# Appendix B - Coordinate Geometry and Lines - B Exercises - Page A15: 12

(a) The triangle with vertices $A(6,-7)$, $B(11,-3)$ and $C(2,-2)$ is a 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

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