The number of straight lines that can be drawn through the point $(2, 5)$ which form a triangle of area $24 \text{ sq. units}$ with the coordinate axes is:

$A$ straight line passing through the origin $O$ intersects the lines $10x - 8y - 10 = 0$ and $\frac{x}{4} - \frac{y}{5} + 1 = 0$ at right angles at points $P$ and $Q$ respectively. Then the ratio in which $O$ divides the line segment $PQ$ is

Two non-parallel sides of a rhombus are parallel to the lines $x+y-1=0$ and $7x-y-5=0$. If $(1,3)$ is the centre of the rhombus and one of its vertices $A(\alpha, \beta)$ lies on $15x-5y=6$,then one of the possible values of $(\alpha+\beta)$ is

If the two lines $x + (a - 1)y = 1$ and $2x + a^2y = 1$ $(a \in R - \{0, 1\})$ are perpendicular,then the distance of their point of intersection from the origin is

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 ............... $sq. \, units$.

Let $P$ be the point of intersection of the lines $L_1 \equiv x-y-7=0$ and $L_2 \equiv x+y-5=0$. $A(x_1, y_1)$ and $B(x_2, y_2)$ are points on the lines $L_1=0$ and $L_2=0$ respectively such that $PA=3\sqrt{2}$,$PB=\sqrt{2}$,$x_1, y_1 \geq 0$,$x_2, y_2 \geq 0$. Then the angle made by the line segment $AB$ at the origin is