The diagonals of a parallelogram $PQRS$ are along the lines $x + 3y = 4$ and $6x - 2y = 7$. Then $PQRS$ must be a

Three lines $x + 2y + 3 = 0 ; x + 2y - 7 = 0$ and $2x - y - 4 = 0$ form the three sides of two squares. The equation to the fourth side of each square 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 the middle points of the sides $BC,\, CA$ and $AB$ of the triangle $ABC$ be $(1, 3), \,(5, 7)$ and $(-5, 7)$, then the equation of the side $AB$ 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

The co-ordinates of the vertices $A$ and $B$ of an isosceles triangle $ABC (AC = BC)$ are $(-2,3)$ and $(2,0)$ respectively. $A$ line parallel to $AB$ and having a $y$ -intercept equal  to $\frac{43}{12}$ passes through $C$, then the co-ordinates of $C$ are :-