Name plane K using three noncollinear points.
1 answer:
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Answer:
A) (5, 7π/4): (5, 15π/4) and (-5, 3π/4)
B) (−6, π/2): (−6, 5π/2) and (6, -π/2)
C) (5, −2): (5, 2π-2) and (-5, -π-2)
Step-by-step explanation:
To find pair of coordinates for (r>0), add 2π to corresponding θ. For (r<0) subtract π from given angle to find second pair of coordinate.
A) (5, 7π/4)
For (r>0)
θ = 7π/4 + 2π
θ = 15π/4
Point is (5, 15π/4)
For (r<0)
θ = 7π/4 - π
θ = 3π/4
Point is (-5, 3π/4)
As it can be seen in Fig 1
(5, 7π/4)=(5, 15π/4)=(-5, 3π/4)
B) (−6, π/2)
For (r>0)
θ = π/2 + 2π
θ = 5π/2
Point is (6, 5π/2)
For (r<0)
θ = π/2 - π
θ = -π/2
Point is (-6, -π/2)
As shown in fig 2
(-6, π/2) = (6, 5π/2) = (-6, -π/2)
C) (5, −2)
For (r>0)
θ = -2 + 2π
Point is (5, 2π-2)
For (r<0)
θ = -2 - π
Point is (-5, -π)
As shown in Fig. 3
(5, −2) = (5, 2π-2) = (-5, -π)
