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

If the lines drawn along the diagonals of the two squares formed by two pairs of lines $x^2-3|x|+2=0$ and $y^2-3y+2=0$ form a square $ABCD$,then the equations of two adjacent sides of the square $ABCD$ are

One side of a rectangle lies along the line $4x + 7y + 5 = 0.$ Two of its vertices are $(-3, 1)$ and $(1, 1).$ Then the equations of the other three sides are

$A$ straight line passing through $P(3, 1)$ meets the coordinate axes at $A$ and $B$. It is given that the distance of this straight line from the origin $O$ is maximum. The area of triangle $OAB$ is equal to

The lines $x + y = |a|$ and $ax - y = 1$ intersect each other in the first quadrant. Then,in which interval do all possible values of $a$ lie?

Suppose that the points $(h, k)$,$(1, 2)$,and $(-3, 4)$ lie on the line $l_1$. If a line $l_2$ passing through the points $(h, k)$ and $(4, 3)$ is perpendicular to $l_1$,then $\left(\frac{k}{h}\right)$ equals