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

A straight the through a fixed point $(2, 3)$ intersects the coordinate axes at distinct points $P$ and $Q.$ If $O$ is the origin and the rectangle $OPRQ$ is completed, then the locus of $R$ is:

The $x -$ co-ordinates of the vertices of a square of unit area are the roots of the equation $x^2 - 3 |x| + 2 = 0$ and the $y -$ co-ordinates of the vertices are the roots of the equation $y^2 - 3y + 2 = 0$ then the possible vertices of the square is/are :

The equation of perpendicular bisectors of the sides $AB$ and $AC$ of a triangle $ABC$ are $x - y + 5 = 0$ and $x + 2y = 0$ respectively. If the point $A$ is $(1,\; - \;2)$, then the equation of line $BC$ is

The sides of a rhombus $ABCD$ are parallel to the lines, $x - y + 2\, = 0$ and $7x - y + 3\, = 0$. If the diagonals of the rhombus intersect at $P( 1, 2)$ and the vertex $A$ ( different from the origin) is on the $y$ axis, then the ordinate of $A$ is

A point moves so that square of its distance from the point $(3, -2)$ is numerically equal to its distance from the line $5x - 12y = 13$. The equation of the locus of the point is