The diagonals of a parallelogram $PQRS$ are along the lines $x + 3y = 4$ and $6x - 2y = 7$. Then $PQRS$ must be a

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

$A$ square of side length $a$ has one vertex at the origin and one side along the $x$-axis. The side passing through the origin makes an angle $\alpha$ $(0 < \alpha < \pi/4)$ with the positive $x$-axis. Find the equation of the diagonal that does not pass through the origin.

The points $\left( \frac{a}{\sqrt{3}}, a \right)$,$\left( \frac{2a}{\sqrt{3}}, 2a \right)$,and $\left( \frac{a}{\sqrt{3}}, 3a \right)$ are the vertices of:

If $x_1, x_2, x_3$ and $y_1, y_2, y_3$ are in $G.P.$ with the same common ratio,then the points $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ are:

Two consecutive sides of a parallelogram are $4x + 5y = 0$ and $7x + 2y = 0$. If the equation to one diagonal is $11x + 7y = 9$,then the equation to the other diagonal is: