $P(1, 2)$ gets reflected from the point $Q$ on the $x$-axis and then passes through the point $R(4, 3)$. If the point $S(h, k)$ is such that $PQRS$ is a parallelogram,then $hk^2$ is equal to :

$A$ rectangle $ABCD$ has its side parallel to the line $y=2x$ and vertices $A, B, D$ are on lines $y=1, x=1$ and $x=-1$ respectively. The coordinate of $C$ can be

The quadrilateral formed by the vertices $A(-1, 1)$,$B(0, -3)$,$C(5, 2)$,and $D(4, 6)$ is:

The points $A(2a, 4a)$,$B(2a, 6a)$,and $C(2a + \sqrt{3}a, 5a)$,where $a > 0$,are the vertices of:

$L \equiv 7x - y + 8 = 0$ is one of the diagonals of a square for which $(-4, 5)$ and $(3, 4)$ are two vertices. Find the coordinates of the two vertices lying on the diagonal $L = 0$.

Let $\left(5, \frac{a}{4}\right)$ be the circumcenter of a triangle with vertices $A(a, -2)$,$B(a, 6)$,and $C\left(\frac{a}{4}, -2\right)$. Let $\alpha$ denote the circumradius,$\beta$ denote the area,and $\gamma$ denote the perimeter of the triangle. Then $\alpha + \beta + \gamma$ is