Question

The mid points $D,E,F$ of the sides of a triangle $ABC$ are $(3,4),(8,9)and(6,7)$ respectively. Find the vertices of the triangle.

Answer 1

Let $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ be the vertices of $△ABC$

We have $D$ is the midpoint of $AB\Rightarrow (3,4)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$\Rightarrow x_1+x_2=6$ ........$\left(1\right)$

and $y_1+y_2=8$ ........$\left(2\right)$

$E$ is the midpoint of $BC\Rightarrow (8,9)=(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})$

$\Rightarrow x_2+x_3=16$ ........$\left(3\right)$

and $y_2+y_3=18$ ........$\left(4\right)$

$F$ is the midpoint of $AC\Rightarrow (6,7)=(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})$

$\Rightarrow x_1+x_3=12$ ........$\left(5\right)$

and $y_1+y_3=14$ ........$\left(6\right)$

Equation $\left(1\right)-\left(3\right)$ we get

$x_1+x_2-x_2-x_3=6-16$

$\Rightarrow x_1-x_3=-10$ .......$\left(7\right)$

Equation $\left(5\right)+\left(7\right)$ we get

$x_1+x_3+x_1-x_3=12-10=2$

$\Rightarrow 2x_1=2$ or $x_1=1$

Substituting the value of $x_1=1$ in eqn$\left(1\right)$ we get

$x_1+x_2=6$ or $x_2=6-1=5$

Substituting the value of $x_1=1$ in eqn$\left(5\right)$ we get

$x_1+x_3=12$ or $x_3=12-1=11$

Equation $\left(2\right)-\left(4\right)$ we get

$y_1+y_2-y_2-y_3=8-18$

$\Rightarrow y_1-y_3=-10$ .......$\left(8\right)$

Add equations $\left(8\right)$ and $\left(6\right)$ we get

$y_1-y_3+y_1+y_3=-10+14$

$\Rightarrow y_1=2$

From $\left(2\right)y_1+y_2=8$ or $y_2=8-2=6$

From $\left(4\right)y_1+y_3=18$ or $y_3=18-2=16$

$\therefore$ the co-ordinates of vertices of $△ABC$ is $A(1,2),(5,6),C(11,16)$