Question

Given that P lies on the unit circle with an x-coordinate of -sqrt(13)/4 and is below the x-axis, find P(x, y).

Answer 1

The coordinates of point P on the unit circle with an x-coordinate of −√13/4 and below the x-axis are P(-√13/4, -√3/4).

Step-by-step explanation:

The point P lies on the unit circle where the x-coordinate is given as −√13/4. Since point P is on the unit circle, the sum of the squares of the x-coordinate and y-coordinate should equal 1 according to the unit circle equation x² + y² = 1. We know that P is below the x-axis, meaning the y-coordinate must be negative. Inserting the value for x, we get:

(−√13/4)² + y² = 1

13/16 + y² = 1

y² = 1 − 13/16

y² = (16/16) − (13/16)

y² = 3/16

y = ±√(3/16)

Since P is below the x-axis, we choose the negative square root for y:

y = −√(3/16)

y = −√3/4

Therefore, the coordinates of P are P(-√13/4, -√3/4).