The lines $x-y-2=0$,$x+y-4=0$,and $x+3y=6$ meet at a common point:

$A$ 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.

Let $C$ be the centroid of the triangle with vertices $(3, -1), (1, 3),$ and $(2, 4).$ Let $P$ be the point of intersection of the lines $x + 3y - 1 = 0$ and $3x - y + 1 = 0.$ Then the line passing through the points $C$ and $P$ also passes through the point

The family of straight lines $4ax + 3by + c = 0$ such that $a + b + c = 0$ (where $a, b, c$ are real constants) are concurrent at the point...

The line parallel to the $x$-axis and passing through the intersection of the lines $ax + 2by + 3b = 0$ and $bx - 2ay - 3a = 0$,where $(a, b) \ne (0, 0)$ is

The equations $(b - c)x + (c - a)y + (a - b) = 0$ and $(b^3 - c^3)x + (c^3 - a^3)y + (a^3 - b^3) = 0$ will represent the same line,if