Find the equation of the line passing through the intersection of the lines $x - 3y + 1 = 0$ and $2x + 5y - 9 = 0$ and whose distance from the origin is $\sqrt{5}$.

$A$ straight line $L \equiv 0$ passing through the point $A=(-5,-4)$ and having slope $\tan \theta$ meets the lines $x+3y+2=0$ and $2x+y+4=0$ respectively at the points $B$ and $C$. If $\frac{100}{AC^2}-\frac{225}{AB^2}=4 \cos 2\theta+\sin 2\theta$,then the slope of the line $L \equiv 0$ is

If $\alpha$ is the angle made by the perpendicular drawn from the origin to the line $3x - 4y + 5 = 0$ with the positive $X$-axis in the positive direction,and $ax + by = 1$ is the equation of a line passing through the point $(1, -1)$ with $\tan \alpha$ as its slope,then $a + ab + b =$

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 straight lines $l_1$ and $l_2$ pass through the origin and trisect the line segment of the line $L: 9x + 5y = 45$ between the axes. If $m_1$ and $m_2$ are the slopes of the lines $l_1$ and $l_2$,then the point of intersection of the line $y = (m_1 + m_2)x$ with $L$ lies on

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: