Question

In the triangle $ABC,A=(1,10)$, circumcentre $=(-\frac{1}{3},\frac{2}{3})$ and orthocentre $=(\frac{11}{3},\frac{4}{3})$ 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 $(1,-\frac{11}{3})$

Given:$A=(1,10)$,Orthocentre $H=(\frac{11}{3},\frac{4}{3})$ and circumcentre $O=(-\frac{1}{3},\frac{2}{3})$

We know that centroid divides line segment joining $O$ and $H$ in the ratio $1:2$

Now,$G=(\frac{1\times \frac{11}{3}+2\times -\frac{1}{3}}{1+2},\frac{1\times \frac{4}{3}+2\times \frac{2}{3}}{1+2})$

$G=(\frac{11-2}{9},\frac{4+4}{9})=(1,\frac{8}{9})$

If $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ then its centroid $G=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

Here $x_1=1,y_1=10$

Hence let us find the mid point of $BC$.

Midpoint $M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Comparing the coordinates of G,we get

$\frac{x_1+x_2+x_3}{3}=1$ and $\frac{y_1+y_2+y_3}{3}=\frac{8}{9}$

Now substituting the values of $A$ we get

$\frac{1+x_2+x_3}{3}=1$ and $\frac{10+y_2+y_3}{3}=\frac{8}{9}$

$\frac{1+x_2+x_3}{1}=3$ and $\frac{10+y_2+y_3}{1}=\frac{8}{3}$

$x_2+x_3=3-1=2$ and $y_2+y_3=\frac{8}{3}-10$

$x_2+x_3=2$ and $y_2+y_3=-\frac{22}{3}$

$\frac{x_2+x_3}{2}=1$ and $\frac{y_2+y_3}{2}=-\frac{11}{3}$

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

Hence the midpoint is $(1,-\frac{11}{3})$