If the lines $x + 3y = 4$ and $6x - 2y = 7$ are the diagonals of a parallelogram $PQRS$,then $PQRS$ is a:

The equations of two sides $AB$ and $AC$ of a triangle $ABC$ are $4x + y = 14$ and $3x - 2y = 5$,respectively. The point $\left(2, -\frac{4}{3}\right)$ divides the third side $BC$ internally in the ratio $2:1$. The equation of the side $BC$ is:

$A$ pair of straight lines $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$ form a square. The coordinates of the centre of the circle inscribed in the square are

If the points $(-5, 1), (p, 5)$ and $(10, 7)$ are collinear,then the value of $p$ will be

The area of a parallelogram formed by the lines $ax \pm by \pm c = 0$ is

The triangle with vertices $A(2, 4)$,$B(2, 6)$,and $C(2 + \sqrt{3}, 5)$ is a . . . .