The equations of the lines which pass through the point of intersection of the lines $4x - 3y - 1 = 0$ and $2x - 5y + 3 = 0$ and are equally inclined to the axes are

Find the equation of the line which passes through $(0, 0)$ with slope $m$.

$A$ 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

$A$ line passing through the point $P(a, 0)$ makes an acute angle $\alpha$ with the positive $x$-axis. Let this line be rotated about the point $P$ through an angle $\frac{\alpha}{2}$ in the clockwise direction. If in the new position,the slope of the line is $2-\sqrt{3}$ and its distance from the origin is $\frac{1}{\sqrt{2}}$,then the value of $3a^2 \tan^2 \alpha - 2\sqrt{3}$ is

The point $P(a, b)$ lies on the straight line $3x + 2y = 13$ and the point $Q(b, a)$ lies on the straight line $4x - y = 5$. Then,the equation of the line $PQ$ is:

$A$ line cuts off equal intercepts on the coordinate axes. The angle made by this line with the positive direction of the $X$-axis is: (in $^{\circ}$)