The line $3x + 2y = 24$ meets the $y$-axis at $A$ and the $x$-axis at $B$. The perpendicular bisector of $AB$ meets the line through $(0, -1)$ parallel to the $x$-axis at $C$. The area of the triangle $ABC$ is ............... $sq. \, units$.

If $P(x_1, y_1)$ and $Q(x_2, y_2)$ are points on $2x + 3y + 1 = 0$ such that $|PA - PB|$ is maximum and $|QA - QB|$ is minimum,where $A(2, 0)$ and $B(0, 2)$,then the value of $x_1 - y_1 + x_2 - y_2$ is -

If $p_1$ and $p_2$ are the lengths of the perpendiculars from the origin to the lines $x \sec \alpha + y \csc \alpha = 2a$ and $x \cos \alpha - y \sin \alpha = a \cos 2\alpha$ respectively,then what is the value of $\left( \frac{p_1}{p_2} + \frac{p_2}{p_1} \right)^2$?

The line $x + y = p$ meets the $x$ and $y$ axes at $A$ and $B$ respectively. $A$ triangle $APQ$ is inscribed in the triangle $OAB$,where $O$ is the origin,with a right angle at $Q$. $P$ and $Q$ lie on $OB$ and $AB$ respectively. If the area of triangle $APQ$ is $3/8$ of the area of triangle $OAB$,then $\frac{AQ}{BQ}$ is equal to:

Suppose $P$ and $Q$ lie on $3x + 4y - 4 = 0$ and $5x - y - 4 = 0$ respectively. If the mid-point of $PQ$ is $(1, 5)$,then the slope of the line passing through $P$ and $Q$ is

The point on the line $3x + 4y = 5$ which is equidistant from $(1, 2)$ and $(3, 4)$ is