If $A(4,7)$,$B(-7,8)$,and $C(1,2)$ are the vertices of $\triangle ABC$,then the equation of the perpendicular bisector of the side $AB$ is

The equation of a straight line,perpendicular to $3x - 4y = 6$ and forming a triangle of area $6 \text{ sq. units}$ with the coordinate axes,is

Find the area of the quadrilateral formed by the lines $4y - 3x = 1, 4y - 3x - 3 = 0, 3y - 4x + 1 = 0,$ and $3y - 4x + 2 = 0$.

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 the points $(a, b)$,$(a', b')$ and $(a - a', b - b')$ are collinear,then:

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 :