If the coordinates of the mid-points of the sides of a triangle be
Question:

If the coordinates of the mid-points of the sides of a triangle be (3, −2), (−3, 1) and (4, −3), then find the coordinates of its vertices.

Solution:

The co-ordinates of the midpoint $\left(x_{m}, y_{m}\right)$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by,

$\left(x_{m}, y_{m}\right)=\left(\left(\frac{x_{1}+x_{2}}{2}\right),\left(\frac{y_{1}+y_{2}}{2}\right)\right)$

Let the three vertices of the triangle be $A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right)$ and $C\left(x_{C}, y_{C}\right)$.

The three midpoints are given. Let these points be $M_{A \delta}(3,-2), M_{B C}(-3,1)$ and $M_{C C}(4,-3)$.

Let us now equate these points using the earlier mentioned formula,

$(3,-2)=\left(\left(\frac{x_{A}+x_{A}}{2}\right),\left(\frac{y_{A}+y_{B}}{2}\right)\right)$

Equating the individual components we get,

$x_{A}+x_{B}=6$

$y_{A}+y_{B}=-4$

Using the midpoint of another side we have,

$(-3,1)=\left(\left(\frac{x_{B}+x_{C}}{2}\right),\left(\frac{y_{B}+y_{C}}{2}\right)\right)$

Equating the individual components we get,

$x_{B}+x_{C}=-6$

 

$y_{B}+y_{C}=2$

Using the midpoint of the last side we have,

$(4,-3)=\left(\left(\frac{x_{A}+x_{C}}{2}\right),\left(\frac{y_{A}+y_{C}}{2}\right)\right)$

Equating the individual components we get,

$x_{A}+x_{C}=8$

 

$y_{A}+y_{C}=-6$

Adding up all the three equations which have variable ‘x’ alone we have,

$x_{4}+x_{B}+x_{3}+x_{f}+x_{4}+x_{f}=6-6+8$

$2\left(x_{A}+x_{B}+x_{C}\right)=8$

$x_{A}+x_{B}+x_{C}=4$

Substituting $x_{B}+x_{C}=-6$ in the above equation we have,

$x_{1}+x_{3}+x_{c}=4$

$x_{1}-6=4$

$x_{A}=10$

Therefore,

$x_{A}+x_{C}=8$

$x_{C}=8-10$

$x_{C}=-2$

And

$x_{A}+x_{B}=6$

$x_{s}=6-10$

$x_{B}=-4$

Adding up all the three equations which have variable ‘y’ alone we have,

$y_{A}+y_{B}+y_{B}+y_{C}+y_{A}+y_{C}=-4+2-6$

$2\left(y_{A}+y_{B}+y_{C}\right)=-8$

$y_{A}+y_{B}+y_{C}=-4$

Substituting $y_{B}+y_{C}=2$ in the above equation we have,

$y_{A}+y_{B}+y_{C}=-4$

$y_{A}+2=-4$

$y_{A}=-6$

Therefore,

$y_{A}+y_{C}=-6$

$y_{c}=-6+6$

$y_{C}=0$

And

$y_{A}+y_{B}=-4$

$y_{B}=-4+6$

$y_{B}=2$

Therefore the co-ordinates of the three vertices of the triangle are $A(10,-6)$ $B(-4,2)$ $C(-2,0)$ .

 

 

 

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