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Question
Verify that the points A(–2, 2), B(2, 2) and C(2, 7) are the vertices of a right-angled triangle.
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Solution
In a right angles triangle ABC, right-angled at B, according to the Pythagoras theorem
AB2 + BC2 = AC2
According to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by
`d = root(2)((a - c)^2 + (b - d)^2`....(1)
For the given points Distance between P and Q is
PQ = `sqrt((-2-2)^2 + (2 - 2)^2) = sqrt(16)`
QR = `sqrt((2-2)^2 + (7 - 2)^2) = sqrt(25)`
PR = `sqrt((-2-2)^2 + (2 - 7)^2) = sqrt(16 + 25) = sqrt(41)`
PQ2 = 16
QR2 = 25
PR2 = 41
As PQ2 + QR2 = PR2
Hence the given points form a right-angled triangle.
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