Coordinates of the vertices of a quadrilateral are $(2, -1), (0, 2), (2, 3)$ and $(4, 0)$. The angle between its diagonals will be

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

The sides $AB,BC,CD$ and $DA$ of a quadrilateral are $x + 2y = 3,\,x = 1,$ $x - 3y = 4,\,$ $\,5x + y + 12 = 0$ respectively. The angle between diagonals $AC$ and $BD$ is ......$^o$

Let two points be $\mathrm{A}(1,-1)$ and $\mathrm{B}(0,2) .$ If a point $\mathrm{P}\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)$ be such that the area of $\Delta \mathrm{PAB}=5\; \mathrm{sq}$ units and it lies on the line, $3 x+y-4 \lambda=0$ then a value of $\lambda$ is

The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points $(a^2 + 1 , a^2 + 1 )$ and $(2a, - 2a)$, $a \ne 0$. Then for any $a$ , the orthocentre of this triangle lies on the line

$A(-1, 1)$, $B(5, 3)$ are opposite vertices of a square in $xy$-plane. The equation of the other diagonal (not passing through $(A, B)$ of the square is given by