Question

Verify that points P(–2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle.

Answer 1

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.