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Find the area of ΔABC whose vertices are A

Question:

Find the area of $\triangle \mathrm{ABC}$ whose vertices are $\mathrm{A}(-3,-5), \mathrm{B}(5,2)$ and $\mathrm{C}(-9,-3)$.

 

Solution:

Given: The vertices of the triangle are $A(-3,-5), B(5,2)$ and $C(-9,-3)$.

Formula: Area of $\triangle A B C=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

Here,

$x_{1}=-3, y_{1}=-5$

$x_{2}=5, y_{2}=2$

$x_{3}=-9, y_{3}=-3$

Putting the values,

Area of $\triangle \mathrm{ABC}=\frac{1}{2}[-3(2+3)+5(-3+5)-9(-5-2)]$

$=\frac{1}{2}[-15+10+63]$

$=29$ square units.

Therefore, the area of $\triangle \mathrm{ABC}$ is 29 square units.

 

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