Quadrilateral 2

Show that the quadrilateral with vertices P1(0,1), P2(4,2) P3(3,6) P4(-5,4) has two right triangles.

Correct result:

x = 2

Solution:

x_{1} = 0 y_{1} = 1 x_{2} = 4 y_{2} = 2 x_{3} = 3 y_{3} = 6 x_{4} = - 5 y_{4} = 4 a = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}} = \sqrt{(0 - 4)^{2} + (1 - 2)^{2}} = \sqrt{17} ≐ 4.1231 b = \sqrt{(x_{2} - x_{3})^{2} + (y_{2} - y_{3})^{2}} = \sqrt{(4 - 3)^{2} + (2 - 6)^{2}} = \sqrt{17} ≐ 4.1231 c = \sqrt{(x_{3} - x_{4})^{2} + (y_{3} - y_{4})^{2}} = \sqrt{(3 - (- 5))^{2} + (6 - 4)^{2}} = 2 \sqrt{17} ≐ 8.2462 d = \sqrt{(x_{4} - x_{1})^{2} + (y_{4} - y_{1})^{2}} = \sqrt{((- 5) - 0)^{2} + (4 - 1)^{2}} = \sqrt{34} ≐ 5.831 u_{1} = \sqrt{(x_{1} - x_{3})^{2} + (y_{1} - y_{3})^{2}} = \sqrt{(0 - 3)^{2} + (1 - 6)^{2}} = \sqrt{34} ≐ 5.831 u_{2} = \sqrt{(x_{2} - x_{4})^{2} + (y_{2} - y_{4})^{2}} = \sqrt{(4 - (- 5))^{2} + (2 - 4)^{2}} = \sqrt{85} ≐ 9.2195 t_{1} = u_{1}^{2} - a^{2} - b^{2} = 5.83 1^{2} - 4.123 1^{2} - 4.123 1^{2} ≐ 1.6222 \cdot 1 0^{- 11} t_{2} = u_{2}^{2} - b^{2} - c^{2} = 9.219 5^{2} - 4.123 1^{2} - 8.246 2^{2} = - 0 t_{3} = u_{1}^{2} - c^{2} - d^{2} = 5.83 1^{2} - 8.246 2^{2} - 5.83 1^{2} = - 68 t_{4} = u_{2}^{2} - d^{2} - a^{2} = 9.219 5^{2} - 5.83 1^{2} - 4.123 1^{2} = 34 t_{1} = 0 = > P_{1} P_{2} P_{3} = 9 0^{\circ} t_{2} = 0 = > P_{2} P_{3} P_{4} = 9 0^{\circ} x = 2 angles = 2

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