The coordinates of the orthocentre of the triangle bounded by the lines $4x - 7y + 10 = 0$,$x + y = 5$,and $7x + 4y = 15$ are:

If the vertices of a triangle are $(2, 1), (5, 2)$ and $(3, 4)$,then its circumcentre 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 $l$)

The vertices of a triangle are at $-\hat{i}+3 \hat{j}$ and $2 \hat{i}+5 \hat{j}$ and its orthocenter is at $\hat{i}+2 \hat{j}$. If the position vector of the third vertex is $a \hat{i}+b \hat{j}$,then $(a, b)=$

The incentre of a triangle with vertices $(7, 1)$,$(-1, 5)$,and $(3 + 2\sqrt{3}, 3 + 4\sqrt{3})$ is

If $P(6,1)$ is the orthocentre of the triangle whose vertices are $A(5,-2)$,$B(8,3)$,and $C(h, k)$,then the point $C$ lies on the circle: