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

Let $b, d>0$. The locus of all points $P(r, \theta)$ for which the line $P$ (where, $O$ is the origin) cuts the line $r \sin \theta=b$ in $Q$ such that $P Q=d$ 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 line $2x + 3y = 12$ meets the $x -$ axis at $A$ and the $y -$ axis at $B$ . The line through $(5, 5)$ perpendicular to $AB$ meets the $x -$ axis, $y -$ axis $\&$ the line $AB$ at $C, D, E$ respectively. If $O$ is the origin, then the area of the $OCEB$ is :

For a point $P$ in the plane, let $d_1(P)$ and $d_2(P)$ be the distance of the point $P$ from the lines $x-y=0$ and $x+y=0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2 \leq d_1(P)+d_2(P) \leq 4$, is

A straight line passing through $P(3, 1)$ meet the coordinates axes at $A$ and $B$. It is given that distance of this straight line from the origin $'O'$ is maximum. Area of triangle $OAB$ is equal to