If the three points $(3q, 0)$, $(0, 3p)$, and $(1, 1)$ are collinear, then which one is true?

The lines $2x + 3y = 6$ and $2x + 3y = 8$ cut the $X$-axis at $A$ and $B$,respectively. $A$ line $L$ drawn through the point $(2, 2)$ meets the $X$-axis at $C$ in such a way that the abscissae of $A, B,$ and $C$ are in arithmetic progression. Then,the equation of the line $L$ is

$A$ line is perpendicular to the line $ax + by + c = 0$ and passes through $(a, b)$. The equation of the line is

If the midpoints of the sides $BC, CA$ and $AB$ of the triangle $ABC$ are $(1, 3), (5, 7)$ and $(-5, 7)$ respectively,then the equation of the side $AB$ is

The equation of the line parallel to the line $2x - 3y = 1$ and passing through the midpoint of the line segment joining the points $(1, 3)$ and $(1, -7)$ is:

Show that two lines $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$,where $b_{1}, b_{2} \neq 0$,are parallel if $\frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}}$.