If the coordinates of the mid-points of the

Question:

If the coordinates of the mid-points of the sides of a triangle are (3, 4) (4, 6) and (5, 7), find its vertices.

Solution:

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 $A(4,5)$ $B(2,3)$ $C(6,9)$ .

 

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