Question

If in a triangle ABC, A $\equiv (1,10)$, circumcenter $\equiv$ $(\frac{-1}{3},\frac{2}{3})$ and orthocenter $\equiv$ $(\frac{11}{3},\frac{4}{3})$ then the coordinates of the midpoint of the side opposite to A is.

• A. $(1,-\frac{11}{3})$
• B. $(1,5)$
• C. $(1,-3)$
• D. $(1,6)$

Answer 1

Answer: (A)

$(1,-\frac{11}{3})$

We know that,

The orthocentre,centroid and circumcentre of any triangle are collinear.And the centroid divides the distance from othocentre to circumcentre in the ratio $2:1$.

Let the centroid be $G(x,y)$ , its coordinates can be found using the section formula.

$\therefore (x,y)\equiv (\frac{\frac{-2}{3}+\frac{11}{3}}{3},\frac{\frac{4}{3}+\frac{4}{3}}{3})\equiv (1,\frac{8}{9})$

Also, the centroid $\left(G\right)$ divides the medians $\left(AD\right)$ in the ratio $2:1$

Let the coordinates of $D$ be $(h,k)$

$\therefore 1=\frac{2h+1}{3}$ and $\frac{8}{9}=\frac{2k+10}{3}$

$h=1$ and $8=6k+30$

and $k=\frac{-11}{3}$

$\therefore D(h,k)=(1,\frac{-11}{3})$