Question

If the midpoints of sides of a triangle are $(3,4),(4,6)$ and $(5,7)$, find the vertices of the triangle.

Answer 1

Let the vertices of triangle be $\left(x_1,y_1\right),\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$.

By using the given information in the question,

$\frac{x_1+x_2}{2}=3\Rightarrow x_1+x_2=6\dots \left(1\right)$

Similarly,

$x_2+x_3=8\dots \left(2\right)x_3+x_1=10\dots \left(3\right)$

By using equation $\left(1\right),$

$x_2=6-x_1$

Substitute the above value of $x_2$ in equation $\left(2\right)$

$6-x_1+x_3=8\Rightarrow x_3-x_1=2\dots \left(4\right)$

Add equations $\left(4\right)$ and $\left(3\right)$,

$2x_3=12\therefore x_3=6$

And,

$x_1=4\left(\text{from}4\right)x_2=6-x_1$

$=6-4$

$=2$

By using the same method value of $y_1=5,y_2=3$ and $y_3=9$ are calculated.

Vertices of triangle are $(4,5),(2,3),(6,9).$