If the coordinates of the mid-points of the sides of a triangle are $(1, 1), (2, -3)$ and $(3, 4)$, find the vertices of the triangle.

 Given:

The coordinates of the mid-points of the sides of a triangle are $(1, 1), (2, -3)$ and $(3, 4)$.

To do:

We have to find the vertices of the triangle.

Solution:

Let $A (x_1, y_1), B (x_2, y_2)$ and $C (x_3, y_3)$ be the vertices of the $\triangle ABC$ and $D(1,1), E(2,-3)$ and $F(3,4)$ are the mid-points of $BC, CA$ and $AB$. 

$D$ is the mid-point of $BC$.

This implies,

\( \frac{x_{2}+x_{3}}{2}=1 \)

\( \Rightarrow x_{2}+x_{3}=2 \).....(i)
\( \frac{y_{2}+y_{3}}{2}=1 \)

\( \Rightarrow y_{2}+y_{3}=2 \)......(a)
Similarly,

\( E \) is the mid-point of \( A C \).
\( \frac{x_{3}+x_{1}}{2}=2 \)

\( \Rightarrow x_{3}+x_{1}=4 \).......(ii)

\( \frac{y_{3}+y_{1}}{2}=-3 \)

\( \Rightarrow y_{3}+y_{1}=-6 \).......(b)
\( \mathrm{F} \) is the mid-point of \( \mathrm{AB} \).
\( \frac{x_{2}+x_{1}}{2}=3 \)

\( \Rightarrow x_{2}+x_{1}=6 \)........(iii)

\( \frac{y_{2}+y_{1}}{2}=4 \)

\( \Rightarrow y_{2}+y_{1}=8 \).......(c)

Adding (i), (ii) and (iii), we get,

\( 2\left(x_{1}+x_{2}+x_{3}\right)=12 \)

\( \Rightarrow x_{1}+x_{2}+x_{3}=6 \)......(iv)

Subtracting (i), (ii) and (iii) from (iv), we get,

\( x_{1}=4, x_{2}=2, x_{3}=0 \)
Similarly, 
Adding (a), (b) and (c), we get,

\( 2\left(y_{1}+y_{2}+y_{3}\right)=4 \)

\( \Rightarrow y_{1}+y_{2}+y_{3}=2 \).......(d)
Subtracting (a),(b) \) and (c) from (d), we get,

\( y_{1}=0 \)
\( y_{2}=8 \)
\( y_{3}=-6 \)
Therefore, vertices of \( \Delta \mathrm{ABC} \) are \( \mathrm{A}(4,0), \mathrm{B}(2,8), \mathrm{C}(0,-6) \)

The vertices of the triangle are $(4,0), (2,8)$ and $(0,-6)$.

Tutorialspoint

Updated on: 10-Oct-2022

26 Views

Kickstart Your Career

Get certified by completing the course

Get Started