The locus of $P$ such that the area of $\Delta PAB = 12 \text{ sq. units}$,where $A(2, 3)$ and $B(-4, 5)$ is:

$A$ variable straight line passes through the point of intersection of the lines $x + 2y = 1$ and $2x - y = 1$ and meets the coordinate axes at $A$ and $B$. The locus of the midpoint of $AB$ is:

Consider the triangles with vertices $A(2,1)$,$B(0,0)$,and $C(t,4)$,where $t \in [0,4]$. If the maximum and the minimum perimeters of such triangles are obtained at $t=\alpha$ and $t=\beta$ respectively,then $6\alpha + 21\beta$ is equal to $.........$.

Through the point $(4, 5)$,a straight line is drawn making positive intercepts on the coordinate axes. The area of the triangle thus formed is least,when the ratio of the intercepts on the $X$ and $Y$ axes is

$A$ quadrilateral $ABCD$ is divided by the diagonal $AC$ into two triangles of equal areas. If $A, B, C$ are respectively $(3, 4), (-3, 6), (-5, 1)$,then the locus of $D$ is

$A$ straight line $L$ cuts both the lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$. The segment of $L$ between the two lines is bisected at the point $(1, 5)$. The equation of $L$ is