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If $ \mathrm{D}\left(\frac{-1}{2}, \frac{5}{2}\right), \mathrm{E}(7,3) $ and $ \mathrm{F}\left(\frac{7}{2}, \frac{7}{2}\right) $ are the midpoints of sides of $ \triangle \mathrm{ABC} $, find the area of the $ \triangle \mathrm{ABC} $.
Given:
$D (\frac{−1}{2}, \frac{5}{2}), E (7, 3)$ and $F (\frac{7}{2}, \frac{7}{2})$ are the mid-points of sides of $\triangle ABC$.
To do:
We have to find the area of $\triangle ABC$.
Solution:
Let \( A=\left(x_{1}, y_{1}\right), B=\left(x_{2}, y_{2}\right) \) and \( C=\left(x_{3}, y_{3}\right) \) are the vertices of the \( \Delta A B C \).
The mid-point of a line segment having points \( (x_{1}, y_{1}) \) and \( (x_{2}, y_{2}) \) is \( (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}) \)
\( D\left(-\frac{1}{2}, \frac{5}{2}\right) \) is the mid-point of \( \mathrm{BC} \).
\( \frac{x_{2}+x_{3}}{2}=-\frac{1}{2} \)
\( \Rightarrow x_{2}+x_{3}=-1 \)......(i)
\( \frac{y_{2}+y_{3}}{2}=\frac{5}{2} \)
\( \Rightarrow y_{2}+y_{3}=5 \).......(a)
Similarly,
\( \mathrm{E}(7,3) \) is the mid-point of \( \mathrm{CA} \).
\( \frac{x_{3}+x_{1}}{2}=7 \)
\( \Rightarrow x_{3}+x_{1}=14 \).......(ii)
\( \frac{y_{3}+y_{1}}{2}=3 \)
\( \Rightarrow y_{3}+y_{1}=6 \).......(b)
\( \mathrm{F}\left(\frac{7}{2}, \frac{7}{2}\right) \) is the mid-point of \( \mathrm{AB} \).
\( \frac{x_{1}+x_{2}}{2}=\frac{7}{2} \)
\( \Rightarrow x_{1}+x_{2}=7 \)......(iii)
\( \frac{y_{1}+y_{2}}{2}=\frac{7}{2} \)
\( \Rightarrow y_{1}+y_{2}=7 \).......(c)
On adding equations (i), (ii) and (iii), we get,
\( 2\left(x_{1}+x_{2}+x_{3}\right)=20 \)
\( \Rightarrow x_{1}+x_{2}+x_{3}=10 \).......(iv)
On subtracting (i), (ii) and (iii) from (iv) respectively, we get,
\( x_{1}=11, x_{2}=-4, x_{3}=3 \)
On adding equations (a), (b) and (c), we get,
\( 2\left(y_{1}+y_{2}+y_{3}\right)=18 \)
\( \Rightarrow y_{1}+y_{2}+y_{3}=9 \)......(d)
On subtracting equations (a), (b) and (c) from (d) respectively, we get,
\( y_{1}=4, y_{2}=3, y_{3}=2 \)
Hence, the vertices of \( \Delta \mathrm{ABC} \) are \( \mathrm{A}(11,4) \) \( \mathrm{B}(-4,3) \text { and } \mathrm{C}(3,2) \)
The area of \( \Delta \mathrm{ABC}=\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})] \)
\( \Delta=\frac{1}{2}[11(3-2)+(-4)(2-4)+3(4-3)] \)
\( =\frac{1}{2}[11 \times 1+(-4)(-2)+3(1)] \)
\( =\frac{1}{2}(11+8+3) \)
\( =\frac{22}{2} \)
\( =11 \)
The area of \( \Delta \mathrm{ABC} \) is \( 11 \) sq.units.
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