Triangle PQR has vertices P(0, 1), Q(0, -4), and R(2, 5). Find the coordinates of R' to the nearest hundredth after rotating triangle PQR counterclockwise about the origin 45º.

Answer 1

the radius to (2,5) is at an angle θ such that tanθ = 5/2

So
cosθ = 2/√29
sinθ = 5/√29

R' is at an angle (θ+45), so
x = √29 cos(θ+45)
y = √29 sin(θ+45)

x = √29 (cosθ cos45 - sinθ sin45)
= √29 (2/√29 * 1/√2 - 5/√29 * 1/√2)
= 2/√2 - 5/√2
= -3/√2

y = √29 (sinθ cos45 + cosθ sin45)
= √29 (5/√29 * 1/√2 + 2/√29 * 1/√2)
= 7/√2

So, R' = (-3/√2, 7/√2)

Answer 2

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