Quadrilateral 2

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


x =  2


x1=0 y1=1 x2=4 y2=2 x3=3 y3=6 x4=5 y4=4 a=(x1x2)2+(y1y2)2=(04)2+(12)2174.1231 b=(x2x3)2+(y2y3)2=(43)2+(26)2174.1231 c=(x3x4)2+(y3y4)2=(3(5))2+(64)22 178.2462 d=(x4x1)2+(y4y1)2=((5)0)2+(41)2345.831 u1=(x1x3)2+(y1y3)2=(03)2+(16)2345.831 u2=(x2x4)2+(y2y4)2=(4(5))2+(24)2859.2195 t1=u12a2b2=5.83124.123124.12312=0 t2=u22b2c2=9.219524.123128.24622=0 t3=u12c2d2=5.83128.246225.8312=68 t4=u22d2a2=9.219525.83124.12312=34 t1=0=>P1P2P3=90 t2=0=>P2P3P4=90 x=2 angles=2x_{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 } \doteq \sqrt{ 17 } \doteq 4.1231 \ \\ b=\sqrt{ (x_{2}-x_{3})^2+(y_{2}-y_{3})^2 }=\sqrt{ (4-3)^2+(2-6)^2 } \doteq \sqrt{ 17 } \doteq 4.1231 \ \\ c=\sqrt{ (x_{3}-x_{4})^2+(y_{3}-y_{4})^2 }=\sqrt{ (3-(-5))^2+(6-4)^2 } \doteq 2 \ \sqrt{ 17 } \doteq 8.2462 \ \\ d=\sqrt{ (x_{4}-x_{1})^2+(y_{4}-y_{1})^2 }=\sqrt{ ((-5)-0)^2+(4-1)^2 } \doteq \sqrt{ 34 } \doteq 5.831 \ \\ u_{1}=\sqrt{ (x_{1}-x_{3})^2+(y_{1}-y_{3})^2 }=\sqrt{ (0-3)^2+(1-6)^2 } \doteq \sqrt{ 34 } \doteq 5.831 \ \\ u_{2}=\sqrt{ (x_{2}-x_{4})^2+(y_{2}-y_{4})^2 }=\sqrt{ (4-(-5))^2+(2-4)^2 } \doteq \sqrt{ 85 } \doteq 9.2195 \ \\ t_{1}=u_{1}^2 - a^2-b^2=5.831^2 - 4.1231^2-4.1231^2=0 \ \\ t_{2}=u_{2}^2 - b^2-c^2=9.2195^2 - 4.1231^2-8.2462^2=-0 \ \\ t_{3}=u_{1}^2 - c^2-d^2=5.831^2 - 8.2462^2-5.831^2=-68 \ \\ t_{4}=u_{2}^2 - d^2-a^2=9.2195^2 - 5.831^2-4.1231^2=34 \ \\ t_{1}=0=> P_{1}P_{2}P_{3}=90^\circ \ \\ t_{2}=0=> P_{2}P_{3}P_{4}=90^\circ \ \\ x=2 \ \text{angles}=2

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