If the point $P\left(-\frac{4}{5}, y\right)$ lies on the unit circle and $P$ is in the third quadrant, what does $y$ equal?
A. $\frac{5}{3}$
B. $\frac{3}{5}$
C. $-\frac{3}{4}$
D. $-\frac{3}{5}$
Answer (2)
Substitute the given x-coordinate into the unit circle equation.
Solve for y 2 .
Take the square root to find the possible values of y .
Choose the negative value for y since the point is in the third quadrant. The final answer is − 5 3 .
Explanation
Problem Analysis The problem states that the point P ( − 5 4 , y ) lies on the unit circle and is in the third quadrant. We need to find the value of y .
Equation of the Unit Circle The equation of the unit circle is x 2 + y 2 = 1 . Since the point P ( − 5 4 , y ) lies on the unit circle, we can substitute the x -coordinate into the equation to find the y -coordinate.
Substitution Substitute x = − 5 4 into the equation x 2 + y 2 = 1 :
( − 5 4 ) 2 + y 2 = 1 25 16 + y 2 = 1
Solving for y^2 Solve for y 2 :
y 2 = 1 − 25 16 y 2 = 25 25 − 25 16 y 2 = 25 9
Solving for y Solve for y :
y = ± 25 9 y = ± 5 3
Determining the Sign of y Since the point P is in the third quadrant, both its x and y coordinates must be negative. Therefore, we choose the negative value for y :
y = − 5 3
Final Answer The value of y is − 5 3 .
Examples
Understanding the unit circle is crucial in various fields like physics and engineering. For instance, when analyzing simple harmonic motion, such as the motion of a pendulum, the coordinates of a point on the unit circle can represent the position of the pendulum at a given time. The x and y coordinates relate to the cosine and sine of the angle, which helps in determining the pendulum's displacement and velocity. This allows engineers to design and optimize systems involving oscillatory motion.
The value of y for the point P ( − 5 4 , y ) on the unit circle in the third quadrant is − 5 3 . Thus, the correct answer is D. − 5 3 .
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