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Question
Verify the following:
(0, 7, –10), (1, 6, –6) and (4, 9, –6) are the vertices of an isosceles triangle.
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Solution
Let the vertices of triangle ABC be A(0, 7, –10), B(1, 6, –6) and C(4, 9, –6).
Now, 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`
= `sqrt(9 +9)`
= `sqrt18`
= `3sqrt2`
CA = `sqrt((0 - 4)^2 + (7 - 9)^2 + (-10 + 6)^2)`
= `sqrt((-4)^2 + (-2)^2 + (-4)^2)`
= `sqrt(16 + 4 + 16)`
= `sqrt36`
= 6
Here, AB = BC ≠ CA
Hence, the given vertices AB = BC are of the isosceles triangle.
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