The equation of a line passing through the point of intersection of the lines $x - 2y + 8 = 0$ and $3x - y + 4 = 0$ and passing through the origin is

Find the equation of the line passing through the intersection of the lines $x + y - 3 = 0$ and $2x - y + 1 = 0$ and the point $(2, -3)$.

Find the value of $p$ so that the three lines $3x + y - 2 = 0$,$px + 2y - 3 = 0$,and $2x - y - 3 = 0$ intersect at a single point.

Let the line $L_1$ passing through the point of intersection of the lines $2x + 3y - 5 = 0$ and $4x - 5y + 7 = 0$ divide the line segment joining the points $(2, 3)$ and $(1, -1)$ in the ratio $2:1$. If the equation of $L_1$ is $ax + by = 1$,then $33(a - b) =$

If $3a + 5b + 6c = 0$,then the family of lines $ax + by + c = 0$ passes through the fixed point:

If $\pi / 3$ is the angle between the straight lines $px + qy + r = 0$ and $x \sin \alpha + y \cos \alpha = r$ $(r \neq 0)$ which meet at a point $A$,and the straight line $x \cos \alpha - y \sin \alpha = 0$ also passes through the point $A$,then: