Question

If $(10,5),(8,4)$ and $(6,6)$ be the mid-points of the sides of a triangle, find the coordinates of the vertices.

Answer 1

Let's say $D,E$ and $F$ are the midpoints of $AB,BC$ and $AC$ and coordinate of vertices $A,B,C$ be $(x,y),(a,b)$ and $(p,q)$ respectively.

For points $(x_1,y_1)$ and $(x_2,y_2)$ we define the midpoint as,

Point $P(X,Y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

For mid point $D$ of $AB,$

$\frac{x+a}{2}=10$

$\Rightarrow x+a=20$

$\frac{y+b}{2}=5$

$\Rightarrow y+b=10$

For midpoint $D$ of $BC,$

$\frac{a+p}{2}=6$

$\Rightarrow a+p=12$

$\frac{b+q}{2}=6$

$\Rightarrow b+q=12$

For midpoint $F$ of $AC,$

$\frac{x+p}{2}=8$

$\Rightarrow x+p=16$

$\frac{y+q}{2}=4$

$\Rightarrow y+q=8$

Consider, $x+a=20$ and $a+p=12$

$x+12-p=20$

$\Rightarrow x-p=8$

Now, we have,

$x-p=8,x+p=16$

On solving;

$x=12$ and $p=4$

Now, we have $x=12$ and $x+a=20$. So,

$12+a=20$

$\Rightarrow a=8$

Consider, $y+q=8$ and $q+b=12$

$y+12-b=8$

$\Rightarrow y-b=-4$

Now, we have,

$y-b=-4,y+b=10$

On solving;

$y=3$ and $b=7$

Now, we have $y=3$ and $y+q=8$. So,

$3+q=8$

$\Rightarrow q=5$

Hence, the coordinates of the vertices will be $A(12,3),B(8,7),C(4,5).$