The point on the line $3x + y + 4 = 0$ which is equidistant from $(-5, 6)$ and $(3, 2)$ is

Let the distance between two parallel lines be $5$ units and a point $P$ lie between the lines at a unit distance from one of them. An equilateral triangle $PQR$ is formed such that $Q$ lies on one of the parallel lines,while $R$ lies on the other. Then $(QR)^2$ is equal to . . . . . . .

If the intercept of a straight line $L$ made between the straight lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1, 5)$,then the equation of $L$ is

The line joining two points $A(2,0)$ and $B(3,1)$ is rotated about $A$ in an anti-clockwise direction through an angle of $15^\circ$. The equation of the line in the new position is:

Let two straight lines drawn from the origin $O$ intersect the line $3x + 4y = 12$ at the points $P$ and $Q$ such that $\triangle OPQ$ is an isosceles triangle and $\angle POQ = 90^{\circ}$. If $l = OP^2 + PQ^2 + QO^2$,then the greatest integer less than or equal to $l$ is:

$A$ straight line through the origin $O(0, 0)$ meets the lines $4x + 3y - 10 = 0$ and $8x + 6y + 5 = 0$ at points $A$ and $B$ respectively. Then $O$ divides the segment $AB$ in the ratio: