Vertices of a right triangle

Show that the points D(2,1), E(4,0), F(5,7) are vertices of a right triangle.

Result

d =  0

Solution:

x1=2 y1=1 x2=4 y2=0 x3=5 y3=7  a=(x1−x2)2+(y1−y2)2=(2−4)2+(1−0)2≐5≐2.2361  b=(x1−x3)2+(y1−y3)2=(2−5)2+(1−7)2≐3 5≐6.7082  c=(x2−x3)2+(y2−y3)2=(4−5)2+(0−7)2≐5 2≐7.0711  c2=a2+b2? d=c2−a2−b2=7.07112−2.23612−6.70822=−0=0x_{1}=2 \ \\ y_{1}=1 \ \\ x_{2}=4 \ \\ y_{2}=0 \ \\ x_{3}=5 \ \\ y_{3}=7 \ \\ \ \\ a=\sqrt{ (x_{1}-x_{2})^2+(y_{1}-y_{2})^2 }=\sqrt{ (2-4)^2+(1-0)^2 } \doteq \sqrt{ 5 } \doteq 2.2361 \ \\ \ \\ b=\sqrt{ (x_{1}-x_{3})^2+(y_{1}-y_{3})^2 }=\sqrt{ (2-5)^2+(1-7)^2 } \doteq 3 \ \sqrt{ 5 } \doteq 6.7082 \ \\ \ \\ c=\sqrt{ (x_{2}-x_{3})^2+(y_{2}-y_{3})^2 }=\sqrt{ (4-5)^2+(0-7)^2 } \doteq 5 \ \sqrt{ 2 } \doteq 7.0711 \ \\ \ \\ c^2=a^2 + b^2 ? \ \\ d=c^2-a^2-b^2=7.0711^2-2.2361^2-6.7082^2=-0=0

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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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