The four points whose coordinates are $(2, 1), (1, 4), (4, 5), (5, 2)$ form:

If the line $3x + 4y - 24 = 0$ intersects $X$ and $Y$ axes at points $A$ and $B$ respectively,then the incenter of the triangle $OAB$,where $O$ is the origin,is:

Let the mirror image of point $A(\alpha, \beta)$ in the line mirror $x + 2y = 3$ be point $B$,and the image of $B$ in the line $3x - 2y = 5$ be $C$. If the origin is the orthocentre of triangle $ABC$ and $P(a, b)$ is a point inside the triangle such that triangles $PAB$,$PBC$,and $PCA$ have the same area,then $3(a + b)$ is:

If a straight line perpendicular to $2x - 3y + 7 = 0$ forms a triangle with the coordinate axes whose area is $3 \text{ sq. units}$,then the equation of the straight line is:

The points $(-a,-b), (a, b), (0,0)$ and $(a^{2}, ab)$ where $a \neq 0, b \neq 0$ are always

The points $A(2a, 4a)$,$B(2a, 6a)$,and $C(2a + \sqrt{3}a, 5a)$,where $a > 0$,are the vertices of: