If the length of the perpendicular to a line from the origin is $2 \sqrt{2}$ units,which makes an angle of $135^{\circ}$ with the $X$-axis,then the equation of the line 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 $y$-intercept of the line passing through $A(6, 1)$ and perpendicular to the line $x - 2y = 4$ is:

The lines $2x + 3y = 6$ and $2x + 3y = 8$ cut the $X$-axis at $A$ and $B$,respectively. $A$ line $L$ drawn through the point $(2, 2)$ meets the $X$-axis at $C$ in such a way that the abscissae of $A, B,$ and $C$ are in arithmetic progression. Then,the equation of the line $L$ is

The equation of the straight line passing through $(-a, 0)$ and forming a triangle with the coordinate axes of area $T$ is

Find the slope of the line making an inclination of $60^{\circ}$ with the positive direction of the $x$-axis.