#### Answer

$ a.\quad$ Please see "step-by-step"
$ b.\quad$ Area = $10$

#### Work Step by Step

a.
Reading the ccoordinates from the graph, we have
$A=(2,2), B=(3, -1)$, and $C=(-3, -3)$.
Calculate the sides' lengths:
$d(A, B)=\sqrt{(3-2)^{2}+(-1-2)^{2}}=\sqrt{1^{2}+(-3)^{2}}=\sqrt{1+9}=\sqrt{10}.$
$d(C, B)=\sqrt{(3-(-3))^{2}+(-1-(-3))^{2}}=\sqrt{6^{2}+2^{2}}=\sqrt{36+4}\\=\sqrt{40}=2\sqrt{10}.$
$d(A, C)=\sqrt{(-3-2)^{2}+(-3-2)^{2}}=\sqrt{(-5)^{2}+(-5)^{2}}\\=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$.
Note that
$[d(A, B)]^{2}+[d(C, B)]^{2}=10+4\cdot 10=50$
$[d(A, C)]^{2}=25\cdot 2=50, $
so,
$[d(A, B)]^{2}+[d(C, B)]^{2}=[d(A, C)]^{2}$.
By the converse of the Pythagorean Theorem, we conclude that the triangle is a right triangle.
b.
The area of the triangle is
$\displaystyle \frac{1}{2}\cdot d(C, B)\cdot d(A, B)=\frac{1}{2}\cdot\sqrt{10}\cdot 2\sqrt{10}=10$.