The equation of the line whose midpoint between the axes is $({x_1}, {y_1})$ is:

If the transversals $y = m_r x; r = 1, 2, 3$ cut off equal intercepts on the line $x + y = 1$,then $1 + m_1, 1 + m_2, 1 + m_3$ are in

Find the equation of the line passing through the point $(2, -3)$ such that the sum of its intercepts on the axes is $-2$.

The equations of the sides $AB$,$AC$,and $BC$ of a $\triangle ABC$ are respectively $x-3y=0$,$3x-y=0$,and $x+y+4=0$. If $P$ and $Q$ are points on the line $3x-y+k=0$ passing through $B$ such that $PB:BQ=1:1$,then $k=$

$A$ straight line $L$ at a distance of $4$ units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of $60^o$ with the line $x + y = 0$. Then an equation of the line $L$ is

$A$ straight line passes through the point $A(3, 4)$ such that its intercept between the axes is bisected at $A$. Find its equation.