Quadrilateral 2

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

Result

x =  2

Solution:

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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Tips to related online calculators
For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc.
Do you want to convert length units?
Pythagorean theorem is the base for the right triangle calculator.
See also our trigonometric triangle calculator.

 

 

 

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