Name the type of triangle PQR formed by the p | vedclass

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(A) Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$PQ = \sqrt{(-\sqrt{2} - \sqrt{2})^2 + (-\sqrt{2} - \sqrt{2})^2} = \sqrt{(-2

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Equilateral triangle

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(A) Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$PQ = \sqrt{(-\sqrt{2} - \sqrt{2})^2 + (-\sqrt{2} - \sqrt{2})^2} = \sqrt{(-2\sqrt{2})^2 + (-2\sqrt{2})^2} = \sqrt{8 + 8} = \sqrt{16} = 4$$PR = \sqrt{(-\sqrt{6} - \sqrt{2})^2 + (\sqrt{6} - \sqrt{2})^2} = \sqrt{(6 + 2 + 2\sqrt{12}) + (6 + 2 - 2\sqrt{12})} = \sqrt{8 + 8} = \sqrt{16} = 4$$RQ = \sqrt{(-\sqrt{6} - (-\sqrt{2}))^2 + (\sqrt{6} - (-\sqrt{2}))^2} = \sqrt{(-\sqrt{6} + \sqrt{2})^2 + (\sqrt{6} + \sqrt{2})^2} = \sqrt{(6 + 2 - 2\sqrt{12}) + (6 + 2 + 2\sqrt{12})} = \sqrt{8 + 8} = \sqrt{16} = 4$Since $PQ = PR = RQ = 4$,all three sides are equal. Therefore,the triangle $PQR$ is an equilateral triangle.

Name the type of triangle $PQR$ formed by the points $P(\sqrt{2}, \sqrt{2})$,$Q(-\sqrt{2}, -\sqrt{2})$,and $R(-\sqrt{6}, \sqrt{6})$.

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