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

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

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:

Consider the fourteen lines in the plane given by $y=x+r$ and $y=-x+r$,where $r \in \{0, 1, 2, 3, 4, 5, 6\}$. The number of squares formed by these lines,whose sides are of length $\sqrt{2}$,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

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