The four points whose co-ordinates are $(2, 1), (1, 4), (4, 5), (5, 2)$ form :

Two vertices of a triangle are $(5, - 1)$ and $( - 2,3)$. If orthocentre is the origin then coordinates of the third vertex are

Let $PS$ be the median of the triangle with vertices $P(2,2) , Q(6,-1) $ and $R(7,3) $. The equation of the line passing through $(1,-1) $ and parallel to $PS $ is :

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

Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.

The locus of the mid-points of the perpendiculars drawn from points on the line, $\mathrm{x}=2 \mathrm{y}$ to the line $\mathrm{x}=\mathrm{y}$ is