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...

Let $\alpha, \beta, \gamma$ be three non-zero real constants and $a, b, c$ be three arbitrary real numbers which satisfy $\alpha a + \beta b + \gamma c = 0$. Then the point of concurrence of the family of lines $ax + by + c = 0$ is

For $a \neq b \neq c$,if the lines $x+2ay+a=0$,$x+3by+b=0$ and $x+4cy+c=0$ are concurrent,then $a, b, c$ are in

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

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.

Given the four lines with equations $x + 2y = 3,$ $3x + 4y = 7,$ $2x + 3y = 4,$ and $4x + 5y = 6,$ these lines are: