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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Updated on: 10-Oct-2022

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