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Question
Show that the points (−2, 3), (8, 3) and (6, 7) are the vertices of a right triangle ?
Solution
The given points are A (−2, 3), B (8, 3) and C (6, 7).
By, distance formula, AB `sqrt{{8-(-2)}^2+(3-3)^2`
`rArr AB^2=10^2+0`
`rArrBC^2=100`
`BC=sqrt((6-8)^2+(7-3)^2)`
`rArrBC^2=(-2)^2+4^2`
`rArrBC^2=4+16`
`rArrBC^2=20`
`BC=sqrt((-2-6)^2+(3-7)^2)`
`rArrCA^2=(-8)^2+(-4)^2`
`rArrCA^2=64+16`
`rArrCA^2=80`
It is observed that BC2 + CA2 = 20 + 80 = 100 = AB2
So, by the converse of Pythagoras Theorem, ΔABC is right angled at C.
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