If the line $3x + 3y -24 = 0$ intersects the $x-$ axis at the point $A$ and the $y-$ axis at the point $B$, then the incentre of the triangle $OAB$, where $O$ is the origin, is

A point moves so that square of its distance from the point $(3, -2)$ is numerically equal to its distance from the line $5x - 12y = 13$. The equation of the locus of the point is

Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on $y -2 x =2$ such that $\triangle ABC$ is an equilateral triangle. Then, the area of the $\triangle ABC$ is

Let $A (-3, 2)$ and $B (-2, 1)$ be the vertices of a triangle $ABC$. If the centroid of this triangle lies on the line $3x + 4y + 2 = 0$, then the vertex $C$ lies on the line

The equation of the lines on which the perpendiculars from the origin make ${30^o}$ angle with $x$-axis and which form a triangle of area $\frac{{50}}{{\sqrt 3 }}$ with axes, are

The medians $AD$ and $BE$ of a triangle with vertices $A\;(0,\;b),\;B\;(0,\;0)$ and $C\;(a,\;0)$ are perpendicular to each other, if