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 $x^2-y^2+2 h x y+2 g x+2 f y+c=0$ is the locus of a point, which moves such that it is always equidistant from the lines $x+2 y+7=0$ and $2 x-y$ $+8=0$, then the value of $\mathrm{g}+\mathrm{c}+\mathrm{h}-\mathrm{f}$ equals

The area of triangle formed by the lines $x + y - 3 = 0 , x - 3y + 9 = 0$ and $3x - 2y + 1= 0$

Two consecutive sides of a parallelogram are $4x + 5y = 0$ and $7x + 2y = 0$. If the  equation to one diagonal is $11x + 7y = 9$, then the equation to the other diagonal is :-

A straight line cuts off the intercepts $OA = a$ and $OB = b$ on the positive directions of $x$-axis and $y -$ axis respectively. If the perpendicular from origin $O$ to this line makes an angle of $\frac{\pi}{6}$ with positive direction of $y$-axis and the area of $\triangle OAB$ is $\frac{98}{3} \sqrt{3}$, then $a ^2- b ^2$ is equal to:

The area of a parallelogram formed by the lines $ax \pm by \pm c = 0$, is