Quadrilateral 2

Show that the quadrilateral with vertices A(0,1), B(4,2), C(3,6) D(-5,4) has two right triangles.

Final Answer:

x = 2

A = (0, 1) B = (4, 2) C = (3, 6) D = (- 5, 4) a = (A_{x} - B_{x})^{2} + (A_{y} - B_{y})^{2} = (0 - 4)^{2} + (1 - 2)^{2} = 17 ≐ 4.1231 b = (B_{x} - C_{x})^{2} + (B_{y} - C_{y})^{2} = (4 - 3)^{2} + (2 - 6)^{2} = 17 ≐ 4.1231 c = (C_{x} - D_{x})^{2} + (C_{y} - D_{y})^{2} = (3 - (- 5))^{2} + (6 - 4)^{2} = 217 ≐ 8.2462 d = (D_{x} - A_{x})^{2} + (D_{y} - A_{y})^{2} = ((- 5) - 0)^{2} + (4 - 1)^{2} = 34 ≐ 5.831 u_{1} = (A_{x} - C_{x})^{2} + (A_{y} - C_{y})^{2} = (0 - 3)^{2} + (1 - 6)^{2} = 34 ≐ 5.831 u_{2} = (B_{x} - D_{x})^{2} + (B_{y} - D_{y})^{2} = (4 - (- 5))^{2} + (2 - 4)^{2} = 85 ≐ 9.2195 t_{1} = u_{1}^{2} - a^{2} - b^{2} = 5.83 1^{2} - 4.123 1^{2} - 4.123 1^{2} = 0 t_{2} = u_{2}^{2} - b^{2} - c^{2} = 9.219 5^{2} - 4.123 1^{2} - 8.246 2^{2} = 2.558 \cdot 1 0^{- 13} 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} = 90 ° t_{2} = 0 = > P_{2} P_{3} P_{4} = 90 ° x = 2 angles = 2

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