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Show that the point is on the unit circle

a. (5/13, 12/13)
b. (√pi/6, 5/6)

User David Dean
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1 Answer

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Final answer:

To show that a point is on the unit circle, we need to check if its coordinates satisfy the equation x^2 + y^2 = 1. The point (5/13, 12/13) is on the unit circle, while (√pi/6, 5/6) is not.

Step-by-step explanation:

To show that a point is on the unit circle, we need to check if its coordinates satisfy the equation x^2 + y^2 = 1. Let's consider the given points:

(a) (5/13, 12/13)

Substituting the values into the equation, we get (5/13)^2 + (12/13)^2 = 25/169 + 144/169 = 169/169 = 1.

Therefore, the point (5/13, 12/13) is on the unit circle.

(b) (√pi/6, 5/6)

Substituting the values into the equation, we get (√pi/6)^2 + (5/6)^2 = pi/36 + 25/36 = (pi + 25)/36.

Since (pi + 25)/36 is not equal to 1, the point (√pi/6, 5/6) is not on the unit circle.

User Mehdi Bugnard
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7.7k points