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Question
By using the distance formula prove that each of the following sets of points are the vertices of a right angled triangle.
(i) (6, 2), (3, -1) and (- 2, 4)
(ii) (-2, 2), (8, -2) and (-4, -3).
Solution
Let A(6, 2), B(3, -1) and C(-2, 4) be the given points
AB = `sqrt((6 - 3)^2 + (2 + 1)^2)`
= `sqrt(9 + 9) = sqrt(18) = 3sqrt(2)"units"`.
BC = `sqrt((3 + 2)^2 + (-1 -4)^2)`
= `sqrt(25 + 25) = 5sqrt(2)"units"`.
AC = `sqrt((6 + 2)^2 + (2 - 4)^2)`
= `sqrt(64 + 4)`
= `sqrt(68)"units"`.
Now AB2 + BC2 = `(3sqrt(2))^2 + (5sqrt(2))^2`
= 18 + 50 = 68 = AC2
AB2 + BC2 = AC2
Hence, the triangle is right angled at A.
Hence proved.
(ii) Let A (-2, 2), B(8, -2) and C(-4, -3) be the given points
AB = `sqrt((8 + 2)^2 + (-2 -2)^2)`
= `sqrt(116)"units"`.
BC = `sqrt((8 + 4)^2 + (-2 + 3)^2)`
= `sqrt(145)"units"`.
AC = `sqrt((-2 + 4)^2 + (2 + 3)^2)`
= `sqrt(29)"units"`.
Now AB2 + AC2
= 116 + 29
= 145
= BC2
Since AB + AC2 = BC2
Hence, the triangle is right angled at A.
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