Problem 10 - Algebra and Trigonometry

Question

\(5-10\) . Determine whether the given points are on the graph of the equation. $$ x^{2}+y^{2}=1 ; \quad(0,1),\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) $$

Step-by-Step Solution

Answer

All given points are on the graph of the equation.

Step 1: Understanding the Equation

The equation given is \(x^2 + y^2 = 1\). This represents a circle centered at the origin (0,0) with a radius of 1. Points on this circle will satisfy this equation.

Step 2: Check Point (0,1)

Substitute \(x = 0\) and \(y = 1\) into the equation \(x^2 + y^2 = 1\). Calculate: \(0^2 + 1^2 = 0 + 1 = 1\).Since the left-side equals the right-side of the equation, (0,1) is on the graph.

Step 3: Check Point \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)

Substitute \(x = \frac{1}{\sqrt{2}}\) and \(y = \frac{1}{\sqrt{2}}\) into the equation \(x^2 + y^2 = 1\). Calculate:\(\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} + \frac{1}{2} = 1\).Since the left-side equals the right-side of the equation, \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) is on the graph.

Step 4: Check Point \(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)

Substitute \(x = \frac{\sqrt{3}}{2}\) and \(y = \frac{1}{2}\) into the equation \(x^2 + y^2 = 1\). Calculate:\(\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1\).Since the left-side equals the right-side of the equation, \(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\) is on the graph.

Key Concepts

Coordinate Geometry

Coordinate geometry is a branch of mathematics that specifically handles the analysis of geometry using a coordinate system. It's like a toolkit for solving geometrical problems by representing geometric figures algebraically. In coordinate geometry, every point on a plane is defined by its coordinates, which are a set of numbers that provide its exact location.
In the context of a circle, such as the one described by the equation \(x^2 + y^2 = 1\), each point on the circle is a distance (radius) of 1 unit away from the center (which in this case is at the origin point \((0,0)\)).
This equation is typical for a simple circle in a 2-dimensional plane and is also known as a unit circle because of its radius of 1. By providing the coordinates \((x, y)\), you effectively map where a point is located in relation to the center of the circle. If a point's coordinates satisfy the equation, the point lies on the circle; otherwise, it does not. It's a straightforward way to determine position and relationship in a plane.

Unit Circle

The unit circle is a fundamental concept in trigonometry and coordinate geometry. It is a circle with a radius of 1 and is centered at the origin point \((0,0)\). The standard equation of the unit circle is \(x^2 + y^2 = 1\).
Understanding the unit circle is crucial because it simplifies numerous mathematical problems, especially those involving trigonometric functions. Trigonometric functions like sine, cosine, and tangent can be directly derived from the coordinates of points along the unit circle.

• The cosine of an angle corresponds to the x-coordinate.
• The sine of an angle corresponds to the y-coordinate.

The unit circle also reflects symmetry, given that its equation does not change when you switch the values of x and y. The circle's simplicity makes it an invaluable tool for analyzing functions and plotting points during both basic and complex calculations.

Substitution Method

The substitution method is a valuable algebraic tool for verifying whether a particular point lies on a graph defined by an equation. It involves replacing variables in the equation with the coordinates of the point in question.
Here's the straightforward process using the circle's equation \(x^2 + y^2 = 1\):

• Take the x-coordinate of the point and substitute it into the equation in place of x.
• Take the y-coordinate of the point and substitute it into the equation in place of y.
• Perform the algebraic calculations. If the equation holds (i.e., both sides are equal), then the point lies on the circle. If not, the point is outside or inside but not precisely on the circle.

In the given exercise, substituting the coordinates of each point individually into the equation allows one to see if the sum is exactly 1. This method is efficient and provides clear confirmation of whether each point is on the circle.