A point (x, y) is on the unit circle. If its y-coordinate is 1/2 and the point lies in Quadrant I, what is the other coordinate?

Question

A point (x, y) is on the unit circle. If its y-coordinate is 1/2 and the point lies in Quadrant I, what is the other coordinate?

Express your answer as a fraction, NO DECIMALS!

Answer 1

Remember that the unit circle has a radius of 1. In other words, every point on the unit circle is a distance of exactly 1 unit away from the origin, (0,0). Knowing that, you can use the distance formula to find your missing coordinate:

 

√[(x₁ - x₂)² + (y₁ - y₂)²] = distance

 

So, the distance from (0,0) to (x,¹/₂) is:

 

√[(0 - x)² + (0 - ¹/₂)²] = 1                               SQUARE EACH SIDE

(0 - x)² + (0 - ¹/₂)² = 1                                    SUBTRACT

(-x)² + (-¹/₂)² = 1                                           SQUARE -x AND -¹/₂

x² + ¹/₄ = 1                                                    SUBTRACT ¹/₄ FROM EACH SIDE

x² = ³/₄                                                                             TAKE THE SQUARE ROOT OF EACH SIDE

x = √(³/₄)

 

Let's simplify √(³/₄) just a bit by distributing the square root sign:

 

^(√3)/_(√4) = ^(√3)/₂

 

So, your x coordinate is ^(√3)/₂.

Answer 2

We can also use the equation for a circle.

 

x² + y² = r²

 

when the center of the circle is located at the origin.  r is the radius of the circle.

 

Since this is a unit circle, the radius is one.

 

 

Plug in y into the formula.

 

x² + (1/2)² = 1

 

x² + 1/4 = 1

 

x² = 1 - (1/4)

 

x² = 3/4

 

x = ±√(3/4)

 

x = ±√(3) / 2

 

x = √(3) / 2           and       x = -√(3) / 2

 

 

When we solve for a square-root, generally there is a positive solution and a negative solution.  However, the point needed here is in Quadrant I.  Positive x-coordinates lie between Quadrant I and Quadrant IV.  Therefore, we accept the positive value of x.

 

x = √(3) / 2