Answer
$d(A,B)=4$
$d(B,C)=\sqrt{41}$
$d(A,C)=5$
Since $(4)^{2}+(5)^{2}=(\sqrt{41})^{2}$,
the triangle is a right triangle, hypotenuse is $\overline{BC}.$
Area =$10$ sq. unit
Work Step by Step
$d(P_{1},P_{2})=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
In a right triangle with legs a and b, hypotenuse c,
$ a^{2}+b^{2}=c^{2}\quad$ (Pythagorean Th.)
Area = $\displaystyle \frac{1}{2}ab$
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$\begin{aligned}d(A,B)&=\sqrt{(0-4)^{2}+(-3-(-3))^{2}}\\&=\sqrt{(-4)^{2}+0^{2}}\\&
=\sqrt{16+0}\\&=\sqrt{16}\\&=4\end{aligned}$
$\begin{aligned}d(B,C)&=\sqrt{(4-0)^{2}+(2-(-3))^{2}}\\&=\sqrt{4^{2}+5^{2}}\\&
=\sqrt{16+25}\\&=\sqrt{41}\end{aligned}$
$\begin{aligned}d(A,C)&=\sqrt{(4-4)^{2}+(2-(-3))^{2}}\\&=\sqrt{0^{2}+5^{2}}\\
&=\sqrt{0+25}\\&=\sqrt{25}\\&=5\end{aligned}$
Since
$a^{2}+b^{2}=(4)^{2}+(5)^{2}=41$
$c^{2}=(\sqrt{41})^{2}=41$
the triangle is a right triangle, hypotenuse is $\overline{BC}.$
Area = $\displaystyle \frac{1}{2}ab=\frac{1}{2}(4)(5)=10$
