Question

If in a triangle $A=(1,10),$ circumcentre $=(-\frac{1}{3},\frac{2}{3})$ and orhocentre $=(\frac{11}{3},\frac{4}{3})$ then the co-ordinate of mid-point of side opposite to $A$ is

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

Answer 1

Answer: (D)

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

Given points are A(1,1,0) and Circumcenter(C) $(\frac{-1}{3},\frac{2}{3})$ and orthocenter(O) $(\frac{11}{3},\frac{4}{3})$

Using mid point theorem, to find the centre(G) of the circumcenter and the orthocenter,

we get

$G(1,\frac{8}{9})$

Now let D be the mid point of side BC (oppposite to A), since G is the mid point of AD where coordinates of $A(1,1,0)$ and $D(x,y)$. So, we find the coordinate of $D(x,y)$

$\frac{2x+1}{3}=1,$

$\frac{2y+10}{3}=\frac{8}{9}$

$⟹x=1,y=\frac{-11}{3}$

$⟹D(x,y)=(1,\frac{-11}{3})$

which is the mid point of the side opposite to A.

hence, option (D) is correct.