The equation of a line perpendicular to $x = c$ is

Find the equation of the line whose perpendicular distance from the origin is $4$ units and the angle which the normal makes with the positive direction of the $x$-axis is $15^{\circ}$.

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

Find the equation of the line perpendicular to the line $x-7y+5=0$ and having $x$-intercept $3$.

The straight line $2x - 3y + 17 = 0$ is perpendicular to the line passing through the points $(7, 17)$ and $(15, \beta)$. Then,$\beta$ equals:

Consider the equation $y-y_1=m(x-x_1)$. If $m$ and $x_1$ are fixed and different lines are drawn for different values of $y_1$, then