The points $A(2a, 4a)$,$B(2a, 6a)$,and $C(2a + \sqrt{3}a, 5a)$,where $a > 0$,are the vertices of:

The equations of the sides of a triangle are $x - 3y = 0$,$4x + 3y = 5$,and $3x + y = 0$. The line $3x - 4y = 0$ passes through:

The area of the parallelogram formed by the lines $y = mx, y = mx + 1, y = nx, y = nx + 1$ is:

For what value of $k$ are the points $(k, 2 - 2k)$,$(1 - k, 2k)$,and $(-4 - k, 6 - 2k)$ collinear?

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 of the other diagonal is

If $P(\sin \alpha, \cos \alpha)$ lies inside the triangle formed by the vertices $(0,0), \left(\frac{\sqrt{3}}{2}, 0\right)$ and $\left(0, \frac{\sqrt{3}}{2}\right)$,then $\alpha$ lies in the interval