If the lines $ax + 2y + 1 = 0$,$bx + 3y + 1 = 0$,and $cx + 4y + 1 = 0$ are concurrent,then $a, b, c$ are in:

By using the concept of the equation of a line,prove that the three points $(3,0), (-2,-2),$ and $(8,2)$ are collinear.

If the lines $ax + y + 1 = 0$,$x + by + 1 = 0$ and $x + y + c = 0$ (where $a, b$ and $c$ are distinct and different from $1$) are concurrent,then the value of $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = $

The lines $(a+2b)x + (a-3b)y = a-b$ for different values of $a$ and $b$ pass through a fixed point whose coordinates are:

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:

What is the nature of the points $(-a, -b)$,$(0, 0)$,$(a, b)$,and $(a^2, ab)$?