The three points $(-2, 2)$,$(8, -2)$,and $(-4, -3)$ are the vertices of:

Straight lines $x-y=7$ and $x+4y=2$ intersect at $B$. Points $A$ and $C$ are so chosen on these two lines such that $AB=AC$. The equation of line $AC$ passing through $(2,-7)$ is

If the extremities of the base of an isosceles triangle are the points $(2a, 0)$ and $(0, a)$ and the equation of one of the sides is $x = 2a$,then the area of the triangle is

In an equilateral triangle $PQR$, let the vertex $P$ be at $(3, 5)$ and the side $QR$ be along the line $x + y = 4$. If the orthocentre of the triangle $PQR$ is $(\alpha, \beta)$, then $9(\alpha + \beta)$ is equal to:

In an isosceles triangle $ABC$,the vertex $A$ is $(6,1)$ and the equation of the base $BC$ is $2x + y = 4$. Let the point $B$ lie on the line $x + 3y = 7$. If $(\alpha, \beta)$ is the centroid of $\triangle ABC$,then $15(\alpha + \beta)$ is equal to

The points $(0, 0)$,$(a, 0)$,and $\left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$ are vertices of: