For all values of $a$ and $b$,the line $(a+2b)x + (a-b)y + (a+5b) = 0$ passes through a fixed point. Find that point.

For what values of $\theta$ are the points $(1, 1), (0, \sec^{2}\theta), (\csc^{2}\theta, 0)$ collinear?

$(a, b)$ is the point of concurrency of the lines $x-3y+3=0$,$kx+y+k=0$,and $2x+y-8=0$. If the perpendicular distance from the origin to the line $L \equiv ax-by+2k=0$ is $p$,then the perpendicular distance from the point $(2, 3)$ to $L=0$ is

If $\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$, then the lines $a_i x + b_i y + c_i = 0$ $(i = 1, 2, 3)$ represent:

The equation of the line passing through the point of intersection of the lines $x + 2y + 6 = 0$ and $2x - y = 2$ and making an intercept $5$ on the $y$-axis is

$A$ line passes through the point of intersection of $2x + y = 5$ and $x + 3y + 8 = 0$ and is parallel to the line $3x + 4y = 7$. Find the equation of this line.