If $P(1, 2), Q(4, 6), R(5, 7)$ and $S(a, b)$ are the vertices of a parallelogram $PQRS$,then:

The points $A(2, 1)$,$B(3, -2)$ and $C(a, b)$ are vertices of the rectangle $ABCD$. If the point $P(3, 4)$ lies on the line $CD$,then $5a + 10b = $

Let a point $A$ lie between the parallel lines $L_1$ and $L_2$ such that its distances from $L_1$ and $L_2$ are $6$ and $3$ units,respectively. Then the area (in sq. units) of the equilateral triangle $ABC$,where the points $B$ and $C$ lie on the lines $L_1$ and $L_2$ respectively,is:

Let $ABCD$ be a square of side length $1$. Let $P, Q, R, S$ be points in the interiors of the sides $AD, BC, AB, CD$ respectively,such that $PQ$ and $RS$ intersect at right angles. If $PQ = \frac{3\sqrt{3}}{4}$,then $RS$ equals

The area (in sq units) of the quadrilateral formed by the lines $2x + 3y + 6 = 0$,$2x - 3y + 6 = 0$,$2x + 3y - 6 = 0$,and $2x - 3y - 6 = 0$ is

If the extremities of the base of an isosceles triangle are the points $(2a, 0)$ and $(0, a)$ and the equation of one of the sides is $x = 2a$,then the area of the triangle is