The equation of the straight line which passes through the point $(1, -2)$ and cuts off equal intercepts from the axes is:

Find the equation of the line that bisects the line segment joining the points $(5, 3)$ and $(4, 4)$ and makes an angle of $45^{\circ}$ with the positive direction of the $x$-axis.

$A$ line passes through the point $(3, 4)$ and cuts off intercepts from the coordinate axes such that their sum is $14$. The equation of the line is

If a straight line passing through the point $P(3,4)$ makes an angle $\frac{\pi}{6}$ with the $X$-axis and meets the line $12x+5y+10=0$ at $Q$,then the length of $PQ$ is

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

The line $(3x - y + 5) + \lambda (2x - 3y - 4) = 0$ will be parallel to the $y$-axis,if $\lambda$ =