Comprehensive Notes on Rectangular Coordinate Systems, Graphing, and Linear Equations

Foundations of Equations, Inequalities, and Coordinate Systems

• Agricultural and Spatial Applications of Coordinate Systems:

  • Center-pivot irrigation systems create circular crop patterns visible from the air, appearing random at first glance but precisely designed through principles of agricultural science, engineering, and mathematics.

  • Dividing arable land into geometric grid portions enables farm operators and irrigation engineers to optimize pipeline networks, pumps, and conduits across large-scale farming plots, maximizing agricultural yield while managing the high financial cost of materials and water distribution.

• Chapter Scope and Structure:

  • Section 2.1: The Rectangular Coordinate Systems and Graphs

  • Section 2.2: Linear Equations in One Variable

  • Section 2.3: Models and Applications

  • Section 2.4: Complex Numbers

  • Section 2.5: Quadratic Equations

  • Section 2.6: Other Types of Equations

  • Section 2.7: Linear Inequalities and Absolute Value Inequalities

• Core Learning Objectives for Section 2.1:

  • Plot ordered pairs accurately within the Cartesian coordinate framework.

  • Graph algebraic equations in two variables by compiling tables and plotting coordinate points.

  • Utilize graphing utilities to graph equations and configure window viewports.

  • Determine algebraic $x$-intercepts and $y$-intercepts of graphs.

  • Apply the distance formula to compute lengths between coordinates in a plane.

  • Apply the midpoint formula to determine centers of segments and circular diameters.

The Cartesian Coordinate System

• Historical Development:

  • Developed by seventeenth-century French philosopher and mathematician René Descartes.

  • While bedridden, Descartes observed an insect crawling across a ceiling and realized its exact location could be described relative to the perpendicular boundaries formed by the intersecting adjacent walls of the room.

  • By treating the intersecting boundaries as horizontal and vertical reference lines with evenly spaced units, any location on a two-dimensional surface could be definitively pinpointed with two numbers.

• Structural Components of the Rectangular Plane:

  • Axes:

  • The horizontal reference line is designated as the $x$-axis.

  • The vertical reference line is designated as the $y$-axis.

  • The two axes intersect at a right angle (90^\n\ncirc), dividing the infinite two-dimensional plane into four distinct regions.

  • The Origin:

  • The intersection point of the horizontal and vertical axes.

  • Has the coordinates: $(0, 0)$.

  • Quadrants:

  • The four regions created by the axes are labeled using Roman numerals in a counterclockwise order starting from the upper-right region:

    • Quadrant I: Upper-right quadrant; both coordinates are positive ($x > 0$, $y > 0$).

    • Quadrant II: Upper-left quadrant; horizontal coordinates are negative, vertical coordinates are positive ($x < 0$, $y > 0$).

    • Quadrant III: Lower-left quadrant; both coordinates are negative ($x < 0$, $y < 0$).

    • Quadrant IV: Lower-right quadrant; horizontal coordinates are positive, vertical coordinates are negative ($x > 0$, $y < 0$).

• Axis Directionality and Scaling:

  • From the origin, values increase positively to the right on the $x$-axis and upward on the $y$-axis.

  • Values decrease negatively to the left on the $x$-axis and downward on the $y$-axis.

  • Arrowheads on the ends of the axes indicate that the coordinate grid extends toward positive and negative infinity ($\infty$ and $-\infty$).

  • The $x$-axis and $y$-axis do not require identical scaling; increments may be unit integers ($1, 2, 3, \dots$), or custom gradations such as increments of $2$, $10$, or $100$.

  • Axes can represent disparate real-world units, such as elapsed years versus bank balances, or product quantity versus production cost.

• Ordered Pairs and Coordinate Mapping:

  • A specific location on the plane is written as an ordered pair (or coordinate pair) in the form $(x, y)$.

  • The first value, $x$, indicates the directed horizontal distance from the origin.

  • The second value, $y$, indicates the directed vertical distance from the origin.

  • For the point $(3, -1)$, begin at the origin $(0, 0)$, traverse $3$ units to the right along the positive $x$-axis, and move $1$ unit downward parallel to the negative $y$-axis.

• Worked Example: Plotting Points in the Coordinate System:

  • Point $(-2, 4)$:

  • Start at the origin $(0, 0)$.

  • The horizontal component is $-2$; move $2$ units to the left along the horizontal axis.

  • The vertical component is $4$; move $4$ units straight up parallel to the vertical axis.

  • The point resides in Quadrant II.

  • Point $(3, 3)$:

  • Start at the origin $(0, 0)$.

  • The horizontal component is $3$; move $3$ units to the right along the horizontal axis.

  • The vertical component is $3$; move $3$ units straight up parallel to the vertical axis.

  • The point resides in Quadrant I.

  • Point $(0, -3)$:

  • Start at the origin $(0, 0)$.

  • The horizontal component is $0$; do not move horizontally along the $x$-axis.

  • The vertical component is $-3$; move $3$ units straight down along the negative $y$-axis.

  • The point resides directly on the negative $y$-axis.

• Positional Rules for Points on the Axes:

  • If the $x$-coordinate of an ordered pair is zero ($x = 0$), the point lies directly on the $y$-axis.

  • If the $y$-coordinate of an ordered pair is zero ($y = 0$), the point lies directly on the $x$-axis.

  • A point lying directly on either axis does not belong to any of the four quadrants.

Graphing Equations by Plotting Points

• Equations and Graphs in Two Variables:

  • An equation containing two distinct variable terms (typically $x$ and $y$) is defined as an equation in two variables.

  • The set of all ordered pairs $(x, y)$ whose coordinates satisfy the algebraic equality represents the graph in two variables.

• Systematic Procedure for Graphing by Point Plotting:

  • Construct a three-column data table:

  • Column 1: Selected values for the independent variable $x$.

  • Column 2: Substitution step evaluating the algebraic equation for $y$.

  • Column 3: Resulting ordered coordinate pair $(x, y)$.

  • Populate the first column with an orderly succession of positive values, negative values, and zero to simplify calculations and detect geometric patterns.

  • Choose values for $x$ that allow simple mental or arithmetic calculation of $y$.

  • Plot each computed ordered pair onto the coordinate grid.

  • Connect the plotted coordinate points with a smooth curve or straight continuous line with arrows at the endpoints.

• Complete Walkthrough: Graphing $y = 2x - 1$:

  • Selecting integer values for $x$ ranging from $-3$ to $3$:

  • For $x = -3$:       $y = 2(-3) - 1 = -6 - 1 = -7$       Ordered pair: $(-3, -7)$

  • For $x = -2$:       $y = 2(-2) - 1 = -4 - 1 = -5$       Ordered pair: $(-2, -5)$

  • For $x = -1$:       $y = 2(-1) - 1 = -2 - 1 = -3$       Ordered pair: $(-1, -3)$

  • For $x = 0$:       $y = 2(0) - 1 = 0 - 1 = -1$       Ordered pair: $(0, -1)$

  • For $x = 1$:       $y = 2(1) - 1 = 2 - 1 = 1$       Ordered pair: $(1, 1)$

  • For $x = 2$:       $y = 2(2) - 1 = 4 - 1 = 3$       Ordered pair: $(2, 3)$

  • For $x = 3$:       $y = 2(3) - 1 = 6 - 1 = 5$       Ordered pair: $(3, 5)$

  • Connecting these points produces a continuous straight line that passes through Quadrants I, III, and IV.

• Worked Example: Graphing $y = -x + 2$:

  • Constructing evaluation table across selected domain inputs:

  • For $x = -5$:       $y = -(-5) + 2 = 5 + 2 = 7$       Ordered pair: $(-5, 7)$

  • For $x = -3$:       $y = -(-3) + 2 = 3 + 2 = 5$       Ordered pair: $(-3, 5)$

  • For $x = -1$:       $y = -(-1) + 2 = 1 + 2 = 3$       Ordered pair: $(-1, 3)$

  • For $x = 0$:       $y = -(0) + 2 = 0 + 2 = 2$       Ordered pair: $(0, 2)$

  • For $x = 1$:       $y = -(1) + 2 = -1 + 2 = 1$       Ordered pair: $(1, 1)$

  • For $x = 3$:       $y = -(3) + 2 = -3 + 2 = -1$       Ordered pair: $(3, -1)$

  • For $x = 5$:       $y = -(5) + 2 = -5 + 2 = -3$       Ordered pair: $(5, -3)$

  • Plotting the points $(-5, 7)$, $(-3, 5)$, $(-1, 3)$, $(0, 2)$, $(1, 1)$, $(3, -1)$, and $(5, -3)$ and drawing a straight line yields a decreasing linear graph.

• Graphing Linear Fractions (Try It Exercise 1):

  • Target equation:     $y = \frac{1}{2}x + 2$

  • To ensure integer calculations for $y$, select even values for $x$:

  • For $x = -4$:       $y = \frac{1}{2}(-4) + 2 = -2 + 2 = 0$       Ordered pair: $(-4, 0)$

  • For $x = -2$:       $y = \frac{1}{2}(-2) + 2 = -1 + 2 = 1$       Ordered pair: $(-2, 1)$

  • For $x = 0$:       $y = \frac{1}{2}(0) + 2 = 0 + 2 = 2$       Ordered pair: $(0, 2)$

  • For $x = 2$:       $y = \frac{1}{2}(2) + 2 = 1 + 2 = 3$       Ordered pair: $(2, 3)$

  • For $x = 4$:       $y = \frac{1}{2}(4) + 2 = 2 + 2 = 4$       Ordered pair: $(4, 4)$

Graphing Utilities and Viewing Windows

• Calculator Input Standards:

  • Most graphing calculator platforms (such as the TI-84 Plus) require equations to be isolated and expressed in terms of the dependent variable:     $y = f(x)$

  • If an equation is in implicit form (such as $Ax + By = C$), it must be algebraically rearranged to explicit slope-intercept form before input.

• Standard Viewing Windows and Boundary Limitations:

  • The default factory viewing window on a standard TI-84 Plus displays:     $\text{Xmin} = -10$     $\text{Xmax} = 10$     $\text{Xscl} = 1$     $\text{Ymin} = -10$     $\text{Ymax} = 10$     $\text{Yscl} = 1$     $\text{Xres} = 1$

  • When entering an equation with large intercepts, such as:     $y = 2x - 20$

  • The standard window clips the graph, showing only a small diagonal segment in the lower right of the screen, hiding the actual $x$- and $y$-intercepts.

• Window Reconfiguration Strategy:

  • To display both axis intersections for $y = 2x - 20$, expand the negative vertical range and positive horizontal range:     $\text{Xmin} = -5$     $\text{Xmax} = 15$     $\text{Xscl} = 1$     $\text{Ymin} = -30$     $\text{Ymax} = 10$     $\text{Yscl} = 1$     $\text{Xres} = 1$

  • In this customized window, the line clearly crosses the horizontal axis at $x = 10$ and the vertical axis at $y = -20$.

• Worked Example: Graphing Rational Coefficients with Technology:

  • Target equation:     $y = -\frac{2}{3}x + \frac{4}{3}$

  • Calculator entry into function menu:     $Y_1 = -(2/3)X + (4/3)$

  • Configuring the window boundaries to view intercepts centered in the display:     $\text{Xmin} = -3$     $\text{Xmax} = 4$     $\text{Ymin} = -3$     $\text{Ymax} = 3$

  • The resulting plot clearly demonstrates a downward-sloping linear graph intersecting the horizontal axis at $x = 2$ and the vertical axis at $y = \frac{4}{3} \approx 1.33$.

Determining x-Intercepts and y-Intercepts

• Definitions of Intercepts:

  • $x$-intercept: The specific coordinate point where a graph crosses or touches the $x$-axis. Because every point along the horizontal axis has zero vertical displacement, the $y$-coordinate of any $x$-intercept is identically zero: $(x, 0)$.

  • $y$-intercept: The specific coordinate point where a graph crosses or touches the $y$-axis. Because every point along the vertical axis has zero horizontal displacement, the $x$-coordinate of any $y$-intercept is identically zero: $(0, y)$.

• Algebraic Method for Computing Intercepts:

  • To solve for the $x$-intercept: Substitute $y = 0$ into the equation and solve the remaining algebraic equation for $x$.

  • To solve for the $y$-intercept: Substitute $x = 0$ into the equation and solve the remaining algebraic equation for $y$.

• Walkthrough: Intercepts of $y = 3x - 1$:

  • Finding the $x$-intercept:     $0 = 3x - 1$     $1 = 3x$     $x = \frac{1}{3}$     $x$-intercept coordinate:     $\left(\frac{1}{3}, 0\right)$

  • Finding the $y$-intercept:     $y = 3(0) - 1$     $y = -1$     $y$-intercept coordinate:     $(0, -1)$

• Worked Example: Intercepts of $y = -3x - 4$:

  • Finding the $x$-intercept by substituting $y = 0$:     $0 = -3x - 4$     $4 = -3x$     $x = -\frac{4}{3}$     $x$-intercept coordinate:     $\left(-\frac{4}{3}, 0\right)$

  • Finding the $y$-intercept by substituting $x = 0$:     $y = -3(0) - 4$     $y = -4$     $y$-intercept coordinate:     $(0, -4)$

  • Graphing by intercepts: Plotting the two calculated points $\left(-\frac{4}{3}, 0\right)$ and $(0, -4)$ and drawing a straight line through them completely defines the linear function.

• Evaluating Intercepts for Fractional Linear Equations (Try It Exercise 2):

  • Given equation:     $y = -\frac{3}{4}x + 3$

  • Finding the $x$-intercept ($y = 0$):     $0 = -\frac{3}{4}x + 3$     $\frac{3}{4}x = 3$     $x = 3 \times \frac{4}{3} = 4$     $x$-intercept: $(4, 0)$

  • Finding the $y$-intercept ($x = 0$):     $y = -\frac{3}{4}(0) + 3 = 3$     $y$-intercept: $(0, 3)$

The Distance Formula

• Geometric Foundation and Derivation:

  • Derived directly from the Pythagorean Theorem for right-angled triangles:     $a^2 + b^2 = c^2$

  • In a right triangle, $a$ and $b$ represent the lengths of the two perpendicular legs adjacent to the 90^\n\ncirc angle, and $c$ represents the length of the hypotenuse opposite the right angle:     $c = \sqrt{a^2 + b^2}$

  • For two arbitrary points in the Cartesian plane, $(x_1, y_1)$ and $(x_2, y_2)$, construct a right triangle whose hypotenuse connects the two points.

  • The third vertex of this right triangle is located at $(x_2, y_1)$.

  • The horizontal leg length is the absolute horizontal displacement:     $a = |x_2 - x_1|$

  • The vertical leg length is the absolute vertical displacement:     $b = |y_2 - y_1|$

  • Substituting these lengths into the Pythagorean Theorem where distance $d = c$:     $d^2 = (|x_2 - x_1|)^2 + (|y_2 - y_1|)^2$

  • Because any real number squared is non-negative ($|u|^2 = u^2$), absolute value brackets are eliminated, resulting in:     $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• Formal Algebraic Statement of the Distance Formula:

  • For any two points $(x_1, y_1)$ and $(x_2, y_2)$ in a Cartesian plane:     $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• Worked Example: Distance Between Two Coordinates:

  • Given endpoints: $(-3, -1)$ and $(2, 3)$.

  • Setting up the right triangle:

  • Horizontal leg $a = |2 - (-3)| = |5| = 5$

  • Vertical leg $b = |3 - (-1)| = |4| = 4$

• Applying the distance formula step-by-step:     $d = \sqrt{(2 - (-3))^2 + (3 - (-1))^2}$     $d = \sqrt{(5)^2 + (4)^2}$     $d = \sqrt{25 + 16}$     $d = \sqrt{41}$

• Exact radical value: $\sqrt{41}$

• Decimal approximation: $\approx 6.403$

  • Worked Example: Applied Multi-Stop Route Analysis (Tracie's Travel):

• Journey description: Tracie sets out from Elmhurst, IL, to Franklin Park, completing errands along street intersections.

• Establishing coordinate grid system:

  • Setting the origin $(0, 0)$ at the Elmhurst starting point.

  • Each unit on the grid represents $1\text{,}000\,\text{ft}$.

  • Stop 1: $1$ block east, $1$ block north: $(1, 1)$

  • Stop 2: $5$ blocks east, $1$ block north: $(5, 1)$

  • Stop 3: $8$ blocks east, $3$ blocks north: $(8, 3)$

  • Final Stop (Franklin Park): $8$ blocks east, $7$ blocks north: $(8, 7)$

• Step-by-step road travel distances along perpendicular streets:

  • From $(0, 0)$ to $(1, 1)$: $1\,\text{unit east} + 1\,\text{unit north} = 1\text{,}000\,\text{ft} + 1\text{,}000\,\text{ft} = 2\text{,}000\,\text{ft}$

  • From $(1, 1)$ to $(5, 1)$: $4\,\text{units east} = 4\text{,}000\,\text{ft}$

  • From $(5, 1)$ to $(8, 3)$: $3\,\text{units east} + 2\,\text{units north} = 3\text{,}000\,\text{ft} + 2\text{,}000\,\text{ft} = 5\text{,}000\,\text{ft}$

  • From $(8, 3)$ to $(8, 7)$: $4\,\text{units north} = 4\text{,}000\,\text{ft}$

• Summary of driving segments:

• Mileage conversion of total driving distance:     $15\text{,}000\,\text{ft} \div 5\text{,}280\,\text{ft/mi} \approx 2.84\,\text{mi}$

• Straight-Line ("As the Crow Flies") Distance from origin $(0, 0)$ to destination $(8, 7)$:     $d = \sqrt{(8 - 0)^2 + (7 - 0)^2}$     $d = \sqrt{8^2 + 7^2}$     $d = \sqrt{64 + 49}$     $d = \sqrt{113} \approx 10.63\,\text{units}$

• Converting straight-line distance to feet and miles:     $10.63014\,\text{units} \times 1\text{,}000\,\text{ft/unit} = 10\text{,}630.14\,\text{ft}$     $10\text{,}630.14\,\text{ft} \div 5\text{,}280\,\text{ft/mi} \approx 2.01\,\text{mi}$

• Direct comparison:

  • The straight diagonal distance ($2.01\,\text{mi}$) is significantly shorter than the street grid path ($2.84\,\text{mi}$) because it corresponds to the hypotenuse of the right triangle whose legs are the horizontal and vertical road spans.

The Midpoint Formula

• Conceptual Definition of Midpoint:

  • The midpoint of a line segment is the exact point situated halfway between the segment's two endpoints.

  • The midpoint divides the line segment into two equal, congruent segments.

• Algebraic Statement of the Midpoint Formula:

  • Given two endpoints $(x_1, y_1)$ and $(x_2, y_2)$, the coordinates of the midpoint $M$ are the arithmetic averages of the respective coordinates:     $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

• Worked Example: Finding the Midpoint of a Segment:

  • Given endpoints: $(7, -2)$ and $(9, 5)$.

  • Computing the coordinates:     $M = \left(\frac{7 + 9}{2}, \frac{-2 + 5}{2}\right)$     $M = \left(\frac{16}{2}, \frac{3}{2}\right)$     $M = \left(8, \frac{3}{2}\right) = (8, 1.5)$

• Worked Example: Locating the Center of a Circle:

  • A circle's diameter is a straight chord passing through the center connecting two boundary points.

  • Therefore, the center of any circle is mathematically identical to the midpoint of its diameter.

  • Given diameter endpoints: $(-1, -4)$ and $(5, -4)$.

  • Computing the center coordinates:     $\text{Center} = \left(\frac{-1 + 5}{2}, \frac{-4 + (-4)}{2}\right)$     $\text{Center} = \left(\frac{4}{2}, \frac{-8}{2}\right)$     $\text{Center} = (2, -4)$

Section Exercises, Methodological Proofs, and Extensions

• Verbal Analysis:

  • Points Outside Quadrants:

  • A point does not lie in any of the four quadrants if either its $x$-coordinate is zero, its $y$-coordinate is zero, or both.

  • When $x = 0$, the point lies on the $y$-axis; when $y = 0$, the point lies on the $x$-axis; when $(0, 0)$, it lies at the origin.

  • Algebraic Determination of Intercepts:

  • To find the $x$-intercept, set $y = 0$ in the equation and isolate $x$.

  • To find the $y$-intercept, set $x = 0$ in the equation and isolate $y$.

  • Geometric Meaning of the $y$-Intercept:

  • The $y$-intercept represents the vertical location where the curve or line crosses the vertical axis; it indicates the baseline value of the dependent variable when the independent variable is zero.

  • Order of Operations in the Distance Formula:

  • In the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:

    1. Perform subtractions inside each parenthesis: $(x_2 - x_1)$ and $(y_2 - y_1)$.

    2. Square each resulting difference: $(x_2 - x_1)^2$ and $(y_2 - y_1)^2$.

    3. Sum the two squared terms together.

    4. Compute the principal (positive) square root of the final sum.

• Algebraic Intercept Calculations:

  • For the fractional linear equation:     $2x - \frac{2}{3} = \frac{3}{4}y + 3$

  • $x$-intercept ($y = 0$):       $2x - \frac{2}{3} = 3$       $2x = 3 + \frac{2}{3} = \frac{11}{3}$       $x = \frac{11}{6}$       Ordered pair: $\left(\frac{11}{6}, 0\right)$

  • $y$-intercept ($x = 0$):       $2(0) - \frac{2}{3} = \frac{3}{4}y + 3$       $-\frac{2}{3} - 3 = \frac{3}{4}y$       $-\frac{11}{3} = \frac{3}{4}y$       $y = -\frac{11}{3} \times \frac{4}{3} = -\frac{44}{9}$       Ordered pair: $\left(0, -\frac{44}{9}\right)$

• Graphical Analysis of Coordinate Points:

  • Points mapped on plane: $A(-3, 2)$, $B(1, 3)$, $C(4, 0)$.

- Point $A(-3, 2)$ lies in Quadrant II.
- Point $B(1, 3)$ lies in Quadrant I.
- Point $C(4, 0)$ lies directly on the $x$-axis.

• Distance and Midpoint Analysis of Plotted Line Segment:

  • Consider the segment connecting endpoints $(-3, 4)$ and $(5, 2)$:

• Distance between endpoints:     $d = \sqrt{(5 - (-3))^2 + (2 - 4)^2} = \sqrt{8^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68} \approx 8.246$

• Coordinates of the segment midpoint:     $M = \left(\frac{-3 + 5}{2}, \frac{4 + 2}{2}\right) = \left(\frac{2}{2}, \frac{6}{2}\right) = (1, 3)$

• Distance from origin $(0, 0)$ to $(-3, 4)$:     $d_1 = \sqrt{(-3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

• Distance from origin $(0, 0)$ to $(5, 2)$:     $d_2 = \sqrt{(5 - 0)^2 + (2 - 0)^2} = \sqrt{25 + 4} = \sqrt{29} \approx 5.385$

• Comparing distances from the origin:

  • Because $5 < 5.385$, point $(-3, 4)$ is closer to the origin than point $(5, 2)$.

  • Calculator Menu Functions for Finding Intercepts:

• Finding $y$-intercept using 2nd CALC 1:value:

  • In the Y= menu, input the linear equation.

  • Press GRAPH.

  • Select 2nd CALC, choose 1:value, and hit ENTER.

  • Enter x = 0 at the prompt; the calculator outputs the exact corresponding $y$-intercept value.

• Finding $x$-intercept using 2nd CALC 2:zero:

  • In the Y= menu, input the linear function.

  • Press 2nd CALC and select 2:zero.

  • Set Left Bound? by moving the cursor to the left of the $x$-intercept and pressing ENTER.

  • Set Right Bound? by moving the cursor to the right of the $x$-intercept and pressing ENTER.

  • Set Guess? by positioning the cursor near the intercept between the bounds and pressing ENTER.

  • The calculator performs numerical convergence and outputs the zero coordinate $(x, 0)$.

  • Geometric Proof: Diagonals of a Rectangle:

• Given a rectangle with vertices located at:     $A(-6, -1)$, $B(10, -1)$, $C(10, 5)$, $D(-6, 5)$.

• Proof that the diagonals are equal in length:

  • Diagonal 1 connects $A(-6, -1)$ to $C(10, 5)$:       $d_{AC} = \sqrt{(10 - (-6))^2 + (5 - (-1))^2} = \sqrt{16^2 + 6^2} = \sqrt{256 + 36} = \sqrt{292}$

  • Diagonal 2 connects $B(10, -1)$ to $D(-6, 5)$:       $d_{BD} = \sqrt{(-6 - 10)^2 + (5 - (-1))^2} = \sqrt{(-16)^2 + 6^2} = \sqrt{256 + 36} = \sqrt{292}$

  • Because $d_{AC} = d_{BD} = \sqrt{292} \approx 17.088$, the diagonals are congruent.

• Midpoints of the Diagonals:

  • Midpoint of diagonal $AC$:       $M_{AC} = \left(\frac{-6 + 10}{2}, \frac{-1 + 5}{2}\right) = \left(\frac{4}{2}, \frac{4}{2}\right) = (2, 2)$

  • Midpoint of diagonal $BD$:       $M_{BD} = \left(\frac{10 + (-6)}{2}, \frac{-1 + 5}{2}\right) = \left(\frac{4}{2}, \frac{4}{2}\right) = (2, 2)$

  • Both diagonals intersect precisely at their shared midpoint $(2, 2)$, proving that the diagonals of a rectangle bisect each other.

  • Applied Real-World Problems:

• Direct Commute Distance vs. Road Travel:

  • A commuter drives $10\,\text{mi}$ east and $5\,\text{mi}$ north.

  • Direct hypotenuse distance:       $d = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} \approx 11.2\,\text{mi}$

  • Distance saved compared to the $15\,\text{mi}$ road commute:       $\Delta d = 15 - 11.2 = 3.8\,\text{mi}$

• Guy Wire Length on a Building:

  • A wire extends from the roof of a $50\,\text{ft}$ tall building to a ground anchor $20\,\text{ft}$ away from the base.

- Applying the distance formula:

      $L = \sqrt{20^2 + 50^2} = \sqrt{400 + 2500} = \sqrt{2900} \approx 53.85\,\text{ft}$ - Rounded to the nearest whole foot, the required guy wire length is $54\,\text{ft}$.

• Rental Truck Linear Cost Model:

  • Daily rental base fee: \n\n75

  • Per-mile operating cost: \n\n0.20/\text{mi}

  • Expressing total daily cost $C$ as a function of driven miles $m$:       $C(m) = 0.20m + 75$

  • Evaluating total daily cost for traveling $m = 70\,\text{mi}$:       $C(70) = 0.20(70) + 75 = 14 + 75 = 89$

  • Total daily cost is \n\n89.

Linear Equations in One Variable Overview

• Section 2.2 Core Learning Objectives:

  • Solve single-variable equations algebraically.

  • Formulate and solve rational equations.

  • Formulate equations of lines given geometric properties.

  • Determine whether pairs of linear equations represent parallel or perpendicular lines.

  • Construct equations of lines parallel or perpendicular to specified lines.

• Application Scenario: Caroline's College Vacation Budget:

  • Caroline is a full-time university student planning a spring break trip costing approximately \n\n2\text{,}500.

  • Available savings: Initial deposit of \n\n400 on January 15.

  • Employment wage: Part-time bank teller earning \n\n15.00/\text{hr}.

  • Target trip date: Spring break begins March 20.

  • Mathematical breakdown of required earnings:

  • Amount needed from wages:       \n\n2\text{,}500 - \n\n400 = \n\n2\text{,}100

  • Total working hours required:       $\frac{2\text{,}100}{15.00} = 140\,\text{hours}$

  • If Caroline works $4\,\text{hours/day}$:       $\frac{140\,\text{hours}}{4\,\text{hours/day}} = 35\,\text{workdays}$

  • These conditions form a linear model relating accumulated earnings over time, establishing the algebraic framework for single-variable linear models.