Answer
$d(A,B)=4$
$d(B,C)=2$
$d(A,C)=2\sqrt{5}$
Since $(4)^{2}+(2)^{2}=(2\sqrt{5})^{2}$,
the triangle is a right triangle, hypotenuse is $\overline{AC}.$
Area =$4$ 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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$A=(4,-3),B=(4,1),C=(2,1)$
$\begin{aligned} d(A, B) &=\sqrt{(4-4)^{2}+(1-(-3))^{2}} \\ &=\sqrt{0^{2}+4^{2}} \\ &=\sqrt{0+16} \\ &=\sqrt{16} \\ &=4 \end{aligned}$
$\begin{aligned}d(B,C)&=\sqrt{(2-4)^{2}+(1-1)^{2}}\\&=\sqrt{(-2)^{2}+0^{2}}=\sqrt{4+0}\\&=\sqrt{4}\\&=2\end{aligned}$
$\begin{aligned}d(A,C)&=\sqrt{(2-4)^{2}+(1-(-3))^{2}}\\&=\sqrt{(-2)^{2}+4^{2}}=\sqrt{4+16}\\&=\sqrt{20}\\&=2\sqrt{5}\end{aligned}$
Since
$a^{2}+b^{2}=(4)^{2}+(2)^{2}=20$
$c^{2}=(2\sqrt{5})^{2}=20$
the triangle is a right triangle, hypotenuse is $\overline{AC}.$
Area = $\displaystyle \frac{1}{2}ab=\frac{1}{2}(4)(2)=4$
