Vertices of RT

Show that the points P1 (5,0), P2 (2,1) & P3 (4,7) are the vertices of a right triangle.

Correct result:

x =  0

Solution:

$$\begin{gathered} x_1=5 \\ {y}_1=0 \\ {x}_2=2 \\ {y}_2=1 \\ {x}_3=4 \\ {y}_3=7 \\ a=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}=\sqrt{(5-2{)}^2+(0-1{)}^2}=\sqrt{10}\doteq 3.1623 \\ b=\sqrt{(x_1-x_3{)}^2+(y_1-y_3{)}^2}=\sqrt{(5-4{)}^2+(0-7{)}^2}=5\sqrt{2}\doteq 7.0711 \\ c=\sqrt{(x_2-x_3{)}^2+(y_2-y_3{)}^2}=\sqrt{(2-4{)}^2+(1-7{)}^2}=2\sqrt{10}\doteq 6.3246 \\ x=b^2-(a^2+c^2)=7.0711^2-(3.1623^2+6.3246^2)=0 \end{gathered}$$

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