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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.
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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