If $ABC$ is an isosceles triangle and the coordinates of the base points are $B(1, 3)$ and $C(-2, 7)$, then the coordinates of $A$ can be:

Consider the triangles with vertices $A(2,1)$,$B(0,0)$,and $C(t,4)$,where $t \in [0,4]$. If the maximum and the minimum perimeters of such triangles are obtained at $t=\alpha$ and $t=\beta$ respectively,then $6\alpha + 21\beta$ is equal to $.........$.

If the line $y = \sqrt{3}x$ cuts the curve $x^3 + 3y^2 + 4x + 5y - 1 = 0$ at the points $A, B, C$,then the product $OA \cdot OB \cdot OC$ is

Let $A$ be the set of all points $(\alpha, \beta)$ such that the area of the triangle formed by the points $(5, 6), (3, 2),$ and $(\alpha, \beta)$ is $12 \text{ square units}.$ Then the least possible length of a line segment joining the origin to a point in $A$ is:

Find the fixed point through which the line $x(a + 2b) + y(a + 3b) = a + b$ always passes for all values of $a$ and $b$.

If the equation of the locus of a point equidistant from the points $(a_1, b_1)$ and $(a_2, b_2)$ is $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$,then the value of $c$ is