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Question
Verify that points P(–2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle.
Solution
Distance between two points = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`
By distance formula,
`"PQ" = sqrt([2 - (-2)]^2 + (2 - 2)^2)`
`"PQ" = sqrt((2 + 2)^2 + (0)^2)`
`"PQ" = sqrt((4)^2)`
PQ = 4 ...(i)
`"QR" = sqrt((2 - 2)^2 + (7 - 2)^2)`
`"QR" = sqrt((0)^2 + (5)^2)`
`"QR" = sqrt((5)^2)`
QR = 5 ...(ii)
`"PR" = sqrt([2 - (-2)]^2 + (7 - 2)^2)`
`"PR" = sqrt((2 + 2)^2 + (5)^2)`
`"PR" = sqrt((4)^2 + (5)^2)`
`"PR" = sqrt(16 + 25)`
`"PR" = sqrt(41)`
Now, PR2 = `(sqrt(41))^2` = 41 ...(iii)
From (i) and (ii),
∴ PQ2 + QR2 = 42 + 52 = 16 + 25 = 41
∴ PR2 = PQ2 + QR2 ...[From (iii)]
∴ ΔPQR is a right angled triangle. ...[Converse of Pythagoras theorem]
∴ Point P, Q, and R are the vertices of a right angled triangle.
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