Show that the area of the triangle formed by the lines

$y=m_{1} x+c_{1}, y=m_{2} x+c_{2}$ and $x=0$ is $\frac{\left(c_{1}-c_{2}\right)^{2}}{2\left|m_{1}-m_{2}\right|}$.

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

Consider the lines $L_1$ and $L_2$ defined by

$L _1: x \sqrt{2}+ y -1=0$ and $L _2: x \sqrt{2}- y +1=0$

For a fixed constant $\lambda$, let $C$ be the locus of a point $P$ such that the product of the distance of $P$ from $L_1$ and the distance of $P$ from $L_2$ is $\lambda^2$. The line $y=2 x+1$ meets $C$ at two points $R$ and $S$, where the distance between $R$ and $S$ is $\sqrt{270}$.

Let the perpendicular bisector of $RS$ meet $C$ at two distinct points $R ^{\prime}$ and $S ^{\prime}$. Let $D$ be the square of the distance between $R ^{\prime}$ and $S ^{\prime}$.

($1$) The value of $\lambda^2$ is

($2$) The value of $D$ is

If in triangle $ABC$ ,$ A \equiv (1, 10) $, circumcentre $\equiv$ $\left( { – \,\,{\textstyle{1 \over 3}}\,\,,\,\,{\textstyle{2 \over 3}}} \right)$ and orthocentre $\equiv$ $\left( {{\textstyle{{11} \over 3}}\,\,,\,\,{\textstyle{4 \over 3}}} \right)$ then the co-ordinates of mid-point of side opposite to $A$ is :

If two vertices of a triangle are $(5, -1)$ and $( – 2, 3)$ and its orthocentre is at $(0, 0)$, then the third vertex