Problem 17 - Algebra and Trigonometry

Question

\(13-18=\) The point \(P\) is on the unit circle. Find \(P(x, y)\) from the given information. The \(x\) -coordinate of \(P\) is \(-\sqrt{2} / 3\) and \(P\) lies below the \(x\) -axis.

Step-by-Step Solution

Answer

The point \(P\) is \(\left(-\frac{\sqrt{2}}{3}, -\frac{\sqrt{7}}{3}\right)\).

Step 1: Recall the Unit Circle Equation

The unit circle is defined by the equation \( x^2 + y^2 = 1 \). In this problem, the \(x\)-coordinate of a point \(P\) on the unit circle is given as \(-\sqrt{2}/3\). We need to use this information to find the corresponding \(y\)-coordinate.

Step 2: Substitute the x-coordinate into the Unit Circle Equation

Substitute \(x = -\sqrt{2}/3\) into the unit circle equation to find \(y^2\): \[ \left(-\frac{\sqrt{2}}{3}\right)^2 + y^2 = 1 \] Calculate \((-\sqrt{2}/3)^2\): \[ \frac{2}{9} + y^2 = 1 \]

Step 3: Solve for y^2

Solve the equation \( \frac{2}{9} + y^2 = 1 \) for \(y^2\):\[ y^2 = 1 - \frac{2}{9} \] \[ y^2 = \frac{9}{9} - \frac{2}{9} \] \[ y^2 = \frac{7}{9} \]

Step 4: Solve for y

Take the square root of both sides to solve for \(y\):\[ y = \pm \sqrt{\frac{7}{9}} \] \[ y = \pm \frac{\sqrt{7}}{3} \]Since the point lies below the \(x\)-axis, choose the negative root: \[ y = -\frac{\sqrt{7}}{3} \]

Step 5: State the Coordinates of P

Combine the \(x\)-coordinate and the \(y\)-coordinate to express the point \(P\): \[ P\left(-\frac{\sqrt{2}}{3}, -\frac{\sqrt{7}}{3}\right) \]

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, allows us to describe geometric figures using an ordered pair of numbers: \(x, y\). Each point in this system corresponds to a pair of coordinates that define its position on the Cartesian plane.
When working with points on a plane:

• The horizontal line is known as the \(x\)-axis.
• The vertical line is the \(y\)-axis.
• The intersection of these axes is called the origin, with coordinates (0,0).
• Any point on the plane can be located using a pair of numbers \((x, y)\).

In the given exercise, the point \(P\) is located on the unit circle, but coordination geometry helps us easily find its position using the information that it lies below the \(x\)-axis.

Unit Circle Equation

In coordinate geometry, the equation of the unit circle is crucial. The unit circle has a radius of 1 and is centered at the origin with an equation given by:\[ x^2 + y^2 = 1 \]This equation states that for any point \((x, y)\) on the circle, the sum of the squares of its coordinates is \1\, reflecting the Pythagorean theorem.

• The unit circle is fundamental for trigonometric functions.
• Its equation helps to find unknown coordinates of points on the circle.

In this exercise, with the \(x\)-coordinate known as \(-\sqrt{2}/3\), we use the unit circle equation to solve for the \(y\)-coordinate.

Solving Equations

Solving equations involves finding the values of the unknown variable that satisfy the equation. In the context of the unit circle:

• We substitute known values into the equation \(x^2 + y^2 = 1\).
• Rearrange the equation to solve for the unknown \(y^2\).

After substituting \(x = -\sqrt{2}/3\), the unit circle equation becomes:\[ \left(-\frac{\sqrt{2}}{3}\right)^2 + y^2 = 1 \]This simplifies to \(y^2 = \frac{7}{9}\). Then, taking the square root helps find potential values for \(y\), specifically \(\pm\frac{\sqrt{7}}{3}\). Since the point \(P\) is below the \(x\)-axis, we select the negative value.

Pythagorean Identity

The Pythagorean identity is a key concept derived from the Pythagorean theorem. It implies the relationship:\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]Where \(\sin(\theta)\) and \(\cos(\theta)\) represent the sine and cosine of an angle \(\theta\). This identity is similar to the unit circle equation and is vital in trigonometry.

• This identity helps verify the unit circle equation.
• It's used to relate angles and coordinates on the circle.

In finding point \(P\), this identity underlies the assumption that any point on the unit circle satisfies \(x^2 + y^2 = 1\), reinforcing the use of the unit circle equation in the exercise.