Question

If in $△ABC$, $A\equiv (1,10)$, circumcentre $\equiv (-\frac{1}{3},\frac{2}{3})$ and orthocentre $\equiv (\frac{11}{3},\frac{4}{3})$, then the coordinates of mid-point of side opposite to $A$ is

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

Answer 1

The correct option is A $(1,-\frac{11}{3})$

Let $D$ be the mid point of side $BC$
Now Euler line, we get $OH=3GO$ and $GH=2GO$
Where $H=$ orthocentre, $G=$ Centroid and $O=$ circumcentre.
Hence, $GH:OG=2:1$
Therefore, the coordinates of $G$ is $(\frac{\frac{-2}{3}+\frac{11}{3}}{3},\frac{\frac{4}{3}+\frac{4}{3}}{3})$ $=(1,\frac{8}{9})$

Therefore, $AD:DG=3:1$ (externally)
Hence, the coordinates of $D$ will be $(\frac{3-1}{2},\frac{\frac{8}{3}-10}{2})$$=(1,\frac{-11}{3})$