If the midpoints of the sides of a triangle are $(-2, 3), (4, -3)$ and $(4, 5)$,then the centroid of the triangle is

Two vertices of a triangle are $(5, 4)$ and $(-2, 4)$. If its centroid is $(5, 6)$,then the third vertex has the coordinates:

If $(\alpha, \beta)$ is the orthocentre of the triangle $ABC$ with vertices $A(3, -7)$,$B(-1, 2)$,and $C(4, 5)$,then $9\alpha - 6\beta + 60$ is equal to:

Let $PQR$ be an acute-angled triangle in which $PQ < QR$. From the vertex $Q$,draw the altitude $QQ_1$,the angle bisector $QQ_2$,and the median $QQ_3$,with $Q_1, Q_2, Q_3$ lying on $PR$. Then,

Assertion $(A)$: If the centroid and circumcenter of a triangle are known,its orthocenter can be found.
Reason $(R)$: The centroid,orthocenter,and circumcenter of a triangle are collinear.

The circumcenter of the triangle formed by the lines $y = x$,$y = 2x$,and $y = 3x + 4$ is