A line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( {\frac{1}{2},0} \right)$ and perpendicular to $L$. Then the area of the triangle formed by the lines $L,L'$ and $y$- axis, is

A straight line through the point $(1, 1)$ meets the $x$-axis at ‘$A$’ and the $y$-axis at ‘$B$’. The locus of the mid-point of $AB$ is

Two lines are drawn through $(3, 4)$, each of which makes angle of $45^\circ$ with the line $x - y = 2$, then area of the triangle formed by these lines is

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

A square of side a lies above the $x$ -axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha ,(0 < \alpha < \frac{\pi }{4})$ with the positive direction of $x$-axis. The equation of its diagonal not passing through the origin is

Let $P$ be a moving point such that sum of its perpendicular distances from $2x + y = 3$ and $x - 2y + 1 = 0$ is always $2\, units$ then area bounded by locus of point $P$ is