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Question
Prove that the points (7 , 10) , (-2 , 5) and (3 , -4) are vertices of an isosceles right angled triangle.
Solution
AB = `sqrt ((7 + 2)^2 + (10 - 5)^2) = sqrt (81 + 25) = sqrt 106` units
BC = `sqrt ((-2-3)^2 + (5 + 4)^2) = sqrt (25 + 81) = sqrt 106` units
AC = `sqrt ((7 - 3)^2 + (10 + 4)^2) = sqrt (16 + 196) = sqrt 212` units
∵ AB = BC
∴ ABC is an isosceleles triangle
AB2 + BC2 = 100 + 106 = 212
AC2 = 212
∵ AB2 + BC2 = AC2
∴ ABC is also a right angled triangle.
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