Math Problem Statement

The point P is on the unit circle. If the x-coordinate of P is 2/7, and P is in quadrant IV , then y = ?

Solution

To solve for the yy-coordinate of the point PP on the unit circle, we use the equation of the unit circle:

x2+y2=1x^2 + y^2 = 1

Given that the xx-coordinate of PP is 27\frac{2}{7} and PP lies in quadrant IV (where the yy-coordinate is negative), we can substitute x=27x = \frac{2}{7} into the equation and solve for yy.

  1. Substitute x=27x = \frac{2}{7} into the equation:

(27)2+y2=1\left( \frac{2}{7} \right)^2 + y^2 = 1

  1. Simplify:

449+y2=1\frac{4}{49} + y^2 = 1

  1. Subtract 449\frac{4}{49} from both sides:

y2=1449=4949449=4549y^2 = 1 - \frac{4}{49} = \frac{49}{49} - \frac{4}{49} = \frac{45}{49}

  1. Take the square root of both sides:

y=±457=±357y = \pm \frac{\sqrt{45}}{7} = \pm \frac{3\sqrt{5}}{7}

Since PP is in quadrant IV, where the yy-coordinate is negative, we choose the negative root:

y=357y = -\frac{3\sqrt{5}}{7}

Thus, the yy-coordinate of point PP is 357-\frac{3\sqrt{5}}{7}.

Would you like more details or have any questions about this solution?

Here are 5 related questions:

  1. What is the equation of the unit circle in Cartesian coordinates?
  2. How do the signs of xx and yy vary in each quadrant?
  3. What is the significance of the unit circle in trigonometry?
  4. How do you find the coordinates of a point given its angle in the unit circle?
  5. Can you derive the distance formula from the equation of the unit circle?

Tip: Remember that in quadrant IV, the xx-coordinate is positive, and the yy-coordinate is negative.

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Math Problem Analysis

Mathematical Concepts

Unit Circle
Trigonometry
Coordinate Geometry

Formulas

x^2 + y^2 = 1

Theorems

Unit Circle Theorem

Suitable Grade Level

Grades 9-11