Two sides of a parallelogram are along the lines, $x + y = 3$ and $x -y + 3 = 0$. If its diagonals intersect at $(2, 4)$, then one of its vertex is

Let $P$ be a moving point such that sum of its perpendicular distances from $2x + y = 3$ and $x - 2y + 1 = 0$ is always $2\, units$ then area bounded by locus of point $P$ is

Without using distance formula, show that points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are vertices of a parallelogram.

If the three lines $x - 3y = p, ax + 2y = q$ and $ax + y = r$ form a right-angled triangle then

The line $2x + 3y = 12$ meets the $x$-axis at $A$ and $y$-axis at $B$. The line through $(5, 5)$ perpendicular to $AB$ meets the $x$- axis , $y$ axis and the $AB$ at $C,\,D$ and $E$ respectively. If $O$ is the origin of coordinates, then the area of $OCEB$ is

The vertex of an equilateral triangle is $(2,-1)$ and the equation of its base in $x + 2y = 1$. The length of its sides is