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 $, is the y-coordinate.</li><li>x-coordinate: Indicates the horizontal distance and direction (right for positive, left for negative) from the y-axis.</li><li>y-coordinate: Indicates the vertical distance and direction (up for positive, down for negative) from the x-axis.</li><li>Quadrants: The four regions into which the coordinate axes divide the plane. They are typically numbered counter-clockwise starting from the top-right quadrant.</li></ul><h4 id="3ddc2d21-13d6-44c7-8f45-28f216477399" data-toc-id="3ddc2d21-13d6-44c7-8f45-28f216477399" collapsed="false" seolevelmigrated="true">The Distance Formula</h4><h5 id="11471c53-4c68-464a-9624-66db4f857311" data-toc-id="11471c53-4c68-464a-9624-66db4f857311" collapsed="false" seolevelmigrated="true">Theorem 1</h5><ul><li>The distance between two points$ 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} $</li></ul><h5 id="001c5282-c343-4df3-bab4-6f58472ef267" data-toc-id="001c5282-c343-4df3-bab4-6f58472ef267" collapsed="false" seolevelmigrated="true">Example 2: Finding the Distance Between Two Points</h5><ul><li>Problem: Find the distance$ d $between the points$ (-4, 5) $and$ (3, 2) $.</li><li>Solution:<ul><li>Let$ P1 = (-4, 5) $and$ P2 = (3, 2) $.</li><li>$ x1 = -4 $,$ y1 = 5 $</li><li>$ x2 = 3 $,$ y2 = 2 $</li><li>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} $</li></ul></li><li>The distance between the points is$ \sqrt{58} $units.</li></ul><h5 id="d1654df4-e2a3-4524-beae-1f1e7a6ed6a7" data-toc-id="d1654df4-e2a3-4524-beae-1f1e7a6ed6a7" collapsed="false" seolevelmigrated="true">Example 3: Analyzing a Triangle</h5><ul><li>Problem: Consider the three points$ A = (-2, 1) $,$ B = (2, 3) $, and$ C = (3, 1) $.<ol><li>Plot each point and form the triangle ABC.</li><li>Find the length of each side of the triangle.</li><li>Show that the triangle is a right triangle.</li><li>Find the area of the triangle.</li></ol></li><li>Solution:<ol><li>Plotting Points: (This step involves drawing on a coordinate plane, which is a visual representation.)</li><li>Length of Each Side (Using Distance Formula):<ul><li>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} $</li><li>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} $</li><li>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 $</li></ul></li><li>Showing it's a Right Triangle:<ul><li>We use the Pythagorean Theorem, which states that for a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides ($ a^2 + b^2 = c^2 $).</li><li>The side lengths are$ \sqrt{20} $,$ \sqrt{5} $, and$ 5 $.</li><li>The longest side is$ 5 $, so this would be the hypotenuse if it's a right triangle.</li><li>Check if$ (\sqrt{20})^2 + (\sqrt{5})^2 = 5^2 $: $ 20 + 5 = 25 $ $ 25 = 25 $</li><li>Since the Pythagorean Theorem holds true, triangle ABC is a right triangle.</li></ul></li><li>Finding the Area of the Triangle:<ul><li>The area of a right triangle is given by the formula:$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $.</li><li>The base and height correspond to the two shorter sides, which form the right angle. In this case, they are$ \sqrt{20} $and$ \sqrt{5} $. (Note: Side AC is horizontal, the line segment from x=-2 to x=3 at y=1. Side AB and BC must form the right angle since AC is the hypotenuse).</li><li>$ \text{Area} = \frac{1}{2} \times \sqrt{20} \times \sqrt{5} $</li><li>$ \text{Area} = \frac{1}{2} \times \sqrt{100} $</li><li>$ \text{Area} = \frac{1}{2} \times 10 $</li><li>$ \text{Area} = 5 $square units.</li></ul></li></ol></li></ul><h4 id="943651a5-0ea6-4183-ad2c-59a94b34ffd0" data-toc-id="943651a5-0ea6-4183-ad2c-59a94b34ffd0" collapsed="false" seolevelmigrated="true">The Midpoint Formula</h4><h5 id="df776ec6-7480-4e2d-a982-dba7ad0fa234" data-toc-id="df776ec6-7480-4e2d-a982-dba7ad0fa234" collapsed="false" seolevelmigrated="true">Theorem 4</h5><ul><li>The midpoint$ 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) $</li></ul><h5 id="6e5c4733-661f-4196-92d8-c5866f808c40" data-toc-id="6e5c4733-661f-4196-92d8-c5866f808c40" collapsed="false" seolevelmigrated="true">Example 5: Finding the Midpoint</h5><ul><li>Problem: Find the midpoint of the line segment from$ P1 = (-5, 5) $to$ P2 = (3, 1) $. Plot the points$ P1 $,$ P2 $, and their midpoint.</li><li>Solution:<ul><li>Let$ P1 = (-5, 5) $and$ P2 = (3, 1) $.</li><li>$ x1 = -5 $,$ y1 = 5 $</li><li>$ x2 = 3 $,$ y2 = 1 $</li><li>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) $</li></ul></li><li>The midpoint of the line segment is$ (-1, 3) $.</li><li>Plotting: (This step involves drawing on a coordinate plane.)</li></ul><h4 id="83063c1e-adbf-44f5-9969-78b7d19d80aa" data-toc-id="83063c1e-adbf-44f5-9969-78b7d19d80aa" collapsed="false" seolevelmigrated="true">Equations in Two Variables and Their Graphs</h4><ul><li>Definition of an Equation in Two Variables: A statement in which two expressions involving two variables, typically$ x $and$ y $, are set equal to each other.<ul><li>Examples:<ul><li>$ x^2 + y^2 = 5 $</li><li>$ 2x - y = 6 $</li><li>$ y = -3x + 2 $</li><li>$ x^2 = y $</li></ul></li></ul></li><li>Graph of an Equation: The graph of an equation in two variables$ x $and$ y $consists of the set of all points$ (x, y) $in the$ xy $-plane whose coordinates satisfy the equation (i.e., make the equation a true statement).</li></ul><h5 id="235bf782-b4d0-4085-92db-3986004228c6" data-toc-id="235bf782-b4d0-4085-92db-3986004228c6" collapsed="false" seolevelmigrated="true">Example 6: Determining if Points are on a Graph</h5><ul><li>Problem: Determine whether the points$ (2, 3) $and$ (2, -2) $are on the graph of the equation$ 2x - y = 6 $.</li><li>Solution:<ul><li>**For point$ (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 $.</li></ul></li></ul><h5 id="9162a963-15a0-4afe-a646-a91ce0a14bf7" data-toc-id="9162a963-15a0-4afe-a646-a91ce0a14bf7" collapsed="false" seolevelmigrated="true">Example 7: Graphing an Equation</h5><ul><li>Problem: Graph the equation:$ y = 2x + 5 $.</li><li>Solution:<ul><li>To graph a linear equation like this, create a table of ordered pairs that satisfy the equation. Choose various values for$ x $and calculate the corresponding$ y $values.</li><li>Sample Points:<ul><li>If$ x = 0 $,$ y = 2(0) + 5 = 5 $(Point:$ (0, 5))
If $x = 1$ , $y = 2(1) + 5 = 7$ (Point: $(1, 7)$ $)</li><li>If$ 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.