2.1 The Rectangular Coordinate System
Section 2.1: The Rectangular Coordinate System
Introduction to Rectangular Coordinates
The Rectangular Coordinate System (also known as the Cartesian Coordinate System) is a plane formed by two perpendicular number lines.
Origin: The point where the x-axis and y-axis intersect, denoted as (0,0).
Coordinate Axes:
The x-axis is the horizontal number line.
The y-axis is the vertical number line.
Ordered Pair (x, y): Represents the unique location of a point in the coordinate plane. The first number, x, is the x-coordinate, and the second number, y, is the y-coordinate.
x-coordinate: Indicates the horizontal distance and direction (right for positive, left for negative) from the y-axis.
y-coordinate: Indicates the vertical distance and direction (up for positive, down for negative) from the x-axis.
Quadrants: The four regions into which the coordinate axes divide the plane. They are typically numbered counter-clockwise starting from the top-right quadrant.
The Distance Formula
Theorem 1
The distance between two points P1 = (x1, y1) and P2 = (x2, y2), denoted by d(P1, P2), is given by the formula:
d(P1, P2) = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}
Example 2: Finding the Distance Between Two Points
Problem: Find the distance d between the points (-4, 5) and (3, 2).
Solution:
Let P1 = (-4, 5) and P2 = (3, 2).
x1 = -4, y1 = 5
x2 = 3, y2 = 2
Substitute these values into the distance formula:
d = \sqrt{(3 - (-4))^2 + (2 - 5)^2}
d = \sqrt{(3 + 4)^2 + (-3)^2}
d = \sqrt{(7)^2 + (-3)^2}
d = \sqrt{49 + 9}
d = \sqrt{58}
The distance between the points is \sqrt{58} units.
Example 3: Analyzing a Triangle
Problem: Consider the three points A = (-2, 1), B = (2, 3), and C = (3, 1).
Plot each point and form the triangle ABC.
Find the length of each side of the triangle.
Show that the triangle is a right triangle.
Find the area of the triangle.
Solution:
Plotting Points: (This step involves drawing on a coordinate plane, which is a visual representation.)
Length of Each Side (Using Distance Formula):
Side AB (distance between A(-2,1) and B(2,3)):
d(A, B) = \sqrt{(2 - (-2))^2 + (3 - 1)^2}
d(A, B) = \sqrt{(4)^2 + (2)^2}
d(A, B) = \sqrt{16 + 4}
d(A, B) = \sqrt{20}Side BC (distance between B(2,3) and C(3,1)):
d(B, C) = \sqrt{(3 - 2)^2 + (1 - 3)^2}
d(B, C) = \sqrt{(1)^2 + (-2)^2}
d(B, C) = \sqrt{1 + 4}
d(B, C) = \sqrt{5}Side AC (distance between A(-2,1) and C(3,1)):
d(A, C) = \sqrt{(3 - (-2))^2 + (1 - 1)^2}
d(A, C) = \sqrt{(5)^2 + (0)^2}
d(A, C) = \sqrt{25}
d(A, C) = 5
Showing it's a Right Triangle:
We use the Pythagorean Theorem, which states that for a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides (a^2 + b^2 = c^2).
The side lengths are \sqrt{20}, \sqrt{5}, and 5.
The longest side is 5, so this would be the hypotenuse if it's a right triangle.
Check if (\sqrt{20})^2 + (\sqrt{5})^2 = 5^2:
20 + 5 = 25
25 = 25Since the Pythagorean Theorem holds true, triangle ABC is a right triangle.
Finding the Area of the Triangle:
The area of a right triangle is given by the formula: \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}.
The base and height correspond to the two shorter sides, which form the right angle. In this case, they are \sqrt{20} and \sqrt{5}. (Note: Side AC is horizontal, the line segment from x=-2 to x=3 at y=1. Side AB and BC must form the right angle since AC is the hypotenuse).
\text{Area} = \frac{1}{2} \times \sqrt{20} \times \sqrt{5}
\text{Area} = \frac{1}{2} \times \sqrt{100}
\text{Area} = \frac{1}{2} \times 10
\text{Area} = 5 square units.
The Midpoint Formula
Theorem 4
The midpoint M = (x, y) of the line segment from P1 = (x1, y1) to P2 = (x2, y2) is given by the formula:
M = \left( \frac{x1 + x2}{2}, \frac{y1 + y2}{2} \right)
Example 5: Finding the Midpoint
Problem: Find the midpoint of the line segment from P1 = (-5, 5) to P2 = (3, 1). Plot the points P1, P2, and their midpoint.
Solution:
Let P1 = (-5, 5) and P2 = (3, 1).
x1 = -5, y1 = 5
x2 = 3, y2 = 1
Substitute these values into the midpoint formula:
M = \left( \frac{-5 + 3}{2}, \frac{5 + 1}{2} \right)
M = \left( \frac{-2}{2}, \frac{6}{2} \right)
M = (-1, 3)
The midpoint of the line segment is (-1, 3).
Plotting: (This step involves drawing on a coordinate plane.)
Equations in Two Variables and Their Graphs
Definition of an Equation in Two Variables: A statement in which two expressions involving two variables, typically x and y, are set equal to each other.
Examples:
x^2 + y^2 = 5
2x - y = 6
y = -3x + 2
x^2 = y
Graph of an Equation: The graph of an equation in two variables x and y consists of the set of all points (x, y) in the xy-plane whose coordinates satisfy the equation (i.e., make the equation a true statement).
Example 6: Determining if Points are on a Graph
Problem: Determine whether the points (2, 3) and (2, -2) are on the graph of the equation 2x - y = 6.
Solution:
**For point (2, 3)(x=2, y=3):
Substitute the coordinates into the equation:
2(2) - 3 = 6
4 - 3 = 6
1 = 6 (False)
Since the statement is false, the point (2, 3) is NOT on the graph of 2x - y = 6.**For point (2, -2)(x=2, y=-2):
Substitute the coordinates into the equation:
2(2) - (-2) = 6
4 + 2 = 6
6 = 6 (True)
Since the statement is true, the point (2, -2) IS on the graph of 2x - y = 6.
Example 7: Graphing an Equation
Problem: Graph the equation: y = 2x + 5.
Solution:
To graph a linear equation like this, create a table of ordered pairs that satisfy the equation. Choose various values for x and calculate the corresponding y values.
Sample Points:
If x = 0, y = 2(0) + 5 = 5 (Point: (0, 5))
If x = 1, y = 2(1) + 5 = 7 (Point: (1, 7))
If x = -2, y = 2(-2) + 5 = 1 (Point: (-2, 1))
Plot these points on the coordinate plane and draw a straight line through them.
Intercepts of a Graph
Definition: The points, if any, at which a graph crosses or touches the coordinate axes are called the intercepts of the graph.
x-intercept(s): The x-coordinate(s) of the point(s) where the graph crosses or touches the x-axis. At x-intercepts, the y-coordinate is $0$. To find x-intercept(s), set y = 0 in the equation and solve for x.
y-intercept(s): The y-coordinate(s) of the point(s) where the graph crosses or touches the y-axis. At y-intercepts, the x-coordinate is $0$. To find y-intercept(s), set x = 0 in the equation and solve for y.
Example 8: Identifying Intercepts from a Graph
Problem: Find the intercepts of the given graph.
**Solution (based on the provided diagram description):
x-intercepts: The graph crosses the x-axis at -3 and 4.5. These correspond to the points (-3, 0) and (4.5, 0).
y-intercepts: The graph crosses the y-axis at 3, -3.5 and -1. (A graph can have multiple y-intercepts if it is not a function.) These correspond to the points (0, 3), (0, -3.5) and (0, -1).
Example 9: Finding Intercepts Algebraically and Graphing
Problem: Find the x-intercept(s) and the y-intercept(s) of the graph of y = x^2 - 4. Then graph y = x^2 - 4 by plotting points.
Solution:
Finding y-intercept(s):
Set x = 0 in the equation:
y = (0)^2 - 4
y = -4
The y-intercept is (0, -4).Finding x-intercept(s):
Set y = 0 in the equation:
0 = x^2 - 4
x^2 = 4
x = \pm \sqrt{4}
x = \pm 2
The x-intercepts are (-2, 0) and (2, 0).Graphing y = x^2 - 4: This is a parabola opening upwards. The intercepts provide three key points. Additional points can be found for more detail:
x=0, y=-4
x=1, y=1^2-4 = -3
x=-1, y=(-1)^2-4 = -3
x=2, y=0
x=-2, y=0
Plot these points (0, -4), (1, -3), (-1, -3), (2, 0), (-2, 0) and draw a smooth curve (parabola) through them.