If the coordinates of the mid-points of the sides of a triangle are (3, 4) (4, 6) and (5, 7), find its vertices.
The co-ordinates of the midpoint $\left(x_{n}, y_{n}\right)$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by,
$\left(x_{w}, y_{m}\right)=\left(\left(\frac{x_{1}+x_{2}}{2}\right),\left(\frac{y_{1}+y_{2}}{2}\right)\right)$
Let the three vertices of the triangle be $A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{n}\right)$ and $C\left(x_{C}, y_{C}\right)$.
The three midpoints are given. Let these points be $M_{A B}(3,4), M_{D C}(4,6)$ and $M_{C i}(5,7)$.
Let us now equate these points using the earlier mentioned formula,
$(3,4)=\left(\left(\frac{x_{A}+x_{n}}{2}\right),\left(\frac{y_{A}+y_{n}}{2}\right)\right)$
Equating the individual components we get,
$x_{4}+x_{B}=6$
$y_{A}+y_{B}=8$
Using the midpoint of another side we have,
$(4,6)=\left(\left(\frac{x_{B}+x_{c}}{2}\right),\left(\frac{y_{B}+y_{c}}{2}\right)\right)$
Equating the individual components we get,
$x_{B}+x_{C}=8$
$y_{B}+y_{C}=12$
Using the midpoint of the last side we have,
$(5,7)=\left(\left(\frac{x_{A}+x_{C}}{2}\right),\left(\frac{y_{A}+y_{C}}{2}\right)\right)$
Equating the individual components we get,
$x_{A}+x_{C}=10$
$y_{A}+y_{C}=14$
Adding up all the three equations which have variable ‘x’ alone we have,
$x_{A}+x_{B}+x_{B}+x_{C}+x_{A}+x_{C}=6+8+10$
$2\left(x_{1}+x_{3}+x_{C}\right)=24$
$x_{A}+x_{B}+x_{C}=12$
Substituting $x_{B}+x_{C}=4$ in the above equation we have,
$x_{A}+x_{B}+x_{C}=12$
$x_{A}+8=12$
$x_{A}=4$
Therefore,
$x_{A}+x_{C}=10$
$x_{C}=10-4$
$x_{C}=6$
And
$x_{A}+x_{B}=6$
$x_{B}=6-4$
$x_{B}=2$
Adding up all the three equations which have variable ‘y’ alone we have,
$y_{A}+y_{B}+y_{B}+y_{C}+y_{A}+y_{C}=8+12+14$
$2\left(y_{A}+y_{B}+y_{C}\right)=34$
$y_{A}+y_{B}+y_{C}=17$
Substituting $y_{B}+y_{C}=12$ in the above equation we have,
$y_{A}+y_{B}+y_{C}=17$
$y_{A}+12=17$
$y_{A}=5$
Therefore,
$y_{A}+y_{C}=14$
$y_{f}=14-5$
$y_{C}=9$
And
$y_{A}+y_{B}=8$
$y_{B}=8-5$
$y_{B}=3$
Therefore the co-ordinates of the three vertices of the triangle are $A(4,5)$ $B(2,3)$ $C(6,9)$ .