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:

$A$ line through $A(-5, -4)$ meets the lines $x + 3y + 2 = 0$,$2x + y + 4 = 0$,and $x - y - 5 = 0$ at $B$,$C$,and $D$ respectively. If $\left( \frac{15}{AB} \right)^2 + \left( \frac{10}{AC} \right)^2 = \left( \frac{6}{AD} \right)^2$,then the equation of the line is

If the coordinates of the points $A, B, C, D$ are $(a, b), (a', b'), (-a, b)$ and $(a', -b')$ respectively,then the equation of the line bisecting the line segments $AB$ and $CD$ is

If we reduce $3x + 3y + 7 = 0$ to the form $x \cos \alpha + y \sin \alpha = p$,then the value of $p$ is

$a$ and $b$ are the intercepts made by a line on the coordinate axes. If $3a = b$ and the line passes through $(1, 3)$,then the equation of the line is

If a straight line passes through the point $(-5, 4)$ and makes an intercept of length $\frac{2}{\sqrt{5}}$ between the lines $x+2y+1=0$ and $x+2y-1=0$,then the equation of that line is