Answer
(a) The triangle with vertices $A(6,-7)$, $B(11,-3)$ and $C(2,-2)$ is a right triangle by using the converse of the Pythagorean Theorem.
(b) The triangle with vertices $A(6,-7)$, $B(11,-3)$ and $C(2,-2)$ is a right triangle.
(c) Area = 20.5 square units
Work Step by Step
(a) As we are given the triangle with vertices $A(6,-7)$, $B(11,-3)$ and $C(2,-2)$.
Consider AB, BC and AC are three sides of a triangle.
Use distance formula $d=\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
For side AB:
$AB=\sqrt {(11-6)^{2}+(-3-(-7))^{2}}=\sqrt {16+25}=\sqrt 41$
For side BC:
$BC=\sqrt {(2-11)^{2}+(-2-(-3))^{2}}=\sqrt {81+1}=\sqrt {82}$
For side AC:
$AC=\sqrt {(2-6)^{2}+(-2-(-7))^{2}}=\sqrt {25+16}=\sqrt {41}$
For a triangle using the converse of the Pythagorean Theorem it must satisfy the condition $BC^{2}=AB^{2}+AC^{2}$
$(\sqrt 41)^{2}+(\sqrt 41)^{2}=(\sqrt 82)^{2}$
$82=82$
Hence, the result is proved.
(b) For a triangle to be right angle triangle, the corresponding sides must be perpendicular.
In order to find this use slope formula $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
For side AB:
$m=\frac{-3-(-7)}{11-6}=\frac{4}{5}$
For side BC:
$m=\frac{-2-(-3)}{2-11}=-\frac{1}{9}$
For side CA:
$m=\frac{-2-(-7)}{2-6}=-\frac{5}{4}$
Thus, product of slopes of AB and AC is -1 this implies that the corresponding sides are perpendicular.
Hence, the triangle is a right triangle.
(c) Area of a triangle$=\frac{1}{2}(AB)(BC)$
$=\frac{1}{2}(\sqrt {41})(\sqrt {41})$
Hence, Area=20.5 square units