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

If the points $(1, 1)$,$(-1, -1)$ and $(-\sqrt{3}, k)$ are vertices of an equilateral triangle,then the value of $k$ will be:

In a $\triangle ABC$,$2x+3y+1=0$ and $x+2y-2=0$ are the perpendicular bisectors of its sides $AB$ and $AC$ respectively. If $A=(3,2)$,then the equation of the side $BC$ is

Let one of the sides of a triangle be $17 \text{ cm}$ and the sum of all the sides of the triangle be $40 \text{ cm}$. If the sum of two adjacent sides is $35 \text{ cm}$, then the area (in $\text{cm}^2$) of the triangle is (in $\sqrt{2}$)

If the sides of a triangle are given by $y = mx + a$,$y = nx + b$,and $x = 0$,then its area is:

Let $ABC$ be an isosceles triangle in which $A$ is at $(-1, 0)$,$\angle A = \frac{2\pi}{3}$,$AB = AC$ and $B$ is on the positive $x$-axis. If $BC = 4\sqrt{3}$ and the line $BC$ intersects the line $y = x + 3$ at $(\alpha, \beta)$,then $\frac{\beta^4}{\alpha^2}$ is :