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 perimeter of a triangle is $16 \text{ cm}$, one of the sides is of length $6 \text{ cm}$. If the area of the triangle is $12 \text{ cm}^2$, then the triangle is:

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:

Let a point $A$ lie between the parallel lines $L_1$ and $L_2$ such that its distances from $L_1$ and $L_2$ are $6$ and $3$ units, respectively. Then the area (in sq. units) of the equilateral triangle $ABC$, where the points $B$ and $C$ lie on the lines $L_1$ and $L_2$ respectively, is:

If a vertex of an equilateral triangle is at the origin and the second vertex is $(4, 0)$,then its third vertex is

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