$A$ 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:

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

The area of the parallelogram formed by the lines $y = mx$,$y = mx + 1$,$y = nx$,and $y = nx + 1$ is equal to

$A$ line passing through $P(2,3)$ and making an angle of $30^{\circ}$ with the positive direction of the $x$-axis meets the curve $x^2-2xy-y^2=0$ at $A$ and $B$. Then the value of $PA \cdot PB$ is

The number of possible distinct straight lines passing through $(2,3)$ and forming a triangle with the coordinate axes whose area is $12$ sq. units is:

$A$ pair of straight lines drawn through the origin form an isosceles right-angled triangle with the line $2x + 3y = 6$. Find the equations of the lines and the area of the triangle thus formed.