Let $PS$ be the median of the triangle with vertices $P(2, 2)$,$Q(6, -1)$,and $R(7, 3)$. The equation of the line passing through $(1, -1)$ and parallel to $PS$ is

The gradient of the line joining the points on the curve $y = x^2 + 2x$ whose abscissae are $1$ and $3$,is

The line $MN$ whose equation is $x-y-2=0$ cuts the $X$-axis at $M$ and the coordinates of $N$ are $(4,2)$. The line $MN$ is rotated about $M$ through $45^{\circ}$ in the anticlockwise direction. What is the equation of the line $MN$ in the new position?

$A$ 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

Reduce the following equation into slope-intercept form and find its slope and the $y$-intercept.
$x+7y=0$

The length of the perpendicular from the origin to a line is $4$ units,and the perpendicular makes an angle of $30^{\circ}$ with the positive direction of the $x$-axis. Find the equation of the line.