Find the equation of a line parallel to $2x - 3y = 4$ which forms a triangle of area $12$ square units with the coordinate axes.

$\Delta ABC$ has integral side lengths. The internal angle bisector of angle $B$ meets side $AC$ at $D$. If $AD = 3$ and $DC = 8$,then the perimeter of $\Delta ABC$ is:

The area of the parallelogram formed by the lines $L_1 \equiv \lambda x+4 y+2=0$,$L_2 \equiv 3 x+4 y-3=0$,$L_3 \equiv 2 x+\mu y+6=0$,and $L_4 \equiv 2 x+y+3=0$,where $L_1$ is parallel to $L_2$ and $L_3$ is parallel to $L_4$,is

In an isosceles $\triangle ABC$,the coordinates of vertices $B$ and $C$ of the base $BC$ are $(3, 2)$ and $(2, 3)$ respectively. If the equation of the line $AB$ is $3y = 2x$,then the equation of the line $AC$ is

The equation of the base $BC$ of an equilateral triangle is $3x + 4y = 1$ and the vertex $A$ is $(-3, 2)$. Then the area of the triangle is-

The equations of the sides of a triangle are $x - 2y = 0$,$4x + 3y = 5$,and $2x + y = 0$. Through which point does the line $3y - 4x = 0$ pass?