Answer
The triangle is a right triangle
with hypotenuse $c=2\sqrt{13}$, and legs $a=6,b=4$.
Work Step by Step
Use the distance formula
$d(P_{1},P_{2})=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
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$P_{1}=(2,1);\quad P_{2}=(-4,1)$
\begin{aligned}
d\left(P_{1},P_{2}\right)&=\sqrt{(-4-2)^{2}+(1-1)^{2}}\\
&=\sqrt{(-6)^{2}+0^{2}}\\&=\sqrt{36}\\&=6\end{aligned}
$P_{1}=(2,1);\quad P_{3}=(-4,-3)$
\begin{aligned}
d\left(P_{1},P_{3}\right)&=\sqrt{(-4-2)^{2}+(-3-1)^{2}}\\
&=\sqrt{(-6)^{2}+(-4)^{2}}\\&=\sqrt{36+16}\\&=\sqrt{52}\\&=2\sqrt{13}\end{aligned}
$P_{2}=(-4,1);\quad P_{3}=(-4,-3)$
\begin{aligned}
d\left(P_{2},P_{3}\right)&=\sqrt{(-4-(-4))^{2}+(-3-1)^{2}}\\
&=\sqrt{0^{2}+(-4)^{2}}\\&=\sqrt{16}\\&=4\end{aligned}
Since all three sides have different sizes,
$2\sqrt{13}=\sqrt{52}$ is the longest, name it c and check whether $a^{2}+b^{2}=c^{2}$
$a=6,b=4,c=2\sqrt{13},$
$a^{2}+b^{2}=6^{2}+4^{2}=36+16=52$
$c^{2}=(\sqrt{52})^{2}=52$
The triangle is a right triangle
with hypotenuse $c=2\sqrt{13}$, and legs $a=6,b=4$.
