5.1 Linear Relations, Rate of Change & Slope

5.1 Linear Relations, Rate of Change & Slope

In the Ontario Grade 9 Mathematics Curriculum (MTH1W), linear relations serve as a cornerstone for algebraic reasoning and contextual modeling. Understanding how quantities change relative to one another allows educators and students to model real-world phenomena—ranging from uniform motion and cell phone pricing to financial depreciation.

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1. The Cartesian Plane & Relations vs. Functions

A relation is any set of ordered pairs $(x, y)$ that associates elements of an independent variable (input, domain) with elements of a dependent variable (output, range).

• Independent Variable ($x$): Plotted on the horizontal axis ($x$-axis). It represents the variable being controlled or varied independently (e.g., time, distance, data usage).
• Dependent Variable ($y$): Plotted on the vertical axis ($y$-axis). Its value depends on or responds to changes in the independent variable (e.g., total cost, height, volume).

In the Cartesian coordinate system, the origin $(0,0)$ is the intersection of the $x$-axis and $y$-axis. Every point is identified uniquely by its coordinates $(x, y)$. While all linear relations covered in Grade 9 math are functions (where every input $x$ corresponds to exactly one output $y$), vertical lines represent relations that are not functions.

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2. First Differences and Linearity

To determine whether a relationship presented in a table of values is linear without plotting a graph, we analyze first differences.

The First Difference Rule

A relation is linear if, for equal increments of the independent variable ($\Delta x$), the first differences of the dependent variable ($\Delta y$) are constant.

Consider the two tables below where $\Delta x = 1$:

Contrast this with a non-linear relation ($y = x^2$):

Because the first differences are not constant, $y = x^2$ is non-linear. The constant second differences indicate a quadratic relation.

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3. Slope as Constant Rate of Change

The slope ($m$) of a line measures its steepness and direction. Mathematically, slope is defined as the constant rate of change of the dependent variable with respect to the independent variable.

$$\text{Rate of Change} = m = \frac{\text{rise}}{\text{run}} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$$

Where $(x_1, y_1)$ and $(x_2, y_2)$ are any two distinct points on the line, with $x_1 \neq x_2$.

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Four Slope Classifications

1. Positive Slope ($m > 0$): The line slants upward from left to right. As $x$ increases, $y$ increases. Example: Hourly earnings.
2. Negative Slope ($m < 0$): The line slants downward from left to right. As $x$ increases, $y$ decreases. Example: Water draining from a reservoir.
3. Zero Slope ($m = 0$): The line is horizontal ($\Delta y = 0$). The value of $y$ remains constant regardless of $x$. Equation form: $y = c$.
4. Undefined Slope: The line is vertical ($\Delta x = 0$). Division by zero is mathematically undefined. Equation form: $x = c$.

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4. Slope-Intercept Form ($y = mx + b$) & Graphing

The slope-intercept form of a linear equation is:

$$y = mx + b$$

• $m$ (Slope / Rate of Change): Tells how much $y$ changes for every 1-unit increase in $x$.
• $b$ (Initial Value / $y$-intercept): The value of $y$ when $x = 0$. The line crosses the $y$-axis at the point $(0, b)$.

Graphing Strategies

1. Using Slope and $y$-intercept:

   • Step 1: Plot the $y$-intercept $(0, b)$ on the Cartesian plane.
   • Step 2: Use the slope $m = \frac{\text{rise}}{\text{run}}$ to locate a second point by moving $\text{rise}$ units vertically and $\text{run}$ units horizontally.
   • Step 3: Draw a straight line through the points extending across the grid.

2. Using $x$- and $y$-intercepts:

   • Find $y$-intercept by substituting $x = 0$ and solving for $y$.
   • Find $x$-intercept by substituting $y = 0$ and solving for $x$.
   • Plot both intercepts and connect them with a straight line.

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5. Step-by-Step Worked Examples

Worked Example 1: Finding Slope and Equation from Two Points

Problem: A straight line passes through the points $A(-3, 11)$ and $B(5, -5)$. Determine the slope, find the $y$-intercept, and write the linear equation in $y = mx + b$ form.

Solution:

• Step 1: Calculate the slope ($m$) using the slope formula.

  $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - 11}{5 - (-3)} = \frac{-16}{5 + 3} = \frac{-16}{8} = -2$$

• Step 2: Calculate the $y$-intercept ($b$). Substitute $m = -2$ and point $A(-3, 11)$ into $y = mx + b$:

  $$11 = (-2)(-3) + b$$ $$11 = 6 + b$$ $$b = 11 - 6 = 5$$

• Step 3: State the final linear equation.

  $$y = -2x + 5$$

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Worked Example 2: Contextual Water Tank Problem

Problem: A municipal water storage tank contains $1,200 \text{ L}$ of water. A valve is opened, draining the water at a constant rate. After $4 \text{ hours}$, $720 \text{ L}$ of water remains in the tank.

1. Calculate the rate of change of water volume.
2. Write a linear equation relating volume ($V$) in liters to time ($t$) in hours.
3. Determine how many hours it takes for the tank to empty completely.

Solution:

• Step 1: Identify coordinates and calculate rate of change. At $t = 0 \text{ hrs}$, $V = 1200 \text{ L} \implies (0, 1200)$. At $t = 4 \text{ hrs}$, $V = 720 \text{ L} \implies (4, 720)$.

  $$\text{Rate of Change} = m = \frac{720 - 1200}{4 - 0} = \frac{-480}{4} = -120 \text{ L/hr}$$

  The tank loses $120 \text{ L}$ per hour.

• Step 2: Write the linear equation. The initial value $b = 1200$.

  $$V(t) = -120t + 1200$$

• Step 3: Determine time when tank is empty ($V = 0$).

  $$0 = -120t + 1200$$ $$120t = 1200$$ $$t = \frac{1200}{120} = 10 \text{ hours}$$

  It takes $10 \text{ hours}$ for the tank to drain completely.

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6. Pedagogical Insights for Ontario Educators

When preparing students for Ontario MPT questions on linear relations:

• Address the common student error of calculating $\frac{\text{run}}{\text{rise}}$ instead of $\frac{\text{rise}}{\text{run}}$. Remind candidates that vertical movement (dependent variable) always forms the numerator.
• Ensure subtraction order consistency in $m = \frac{y_2 - y_1}{x_2 - x_1}$. Subtracting $y_2 - y_1$ in the numerator requires $x_2 - x_1$ in the denominator.