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**Unformatted text preview: **5π
3 θ=
x 5π
4 x Q
P √
P has rectangular coordinates (2, −2 3)
π
P has polar coordinates 4, 53 Q has rectangular coordinates (−3, −3)
√π
Q has polar coordinates 3 2, 54 3. The point R(0, −3) lies along the negative y -axis. While we could go through the usual
computations4 to ﬁnd the polar form of R, in this case we can ﬁnd the polar coordinates of R
using the deﬁnition. Since the pole is identiﬁed with the origin, we can easily tell the point R
is 3 units from the pole, which means in the polar representation (r, θ) of R we know r = ±3.
π
Since we require r ≥ 0, we choose r = 3. Concerning θ, the angle θ = 32 satisﬁes 0 ≤ θ < 2π
3π
with its terminal side along the negative y -axis, so our answer is 3, 2 . To check, we note
π
π
x = r cos(θ) = 3 cos 32 = (3)(0) = 0 and y = r sin(θ) = 3 sin 32 = 3(−1) = −3.
4. The point S (−3, 4) lies in Quadrant II. With x = −3 and y = 4, we get r2 = (−3)2 +(4)2 = 25
so r = ±5. As usual, we choose r = 5 ≥ 0 and proceed to determine θ. We have tan(θ) =
4 Since x = 0, we would have to determine θ ge...

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