Final answer:
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).