The vertices of ΔABC are (−2, 1), (5, 4) and (2, −3) respectively.
Question:

The vertices of ΔABC are (−2, 1), (5, 4)  and (2, −3)  respectively. Find the area of the triangle and the length of the altitude through A.

Solution:

GIVEN: The vertices of triangle ABC are A (−2, 1) and B (5, 4) and C (2, −3)

TO FIND: The area of triangle ABC and length if the altitude through A

PROOF: We know area of triangle formed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by $\Delta=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

Now Area of ΔABC

Taking three points A (−2, 1) and B (5, 4) and C(2, −3)

$\operatorname{Area}(\triangle A B C)=\frac{1}{2}|\{-8-15+2\}-\{5+8+6\}|$

$=\frac{1}{2}|\{-21\}-\{19\}|$

$=\frac{1}{2}|\{-40\}|$

$=\frac{1}{2}(40)$

$=20$

We have

$B C=\sqrt{(5-2)^{2}+(4+3)^{2}}$

$B C=\sqrt{(3)^{2}+(7)^{2}}$

$B C=\sqrt{9+49}$

$B C=\sqrt{58}$

Now,

Area $(\triangle A B C)=\frac{1}{2} \times B C \times$ length of altitude through $\mathrm{A}$

$20=\frac{1}{2} \times \sqrt{58} \times$ length of altitude through A

length of altitude through $\mathrm{A}=\frac{40}{\sqrt{58}}$

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