Given three points $P, Q, R$ with $P(5, 3)$ and $R$ lies on the $x-$ axis. If equation of $RQ$ is $x - 2y = 2$ and $PQ$ is parallel to the $x-$ axis, then the centroid of $\Delta PQR$ lies on the line

Let $m, n$ be real numbers such that $0 \leq m \leq \sqrt{3}$ and $-\sqrt{3} \leq n \leq 0$. The minimum possible area of the region of the plane consisting of points $(x, y)$ satisfying in inequalities $y \geq 0, y-3 \leq m x$, $y -3 \leq nx$, is

The vertices of a triangle are $\mathrm{A}(-1,3), \mathrm{B}(-2,2)$ and $\mathrm{C}(3,-1)$. $A$ new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

Area of the parallelogram formed by the lines ${a_1}x + {b_1}y + {c_1} = 0$,${a_1}x + {b_1}y + {d_1} = 0$and ${a_2}x + {b_2}y + {c_2} = 0$, ${a_2}x + {b_2}y + {d_2} = 0$is

The opposite vertices of a square are $(1, 2)$ and $(3, 8)$, then the equation of a diagonal of the square passing through the point $(1, 2)$, is

Area of the rhombus bounded by the four lines, $ax \pm by \pm c = 0$ is :