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Question
Show that 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`
= `sqrt((2 + 2)^2 + (0)^2`
= `sqrt((4)^2`
= 4 .....(i)
QR = `sqrt((2 - 2)^2 + (7 - 2)^2`
= `sqrt((0)^2 + (5)^2`
= `sqrt((5)^2`
= 5 ......(ii)
PR = `sqrt([2 -(-2)]^2 + (7 - 2)^2`
= `sqrt((2 + 2)^2 + (5)^2`
= `sqrt((4)^2 + (5)^2`
= `sqrt(16 + 25)`
= `sqrt(41)`
Now, PR2 = `(sqrt(41))^2` = 41 ......(iii)
Consider, PQ2 + QR2
= 42 + 52
= 16 + 25
= 41 ......[From (i) and (ii)]
∴ PR2 = PQ2 + QR2 ......[From (iii)]
∴ ∆PQR is a right angled triangle. ......[Converse of Pythagoras theorem]
∴ Points P, Q, and R are the vertices of a right angled triangle.
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