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  • Show that the triangle ABC with vertices A (0, 4, 1), B (2, 3,minus 1) and C (4, 5, 0) is right angled

Show that the triangle ABC with vertices A (0, 4, 1), B (2, 3, – 1) and C (4, 5, 0) is right angled

Answers (1)

Vertices of ?ABC are  –

A(0,4,1), 5(2,3,-1), C(4,5,0)

Thus,

AB = \sqrt{(0-2)^{2}+(4-3)^{2}+(1+1)^{2} }

                  = \sqrt{4+1+1 } = 3

BC = \sqrt{(2-4)^{2}+(3-5)^{2}+(-1-0)^{2} }

                  = \sqrt{4+4+1 } = 3

AC = \sqrt{(0-4)^{2}+(4-5)^{2}+(1-0)^{2} }

                  = \sqrt{16+1+1 } = \sqrt{18}

Thus, it is clear that,

AC2 = AB2 + BC2,

Hence it is a right angled triangle.

Posted by

infoexpert22

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