The diagonals of a parallelogram $PQRS$ are along the lines $x + 3y = 4$ and $6x - 2y = 7$. Then $PQRS$ must be a

$A$ line $L_1$ passing through $A(3,4)$ and having slope $1$ cuts another line $L_2$ passing through $C$ at $B$,such that $AB = AC$. If the equation of line $BC$ is $2x - y + 4 = 0$,then the equation of $AC$ is

If $\Delta_1$ is the area of the triangle formed by the centroid and two vertices of a triangle,and $\Delta_2$ is the area of the triangle formed by the mid-points of the sides of the same triangle,then $\Delta_1 : \Delta_2 =$

The triangle formed by joining the points $P(2, 7)$,$Q(4, -1)$,and $R(-2, 6)$ is:

Let the midpoints of the sides of a triangle $ABC$ be $(\frac{5}{2}, 7)$, $(\frac{5}{2}, 3)$, and $(4, 5)$. If its incentre is $(h, k)$, then $3h + k$ is equal to:

The diagonals of the parallelogram whose sides are $lx + my + n = 0$,$lx + my + n' = 0$,$mx + ly + n = 0$,and $mx + ly + n' = 0$ include an angle of: