$L_1 \equiv 2x+y-3=0$ and $L_2 \equiv ax+by+c=0$ are two equal sides of an isosceles triangle. If $L_3 \equiv x+2y+1=0$ is the third side of this triangle and $(5,1)$ is a point on $L_2=0$,then $\frac{b^2}{|ac|}=$

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

If $(1, 5)$ is the midpoint of the segment of a line between the lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$, then the equation of the line is:

For $\lambda, \mu \in R$,$(x-2y-1)+\lambda(3x+2y-11)=0$ and $(3x+4y-11)+\mu(-x+2y-3)=0$ represent two families of lines. If the equation of the line common to both the families is $ax+by-5=0$,then $2a+b=$

If $l, m, n$ are in arithmetic progression,then the straight line $lx + my + n = 0$ will always pass through the point:

$A$ straight line $L$ is perpendicular to the line $5x - y = 1$ and the area of the triangle formed by the line $L$ and the coordinate axes is $5$ square units. The equation of the line $L$ can be