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Question
Verify the following:
(0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right angled triangle.
Solution
Let (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) be denoted by A, B, and C respectively.
AB = `sqrt((–1 – 0)^2 + (6 – 7)^2 + (6 – 10)^2)`
= `sqrt((–1)^2 + (-1)^2 + (-4)^2)`
= `sqrt(1 + 1 + 16)`
= `sqrt18`
= `3sqrt2`
BC = `sqrt((–4 + 1)^2 + (9 – 6)^2 + (6 – 6)^2)`
= `sqrt((–3)^2 + (3)^2 + (0)^2)`
= `sqrt(9 + 9)`
= `sqrt18`
= `3sqrt2`
CA = `sqrt((0 + 4)^2 + (9 – 7)^2 + (6 – 10)^2)`
= `sqrt((4)^2 + (-2)^2 + (4)^2)`
= `sqrt(16 + 4 + 16)`
= `sqrt36`
= 6
AB2 + BC2 = 18 + 18 = 36
Now, AC2 = 36
∴ AB2 + BC2 = AC2
Hence, the given vertices are of right angled triangle.
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