The centroid of a triangle is $(2, 3)$ and two of its vertices are $(5, 6)$ and $(-1, 4)$. Find the third vertex of the triangle.

The quadratic equation whose roots are the coordinates of the circumcentre of the triangle formed by the points $(-2,-1), (6,-1),$ and $(2,5)$ is

The orthocentre of the triangle with vertices $A(0,0)$,$B(0, \frac{3}{2})$,and $C(-5,0)$ is

$A(1, -2), B(-2, 3), C(-1, -3)$ are the vertices of a triangle $ABC$. $L_1$ is the perpendicular drawn from $A$ to $BC$ and $L_2$ is the perpendicular bisector of $AB$. If $(l, m)$ is the point of intersection of $L_1$ and $L_2$, then $26m - 3 =$ (in $l$)

Let the vertex of a triangle $ABC$ be $A(1, 2)$, and the mid-point of the side $AB$ be $M(5, -1)$. If the centroid of this triangle is $G(3, 4)$ and its circumcenter is $O(\alpha, \beta)$, then $21(\alpha + \beta)$ is equal to:

The vertices of a triangle are $A(0,0), B(0,2)$ and $C(2,0)$. Find the distance between its orthocentre and circumcentre.