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If the coordinates of the mid-points of the sides of a triangle are $(3, 4), (4, 6)$ and $(5, 7)$, find its vertices.
Given:
The coordinates of the mid-points of the sides of a triangle are $(3, 4), (4, 6)$ and $(5, 7)$.
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 $D(3,4), E(4,6)$ and $F(5,7)$ are the mid-points of $BC, CA$ and $AB$.
$D$ is the mid-point of $BC$.
This implies,
\( \frac{x_{2}+x_{3}}{2}=3 \)
\( \Rightarrow x_{2}+x_{3}=6 \).....(i)
\( \frac{y_{2}+y_{3}}{2}=4 \)
\( \Rightarrow y_{2}+y_{3}=8 \)......(a)
Similarly,
\( E \) is the mid-point of \( A C \).
\( \frac{x_{3}+x_{1}}{2}=4 \)
\( \Rightarrow x_{3}+x_{1}=8 \).......(ii)
\( \frac{y_{3}+y_{1}}{2}=6 \)
\( \Rightarrow y_{3}+y_{1}=12 \).......(b)
\( \mathrm{F} \) is the mid-point of \( \mathrm{AB} \).
\( \frac{x_{2}+x_{1}}{2}=5 \)
\( \Rightarrow x_{2}+x_{1}=10 \)........(iii)
\( \frac{y_{2}+y_{1}}{2}=7 \)
\( \Rightarrow y_{2}+y_{1}=14 \).......(c)
Adding (i), (ii) and (iii), we get,
\( 2\left(x_{1}+x_{2}+x_{3}\right)=24 \)
\( \Rightarrow x_{1}+x_{2}+x_{3}=12 \)......(iv)
Subtracting (i), (ii) and (iii) from (iv), we get,
\( x_{1}=6, x_{2}=4, x_{3}=2 \)
Similarly,
Adding (a), (b) and (c), we get,
\( 2\left(y_{1}+y_{2}+y_{3}\right)=34 \)
\( \Rightarrow y_{1}+y_{2}+y_{3}=17 \).......(d)
Subtracting (a), (b) and (c) from (d), we get,
\( y_{1}=9 \) \( y_{2}=5 \) \( y_{3}=3 \)
Therefore, vertices of \( \Delta \mathrm{ABC} \) are \( \mathrm{A}(6,9), \mathrm{B}(4,5), \mathrm{C}(2,3) \)
The vertices of the triangle are $(6,9), (4,5)$ and $(2,3)$.
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