Question

The coordinates of the midpoint of the sides of a $△ABC$ are $D(2,1),E(5,3),F(3,7)$. Find the length and equation of its sides.

Answer 1

Given, the coordinates of the midpoints of $△ABC$ are $D(2,1), E(5,3), F(3,7)$ respectively.

Let the coordinates of the vertices be $A=(x_1,y_1),$

$B=(x_2,y_2),$

$C=(x_3,y_3)$

$\therefore$ Coordinates of $D$ is $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(2,1)$

$\Rightarrow \frac{x_1+x_2}{2}=2$

$\Rightarrow x_1+x_2=4⟶\left(1\right)$

and, $\frac{y_1+y_2}{2}=1$

$\Rightarrow y_1+y_2=2⟶\left(2\right)$

Coordinates of $E$ is $(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})=(5,3)$

$\Rightarrow \frac{x_2+x_3}{2}=5$

$\Rightarrow x_2+x_3=10⟶\left(3\right)$

and, $\frac{y_2+y_3}{2}=3$

$\Rightarrow y_2+y_3=6⟶\left(4\right)$

Coordinates of $F$ is $(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})=(3,7)$

$\Rightarrow \frac{x_1+x_3}{2}=3$

$\Rightarrow x_1+x_3=6⟶\left(5\right)$

and, $\frac{y_1+y_3}{2}=7$

$\Rightarrow y_1+y_3=14⟶\left(6\right)$

Solving equations $\left(1\right)$, $\left(3\right)$, $\left(5\right)$ we get,

$x_1=0,x_2=4,x_3=6$

Solving equations $\left(2\right)$, $\left(4\right)$, $\left(6\right)$ we get,

$y_1=5,y_2=-3,y_3=9$

$\therefore$ coordinates of the vertices will be $A=(0,5)$

$B=(4,-3)$

$C=(6,9)$

$\therefore$ Lengths of $AB=\sqrt{{(-4)}^2+8^2}$

$=2\sqrt{5}$

$BC=\sqrt{{(-2)}^2+{12}^2}$

$=2\sqrt{37}$

$CA=\sqrt{6^2+4^2}$

$=2\sqrt{13}$

The equations of the lines $AB$

$\Rightarrow (y-5)=\frac{-3-5}{4-0}(x-0)$

$\Rightarrow 2x+y=20$

$BC\Rightarrow (y+3)=\frac{9+3}{6-4}(x-4)$

$\Rightarrow 6x-y=27$

$CA\Rightarrow (y-9)=\frac{5-9}{0-6}(x-6)$

$\Rightarrow 2x-3y=-15.$