Question

The mid point of the sides of a triangle are at $(1,4),(4,8)$ and $(5,6)$. Find the coordinates of the vertices of the triangle.

Answer 1

Let, $ABC$ be a triangle with vertices $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ and $P(1,4),Q(4,8),R(5,6)$ be the midpoints of $AB,BC,CA$ respectively.

Then,

$1=\frac{x_1+x_2}{2}\Rightarrow x_1+x_2=2⟶\left(1\right)4=\frac{y_1+y_2}{2}\Rightarrow y_1+y_2=8⟶\left(2\right)4=\frac{x_2+x_3}{2}\Rightarrow x_2+x_3=8⟶\left(3\right)8=\frac{y_2+y_3}{2}\Rightarrow y_2+y_3=16⟶\left(4\right)5=\frac{x_3+x_1}{2}\Rightarrow x_3+x_1=10⟶\left(5\right)6=\frac{y_3+y_1}{2}\Rightarrow y_3+y_1=12⟶\left(6\right)$

Adding equations $\left(1\right),\left(3\right)$ and $\left(5\right)$ we get,

$2(x_1+x_2+x_3)=20\Rightarrow x_1+x_2+x_3=10⟶\left(7\right)$

Subtracting equation $\left(7\right)$ from $\left(1\right)$ $\Rightarrow x_3=8$

Subtracting equation $\left(7\right)$ from $\left(3\right)$ $\Rightarrow x_1=2$

Subtracting equation $\left(7\right)$ from $\left(5\right)$ $\Rightarrow x_2=0$

Now, adding equations $\left(2\right),\left(4\right)$ and $\left(6\right)$ we get,

$2(y_1+y_2+y_3)=36\Rightarrow y_1+y_2+y_3=18⟶\left(8\right)$

Subtracting equation $\left(8\right)$ from $\left(2\right)$ $\Rightarrow y_3=10$

Subtracting equation $\left(8\right)$ from $\left(4\right)$ $\Rightarrow y_1=2$

Subtracting equation $\left(8\right)$ from $\left(6\right)$ $\Rightarrow y_2=6$

$\therefore$ the required coordinates of the vertices of the triangle are

$A=(2,2)B=(0,6)C=(8,10).$