Let $\mathrm{C}$ be the centroid of the triangle with vertices $(3,-1),(1,3)$ and $(2,4) .$ Let $P$ be the point of intersection of the lines $x+3 y-1=0$ and $3 \mathrm{x}-\mathrm{y}+1=0 .$ Then the line passing through the points $\mathrm{C}$ and $\mathrm{P}$ also passes through the point

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 the extremities of the base of an isosceles triangle are the points $(2a,0)$ and $(0,a)$ and the equation of one of the sides is $x = 2a$, then the area of the triangle is

A straight line passing through $P(3, 1)$ meet the coordinates axes at $A$ and $B$. It is given that distance of this straight line from the origin $'O'$ is maximum. Area of triangle $OAB$ is equal to

The locus of the orthocentre of the triangle formed by the lines

$ (1+p) x-p y+p(1+p)=0, $

$ (1+q) x-q y+q(1+q)=0,$

and $y=0$, where $p \neq q$, is

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 of the other diagonal is