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

Let $a, b, c$ and $d$ be non-zero numbers. If the point of intersection of the lines $4ax + 2ay + c = 0$ and $5bx + 2by + d = 0$ lies in the fourth quadrant and is equidistant from the two axes,then:

The equation of the line passing through the point of intersection of the lines $x-3y+2=0$ and $2x+5y-7=0$ and perpendicular to the line $3x+2y+5=0$ is:

Find the equation of a line passing through the point of intersection of the lines $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$ and perpendicular to the line $6x - 7y + 3 = 0$.

If $a + b + c = 0$ and $p \neq 0$,the lines $ax + (b + c)y = p$,$bx + (c + a)y = p$ and $cx + (a + b)y = p$ are:

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