The orthocenter of an equilateral triangle is $(3, -2)$. If one of its sides lies on the $x$-axis,find the vertex of the triangle that does not lie on the $x$-axis.

The incentre of the triangle formed by the straight lines $y=\sqrt{3}x$,$y=-\sqrt{3}x$ and $y=3$ is

The circumcentre of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points $(a^2 + 1, a^2 + 1)$ and $(2a, -2a)$,where $a \ne 0$. Then for any $a$,the orthocentre of this triangle lies on the line:

Let $A(1,1), B(1,-1), C(-1,1)$ be the vertices of $\triangle ABC$. Let $S$ be the circumcentre,$O$ be the orthocentre,and $I$ be the incentre of the $\triangle ABC$. Then $IS + OS =$ ?

Point $R(h, k)$ divides a line segment between the axes in the ratio $1: 2$. Find the equation of the line.

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 $(a, b)=$