Chapter 5: Problem 13
The point \(P\) is on the unit circle. Find \(P(x, y)\) from the given information. The \(x\) -coordinate of \(P\) is \(\frac{4}{5}\) and the \(y\) -coordinate is positive.
Short Answer
Expert verified
The point \(P\) is \(\left(\frac{4}{5}, \frac{3}{5}\right)\)."
Step by step solution
01
Understand the Unit Circle
The unit circle has a radius of 1 and its center is at the origin (0,0) in the coordinate plane. Any point \(P(x, y)\) on the unit circle satisfies the equation \(x^2 + y^2 = 1\).
02
Use the Given Information
We know the \(x\)-coordinate of point \(P\) is \(\frac{4}{5}\). This means we need to find the \(y\)-coordinate that satisfies the equation of the unit circle. Additionally, the \(y\)-coordinate must be positive.
03
Substitute the x-coordinate into the Unit Circle Equation
Substitute \(x = \frac{4}{5}\) into the equation \(x^2 + y^2 = 1\). We have: \(\left(\frac{4}{5}\right)^2 + y^2 = 1\).
04
Calculate the x^2 Value
Calculate \(\left(\frac{4}{5}\right)^2\): \[\left(\frac{4}{5}\right)^2 = \frac{16}{25}\]
05
Solve for \(y^2\)
Substitute \(\frac{16}{25}\) into the equation and solve for \(y^2\):\[\frac{16}{25} + y^2 = 1\]Subtract \(\frac{16}{25}\) from both sides:\[y^2 = 1 - \frac{16}{25}\]
06
Calculate 1 - \(\frac{16}{25}\)
Find \(1 - \frac{16}{25}\). Convert 1 to \(\frac{25}{25}\) for easy subtraction:\[1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25}\]
07
Solve for \(y\)
Since \(y^2 = \frac{9}{25}\), take the square root of both sides to find \(y\):\[y = \sqrt{\frac{9}{25}}\]So, \(y = \frac{3}{5}\) or \(y = -\frac{3}{5}\). However, since \(y\) is positive, we choose \(y = \frac{3}{5}\).
08
Present the Point \(P(x, y)\)
The coordinates of the point \(P(x, y)\) are \(x = \frac{4}{5}\) and \(y = \frac{3}{5}\). Thus, the point \(P\) is \(\left(\frac{4}{5}, \frac{3}{5}\right)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, is a branch of mathematics that uses algebraic equations to describe geometric figures and their properties. The coordinate plane consists of two axes, the x-axis and the y-axis, which intersect at the origin point \(0,0\). Each point on the plane is represented by a pair of numbers, \(x\) and \(y\), known as coordinates. In the context of the unit circle, the point \(P(x,y)\) falls on a circle with a radius of 1 centered at the origin. This simply means that any point on the unit circle is exactly one unit away from the origin.
- The x-coordinate represents the horizontal distance from the origin.
- The y-coordinate represents the vertical distance.
- On the unit circle, these coordinates satisfy the equation \(x^2 + y^2 = 1\).
Trigonometry
Trigonometry is a field of mathematics focused on the study of triangles and the relationships between their angles and sides. In the context of a unit circle, trigonometry helps us understand the connection between angles and the coordinates of points on the circle. The unit circle in trigonometry is a circle with a radius of 1, centered at the origin of a coordinate system.
- The angle \(\theta\) in the unit circle can be understood as the angle between the positive x-axis and a line connecting the origin to a point \(P(x, y)\).
- Key trigonometric functions, such as sine and cosine, are defined using the coordinates of this point: \(\cos(\theta) = x\text{ and }\sin(\theta) = y\).
- The unit circle allows us to visualize these functions and understand their behavior as \(\theta\) changes.
Equation of a Circle
The equation of a circle is a fundamental concept in coordinate geometry. It provides a way to algebraically represent a circle in the coordinate plane. A generic circle's equation with center at \((h, k)\) and a radius \(r\) is given by:\[ (x-h)^2 + (y-k)^2 = r^2 \]The unit circle, however, is a special case where the center is at the origin \( (0,0)\) and the radius \(r\) is 1. Hence, the equation simplifies to:\[ x^2 + y^2 = 1 \]
- Any point \( (x, y)\) lying on this circle must satisfy the simplified equation.
- Solutions to this equation are \( x^2 + y^2 = 1 \).
- This implies that the sum of the squares of the coordinates \(x\) and \(y\) will always equal 1, indicating the consistent distance from the origin.