The Cartesian Plane, Linear Graphs, and Gradient

The Cartesian Plane and Coordinate Geometry

• Structure of the Cartesian Plane:
  • The number plane is formed by two perpendicular intersecting number lines.
  • The horizontal line is designated as the $x$-axis.
  • The vertical line is designated as the $y$-axis.
  • The point where the two axes intersect is called the origin, having coordinates $(0, 0)$.
  • Coordinates are expressed as an ordered pair $(x, y)$, where $x$ represents the horizontal displacement and $y$ represents the vertical displacement relative to the origin.

• Reading Coordinates from the Plane:
  • Coordinates of labeled points on a standard grid:
    • Point $P$ has coordinates $(3, -2)$.
    • Point $Q$ has coordinates $(-2, -1)$.
    • Point $R$ has coordinates $(1, 2)$.
    • Point $S$ has coordinates $(0, -1)$.
    • Point $T$ has coordinates $(-3, 3)$.

• Plotting Given Coordinates:
  • $A(3, 4)$: Located $3$ units along the positive $x$-axis and $4$ units along the positive $y$-axis.
  • $B(4, 3)$: Located $4$ units along the positive $x$-axis and $3$ units along the positive $y$-axis.
  • $C(2, 6)$: Located $2$ units along the positive $x$-axis and $6$ units along the positive $y$-axis.
  • $D(-3, 2)$: Located $3$ units along the negative $x$-axis and $2$ units along the positive $y$-axis.
  • $E(5, -3)$: Located $5$ units along the positive $x$-axis and $3$ units along the negative $y$-axis.
  • $F(-2, -4)$: Located $2$ units along the negative $x$-axis and $4$ units along the negative $y$-axis.
  • $G(0, 5)$: Located on the vertical $y$-axis at $y = 5$.
  • $H(-3, 0)$: Located on the horizontal $x$-axis at $x = -3$.

Linear Relations and Plotting Linear Graphs

• Variables in Linear Relationships:

  • The independent variable is the input value that can be chosen freely, conventionally plotted on the horizontal axis ($x$).
  • The dependent variable is the output value determined by the relationship, conventionally plotted on the vertical axis ($y$).

• Plotting via Table of Values:

  • For the linear equation $y = x - 2$:
    • $x$ is the independent variable.
    • $y$ is the dependent variable.
    • Table of values constructed for $x \in \{-3, -2, -1, 0, 1, 2, 3\}$:
      • When $x = -3$, $y = -3 - 2 = -5$
      • When $x = -2$, $y = -2 - 2 = -4$
      • When $x = -1$, $y = -1 - 2 = -3$
      • When $x = 0$, $y = 0 - 2 = -2$
      • When $x = 1$, $y = 1 - 2 = -1$
      • When $x = 2$, $y = 2 - 2 = 0$
      • When $x = 3$, $y = 3 - 2 = 1$
    • Each ordered pair $(x, y)$ is plotted on the Cartesian plane and joined with a continuous straight line that extends indefinitely in both directions.

• Systematic Evaluation of Elementary Linear Equations:

  • $y = x$:
    • Independent variable: $x$; dependent variable: $y$.
    • Coordinates across domain $x \in \{-3, -2, -1, 0, 1, 2, 3\}$: $(-3, -3)$, $(-2, -2)$, $(-1, -1)$, $(0, 0)$, $(1, 1)$, $(2, 2)$, $(3, 3)$.
  • $y = 2x$:
    • Independent variable: $x$; dependent variable: $y$.
    • Coordinates: $(-3, -6)$, $(-2, -4)$, $(-1, -2)$, $(0, 0)$, $(1, 2)$, $(2, 4)$, $(3, 6)$.
  • $y = \frac{1}{2}x$:
    • Independent variable: $x$; dependent variable: $y$.
    • Coordinates: $(-3, -1.5)$, $(-2, -1)$, $(-1, -0.5)$, $(0, 0)$, $(1, 0.5)$, $(2, 1)$, $(3, 1.5)$.
  • $y = -2x$:
    • Independent variable: $x$; dependent variable: $y$.
    • Coordinates: $(-3, 6)$, $(-2, 4)$, $(-1, 2)$, $(0, 0)$, $(1, -2)$, $(2, -4)$, $(3, -6)$.
  • $y = 2x + 3$:
    • Independent variable: $x$; dependent variable: $y$.
    • Coordinates: $(-3, -3)$, $(-2, -1)$, $(-1, 1)$, $(0, 3)$, $(1, 5)$, $(2, 7)$, $(3, 9)$.
  • $y = -2x + 3$:
    • Independent variable: $x$; dependent variable: $y$.
    • Coordinates: $(-3, 9)$, $(-2, 7)$, $(-1, 5)$, $(0, 3)$, $(1, 1)$, $(2, -1)$, $(3, -3)$.
  • $y = \frac{1}{4}x + 1$:
    • Independent variable: $x$; dependent variable: $y$.
    • Coordinates: $(-3, 0.25)$, $(-2, 0.5)$, $(-1, 0.75)$, $(0, 1)$, $(1, 1.25)$, $(2, 1.5)$, $(3, 1.75)$.
  • $y = -\frac{1}{4}x + 1$:
    • Independent variable: $x$; dependent variable: $y$.
    • Coordinates: $(-3, 1.75)$, $(-2, 1.5)$, $(-1, 1.25)$, $(0, 1)$, $(1, 0.75)$, $(2, 0.5)$, $(3, 0.25)$.

• Effects of Equation Components on Line Behavior:

  • Steepness: Arranging the graphs of $y = x$, $y = 2x$, and $y = \frac{1}{2}x$ in order of increasing steepness gives $y = \frac{1}{2}x$, $y = x$, $y = 2x$. The magnitude of the coefficient of $x$ controls the degree of steepness of the line.
  • Direction of Slope: Comparing the graphs of $y = 2x$ and $y = -2x$ shows that $y = 2x$ slopes forwards (rises from left to right) while $y = -2x$ slopes backwards (falls from left to right). The sign (positive or negative) of the coefficient of $x$ controls whether the line slopes forwards or backwards.
  • Axis Intercept: Comparing the graphs of $y = 2x$, $y = 2x + 3$, and $y = \frac{1}{4}x + 1$ reveals that each line intersects the vertical axis at $y = 0$, $y = 3$, and $y = 1$ respectively. The constant term added or subtracted controls where the graph cuts the $y$-axis.

The Equation of a Line

• Formal Definition:

  • The equation of a line is an equation which connects the $x$ and $y$-coordinates of all points on the line.

• Expressing Equations in Words:

  • $y = 3x + 7$: Connects the $x$ and $y$ values for every point on the line where the $y$-coordinate is three times the $x$-coordinate, plus $7$.
  • $y = x + 3$: The $y$-coordinate is $3$ more than the $x$-coordinate for every point on the line.
  • $y = 5x$: The $y$-coordinate is five times the $x$-coordinate for every point on the line.
  • $y = 2x - 6$: The $y$-coordinate is two times the $x$-coordinate, minus $6$, for every point on the line.
  • $x + y = 5$: The sum of the $x$-coordinate and $y$-coordinate is always equal to $5$ for every point on the line.

• Deducing Equations by Inspection from Tables of Values:

  • Values: $x \in \{1, 2, 3, 4, 5\}$, $y \in \{-3, -6, -9, -12, -15\}$.
    • Each $y$-value is $-3$ times its corresponding $x$-value, so the equation is $y = -3x$.
  • Values: $x \in \{1, 2, 3, 4, 5\}$, $y \in \{5, 4, 3, 2, 1\}$.
    • The sum of each $x$ and $y$-value is always $6$, so the equation is $x + y = 6$ (or $y = 6 - x$).
  • Values: $x \in \{1, 2, 3, 4, 5\}$, $y \in \{4, 8, 12, 16, 20\}$.
    • Each $y$-value is $4$ times its corresponding $x$-value, so the equation is $y = 4x$.
  • Values: $x \in \{2, 3, 4, 5, 6\}$, $y \in \{4, 5, 6, 7, 8\}$.
    • Each $y$-value is $2$ greater than its corresponding $x$-value, so the equation is $y = x + 2$.
  • Values: $x \in \{0, 1, 2, 3\}$, $y \in \{0, -2, -4, -6\}$.
    • Each $y$-value is $-2$ times its corresponding $x$-value, so the equation is $y = -2x$.
  • Values: $x \in \{0, 1, 2, 3\}$, $y \in \{3, 2, 1, 0\}$.
    • The sum of each $x$ and $y$-value is always $3$, so the equation is $x + y = 3$ (or $y = 3 - x$).
  • Values: $x \in \{1, 2, 3, 4, 5\}$, $y \in \{3, 5, 7, 9, 11\}$.
    • Each $y$-value is twice the $x$-value plus $1$, so the equation is $y = 2x + 1$.
  • Values: $x \in \{1, 2, 3, 4, 5\}$, $y \in \{1, 3, 5, 7, 9\}$.
    • Each $y$-value is twice the $x$-value minus $1$, so the equation is $y = 2x - 1$.

• Point Satisfaction on a Line:

  • A coordinate point lies on the graph of a line if and only if substituting its $x$ and $y$ values into the equation produces a true equality.
  • Evaluating point $(2, 5)$ in equation $y = 4x - 3$:
    • $\text{RHS} = 4(2) - 3 = 8 - 3 = 5 = \text{LHS}$
    • Conclusion: Yes, the point satisfies the equation and lies on the line.
  • Evaluating point $(3, -1)$ in equation $y = 2x - 7$:
    • $\text{RHS} = 2(3) - 7 = 6 - 7 = -1 = \text{LHS}$
    • Conclusion: Yes, the point satisfies the equation and lies on the line.
  • Evaluating point $(-4, 2)$ in equation $y = -x + 3$:
    • $\text{RHS} = -(-4) + 3 = 4 + 3 = 7 \neq \text{LHS}$
  • Conclusion: No, the point does not satisfy the equation and does not lie on the line.

• Evaluating point $(-1, 7)$ in equation $y = -2x + 5$:

  • $\text{RHS} = -2(-1) + 5 = 2 + 5 = 7 = \text{LHS}$
  • Conclusion: Yes, the point satisfies the equation and lies on the line.

• Evaluating point $(4, -1)$ in equation $y = -2x + 7$:

  • $\text{RHS} = -2(4) + 7 = -8 + 7 = -1 = \text{LHS}$
  • Conclusion: Yes, the point satisfies the equation and lies on the line.

• Evaluating point $(1, 6)$ in equation $y = -3x + 5$:

  • $\text{RHS} = -3(1) + 5 = -3 + 5 = 2 \neq \text{LHS}$

• Conclusion: No, the point does not satisfy the equation and does not lie on the line.

Gradient or Slope

• Definition and Formula:
  • Gradient is the comparison of the vertical step to the horizontal step.
  • Fundamental definition formula:         $\text{gradient} = \frac{\text{vertical step}}{\text{horizontal step}}$

• Physical and Real-World Interpretation:
  • Inclination of roads serves as a primary practical manifestation of gradient.
  • For a road modeled with a vertical rise of $1$ over a horizontal run of $6$:         $\text{gradient of road} = \frac{1}{6}$

• Horizontal and Vertical Lines:

  • For a horizontal line, the vertical step is $0$. This yields $0$ in the numerator, meaning:         $\text{gradient of a horizontal line} = 0$
  • For a vertical line, the horizontal step is $0$. This yields $0$ in the denominator. Division by zero is undefined, meaning:         $\text{gradient of a vertical line is undefined}$

• Two-Point Gradient Formula:

  • For any two distinct points $A(x_1, y_1)$ and $B(x_2, y_2)$, the gradient of the line passing through them is:         $\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1}$
  • Worked Example: Finding the gradient of the line segment $[PQ]$ connecting $P(1, 3)$ and $Q(4, -2)$:
    • Assign coordinates: $x_1 = 1$, $y_1 = 3$, $x_2 = 4$, $y_2 = -2$.
    • Calculate:             $\text{gradient of } [PQ] = \frac{y_2 - y_1}{x_2 - x_1}$$\text{gradient of } [PQ] = \frac{-2 - 3}{4 - 1}$$\text{gradient of } [PQ] = \frac{-5}{3}$$\text{gradient of } [PQ] = -\frac{5}{3}$

The Gradient-Intercept Form ($y = mx + c$)

• General Formulation:

  • The equation $y = mx + c$ describes a straight line with gradient $m$ and $y$-intercept $c$.
  • The point where a straight line meets the $x$-axis is called its $x$-intercept (where $y = 0$).
  • The point where a straight line meets the $y$-axis is called its $y$-intercept (where $x = 0$).

• Direct Parameter Extraction from Linear Equations:

  • $y = 3x - 2$ has $m = 3$ and $c = -2$, so gradient is $3$ and $y$-intercept is $-2$.
  • $y = 7 - 2x$, rewritten as $y = -2x + 7$, has $m = -2$ and $c = 7$, so gradient is $-2$ and $y$-intercept is $7$.
  • $y = 0$, rewritten as $y = 0x + 0$, has $m = 0$ and $c = 0$, so gradient is $0$ and $y$-intercept is $0$.
  • $y = 4x + 8 \implies m = 4$, $c = 8$.
  • $y = -3x + 2 \implies m = -3$, $c = 2$.
  • $y = 6 - x \implies y = -x + 6 \implies m = -1$, $c = 6$.
  • $y = -2x + 3 \implies m = -2$, $c = 3$.
  • $y = -2 \implies y = 0x - 2 \implies m = 0$, $c = -2$.
  • $y = 11 - 3x \implies y = -3x + 11 \implies m = -3$, $c = 11$.
  • $y = \frac{1}{2}x - 5 \implies m = \frac{1}{2}$, $c = -5$.
  • $y = 3 - \frac{3}{2}x \implies y = -\frac{3}{2}x + 3 \implies m = -\frac{3}{2}$, $c = 3$.
  • $y = \frac{2}{5}x + \frac{4}{5} \implies m = \frac{2}{5}$, $c = \frac{4}{5}$.
  • $y = \frac{x + 1}{2} = \frac{1}{2}x + \frac{1}{2} \implies m = \frac{1}{2}$, $c = \frac{1}{2}$.
  • $y = \frac{2x - 10}{5} = \frac{2}{5}x - 2 \implies m = \frac{2}{5}$, $c = -2$.
  • $y = \frac{11 - 3x}{2} = -\frac{3}{2}x + \frac{11}{2} \implies m = -\frac{3}{2}$, $c = \frac{11}{2}$.

• Constructing Linear Equations Given Gradient and Intercept:

  • Gradient $2$ and $y$-intercept $11 \implies y = 2x + 11$.
  • Gradient $4$ and $y$-intercept $-6 \implies y = 4x - 6$.
  • Gradient $-3$ and $y$-intercept $\frac{1}{2} \implies y = -3x + \frac{1}{2}$.
  • Gradient $-1$ and $y$-intercept $4 \implies y = -x + 4$.
  • Gradient $0$ and $y$-intercept $7 \implies y = 7$.
  • Gradient $\frac{2}{3}$ and $y$-intercept $6 \implies y = \frac{2}{3}x + 6$.

Parallel Lines Investigation and Principles

• Analysis of Coplanar Lines:
  • Consider five distinct straight lines graphed simultaneously:

*   Given the equations (unordered): $y = 3x + 1$, $y = 2x + 3$, $y = 2x + 1$, $y = 2x$, $y = 2x - 2$.
*   Complete identification of lines from graph inspection:
    *   Graph label $a$: Equation $y = 2x$, Gradient $= 2$, $y$-intercept $= 0$.
    *   Graph label $b$: Equation $y = 2x + 3$, Gradient $= 2$, $y$-intercept $= 3$.
    *   Graph label $c$: Equation $y = 2x + 1$, Gradient $= 2$, $y$-intercept $= 1$.
    *   Graph label $d$: Equation $y = 2x - 2$, Gradient $= 2$, $y$-intercept $= -2$.
    *   Graph label $e$: Equation $y = 3x + 1$, Gradient $= 3$, $y$-intercept $= 1$.

• Identification of Non-Parallel Line:

  • Line $e$ is not parallel to the remaining lines.
  • Explanation: Line $e$ has a gradient of $3$, while lines $a$, $b$, $c$, and $d$ each have an identical gradient of $2$. Because line $e$ has a steeper slope, it intersects the other lines rather than remaining equidistant.

• General Rule for Parallel Lines:

  • Observed pattern: All parallel lines share identical values for their gradient ($m$).
  • General algebraic rule: Two straight lines in the form $y = ax + b$ and $y = cx + d$ are parallel if and only if:         $a = c$
  • If $a = c$ and $b \neq d$, the lines are distinct parallel lines.
  • If $a = c$ and $b = d$, the lines are identical (coincident).

• Geometric Justification:

  • The gradient defines the ratio of vertical change per unit horizontal change ($\frac{\text{vertical step}}{\text{horizontal step}}$).
  • If two lines possess the same gradient, they change vertically at the exact same rate over any given horizontal run.
  • Consequently, the vertical distance separating the two lines remains constant across all values of $x$, ensuring that the lines will never intersect in the coordinate plane.

Graphing Techniques for Linear Equations

• Standard Methodologies:

  • To draw the graph of $y = mx + c$, three primary methods exist:
    1. Construct a table of values using at least $3$ points and then graph the points.
    2. Use the $y$-intercept and gradient.
    3. Use the axes intercepts ($x$-intercept and $y$-intercept).

• Graphing via Gradient and $y$-Intercept:

  • Procedural rule: Locate the $y$-intercept $(0, c)$ first. Always let the $x$-step be positive (move rightward) and apply the vertical step upwards if positive, downwards if negative.
  • Worked Example 1 (Positive Gradient): Graphing $y = \frac{2}{3}x + 1$
    • $m = \frac{2}{3}$ and $c = 1$.
    • The $y$-intercept is $1$, plotted at $(0, 1)$.
    • Gradient is $\frac{2}{3} = \frac{y\text{-step}}{x\text{-step}}$.
    • From $(0, 1)$, step $3$ units right ($x$-step $= 3$) and $2$ units up ($y$-step $= 2$) to locate the point $(3, 3)$.
    • Connect the points with a ruled straight line.

• Worked Example 2 (Negative Gradient): Graphing $y = -2x - 3$
  • $m = -2$ and $c = -3$.
  • The $y$-intercept is $-3$, plotted at $(0, -3)$.
  • Gradient is $\frac{-2}{1} = \frac{y\text{-step}}{x\text{-step}}$.
  • From $(0, -3)$, step $1$ unit right ($x$-step $= 1$) and $2$ units down ($y$-step $= -2$) to locate the point $(1, -5)$.
  • Connect the points with a ruled straight line.

*   **Summary of Line Characteristics for Sketching:**
    *   $y = x + 3 \implies \text{slope} = 1, y\text{-intercept} = 3$
    *   $y = -x + 4 \implies \text{slope} = -1, y\text{-intercept} = 4$
    *   $y = 2x + 2 \implies \text{slope} = 2, y\text{-intercept} = 2$
    *   $y = \frac{1}{2}x - 1 \implies \text{slope} = \frac{1}{2}, y\text{-intercept} = -1$
    *   $y = 3x \implies \text{slope} = 3, y\text{-intercept} = 0$
    *   $y = -\frac{1}{2}x \implies \text{slope} = -\frac{1}{2}, y\text{-intercept} = 0$
    *   $y = -2x + 1 \implies \text{slope} = -2, y\text{-intercept} = 1$
    *   $y = 3 - \frac{1}{3}x \implies \text{slope} = -\frac{1}{3}, y\text{-intercept} = 3$
    *   $y = 3 \implies \text{slope} = 0, y\text{-intercept} = 3$

• Graphing via Axes Intercepts:
  • Procedural method:
    • Find the $y$-intercept by substituting $x = 0$ into the equation and solving for $y$.
    • Find the $x$-intercept by substituting $y = 0$ into the equation and solving for $x$.
    • Plot the coordinates $(0, y)$ and $(x, 0)$, drawing a straight line passing through both.
  • Worked Example 1: Graphing $y = 2x - 5$
    • $y$-intercept is $-5$, giving point $(0, -5)$.
    • When $y = 0$:             $2x - 5 = 0$$2x = 5$$x = \frac{5}{2} = 2\frac{1}{2}$
    • $x$-intercept is $2\frac{1}{2}$, giving point $(2\frac{1}{2}, 0)$.

• Worked Example 2: Graphing standard/general form line $2x - 3y = -6$
  • Line cuts the $y$-axis when $x = 0$:         $-3y = -6$$\frac{-3y}{-3} = \frac{-6}{-3}$$y = 2$         Point: $(0, 2)$.
  • Line cuts the $x$-axis when $y = 0$:         $2x = -6$$\frac{2x}{2} = \frac{-6}{2}$$x = -3$         Point: $(-3, 0)$.

• Comprehensive Intercepts Catalog:
  • $y = x + 4 \implies y\text{-int} = 4, x\text{-int} = -4$
  • $y = -x + 3 \implies y\text{-int} = 3, x\text{-int} = 3$
  • $y = 2x + 4 \implies y\text{-int} = 4, x\text{-int} = -2$
  • $y = -2x + 5 \implies y\text{-int} = 5, x\text{-int} = 2\frac{1}{2}$
  • $y = \frac{1}{2}x - 2 \implies y\text{-int} = -2, x\text{-int} = 4$
  • $y = 3 - \frac{1}{2}x \implies y\text{-int} = 3, x\text{-int} = 6$
  • $y = \frac{1}{3}x + 1 \implies y\text{-int} = 1, x\text{-int} = -3$
  • $y = 2 - \frac{x}{3} \implies y\text{-int} = 2, x\text{-int} = 6$
  • $y = \frac{x + 2}{3} \implies y\text{-int} = \frac{2}{3}, x\text{-int} = -2$
  • $x + y = 6 \implies y\text{-int} = 6, x\text{-int} = 6$
  • $2x + y = 4 \implies y\text{-int} = 4, x\text{-int} = 2$
  • $3x - y = 5 \implies y\text{-int} = -5, x\text{-int} = 1\frac{2}{3}$
  • $2x + 3y = 6 \implies y\text{-int} = 2, x\text{-int} = 3$
  • $3x - 4y = 12 \implies y\text{-int} = -3, x\text{-int} = 4$
  • $x + 3y = -6 \implies y\text{-int} = -2, x\text{-int} = -6$
  • $2x - 5y = 10 \implies y\text{-int} = -2, x\text{-int} = 5$
  • $2x + 7y = 14 \implies y\text{-int} = 2, x\text{-int} = 7$
  • $3x - 4y = 8 \implies y\text{-int} = -2, x\text{-int} = 2\frac{2}{3}$

Algebraic Manipulation and Equation Matching

• Rearranging to Gradient-Intercept Form ($y = mx + c$):

  • To determine the gradient and $y$-intercept of an implicit or general linear equation, rearrange the equation to isolate $y$ on the left-hand side.
  • $x + y = 5 \implies y = 5 - x \implies \text{gradient } m = -1$
  • $2x + y = 4 \implies y = 4 - 2x \implies \text{gradient } m = -2$
  • $y - 3x = 5 \implies y = 3x + 5 \implies \text{gradient } m = 3$
  • $2x - 3y = 6 \implies -3y = 6 - 2x \implies y = \frac{2}{3}x - 2 \implies \text{gradient } m = \frac{2}{3}$
  • $2y + x - 6 = 0 \implies 2y = 6 - x \implies y = -\frac{1}{2}x + 3 \implies \text{gradient } m = -\frac{1}{2}$
  • $6 - 2y + 5x = 0 \implies 2y = 5x + 6 \implies y = \frac{5}{2}x + 3 \implies \text{gradient } m = \frac{5}{2}$
  • $3x + 7y = -21 \implies 7y = -3x - 21 \implies y = -\frac{3}{7}x - 3 \implies \text{gradient } m = -\frac{3}{7}$
  • $2x - 5y = 10 \implies -5y = 10 - 2x \implies y = \frac{2}{5}x - 2 \implies \text{gradient } m = \frac{2}{5}$
  • $5x + 3y = 30 \implies 3y = 30 - 5x \implies y = -\frac{5}{3}x + 10 \implies \text{gradient } m = -\frac{5}{3}$

• Matching Equations to Graphed Lines:

  • Equations are matched to lines labeled $A$ through $H$ by identifying their gradient and $y$-intercept:

• Equation $y = x + 1$ has $m = 1$, $c = 1 \implies \text{matches Graph D}$.
• Equation $y = 1 - x$ has $m = -1$, $c = 1 \implies \text{matches Graph H}$.
• Equation $2y = 3x + 4 \implies y = \frac{3}{2}x + 2$ has $m = \frac{3}{2}$, $c = 2 \implies \text{matches Graph E}$.
• Equation $x + y = 5 \implies y = -x + 5$ has $m = -1$, $c = 5 \implies \text{matches Graph B}$.
• Equation $2x + 5y = 10 \implies y = -\frac{2}{5}x + 2$ has $m = -\frac{2}{5}$, $c = 2 \implies \text{matches Graph A}$.
• Equation $4 - 2x = 4y \implies y = -\frac{1}{2}x + 1$ has $m = -\frac{1}{2}$, $c = 1 \implies \text{matches Graph C}$.
• Equation $x - y = 3 \implies y = x - 3$ has $m = 1$, $c = -3 \implies \text{matches Graph G}$.
• Equation $\frac{1}{2}y = \frac{1}{2}x - 1 \implies y = x - 2$ has $m = 1$, $c = -2 \implies \text{matches Graph F}$.

Determining Equations Directly from Graphs

• Methodology:

  • Step 1: Read the $y$-intercept directly from the intersection of the line with the vertical axis; this provides $c$.
  • Step 2: Identify another distinct grid point $(x_2, y_2)$ on the line.
  • Step 3: Calculate the gradient $m$ using $m = \frac{y_2 - y_1}{x_2 - x_1}$.
  • Step 4: Substitute $m$ and $c$ into $y = mx + c$.

• Worked Example:

  • Determining the equation of a line passing through $(0, 3)$ and $(5, 13)$:
    • The $y$-intercept is $3$, so $c = 3$.
    • Compute the gradient:             $m = \frac{13 - 3}{5 - 0} = \frac{10}{5} = 2$
    • The equation of the line is:             $y = 2x + 3$

• Compilation of Derived Graph Equations:

  • Line passing through $(-1, 0)$ and $(0, 1)$: Gradient is $1$, $y$-intercept is $1$, equation is $y = x + 1$.
  • Line passing through $(0, 2)$ and $(4, 3)$: Gradient is $\frac{3 - 2}{4 - 0} = \frac{1}{4}$, $y$-intercept is $2$, equation is $y = \frac{1}{4}x + 2$.
  • Line passing through $(0, 0)$ and $(2, 1)$: Gradient is $\frac{1 - 0}{2 - 0} = \frac{1}{2}$, $y$-intercept is $0$, equation is $y = \frac{1}{2}x$.
  • Line passing through $(0, 2)$ and $(2, 0)$: Gradient is $\frac{0 - 2}{2 - 0} = -1$, $y$-intercept is $2$, equation is $y = -x + 2$.
  • Line passing through $(-1, 3)$ and $(0, 0)$: Gradient is $\frac{0 - 3}{0 - (-1)} = -3$, $y$-intercept is $0$, equation is $y = -3x$.
  • Line passing through $(-2, 2)$ and $(0, -1)$: Gradient is $\frac{-1 - 2}{0 - (-2)} = -\frac{3}{2}$, $y$-intercept is $-1$, equation is $y = -\frac{3}{2}x - 1$.
  • Line passing through $(0, 1)$ and $(3, 0)$: Gradient is $\frac{0 - 1}{3 - 0} = -\frac{1}{3}$, $y$-intercept is $1$, equation is $y = -\frac{1}{3}x + 1$.
  • Line passing through $(0, -1)$ and $(2, 0)$: Gradient is $\frac{0 - (-1)}{2 - 0} = \frac{1}{2}$, $y$-intercept is $-1$, equation is $y = \frac{1}{2}x - 1$.
  • Line passing through $(0, -2)$ and $(4, 0)$: Gradient is $\frac{0 - (-2)}{4 - 0} = \frac{2}{4} = \frac{1}{2}$, $y$-intercept is $-2$, equation is $y = \frac{1}{2}x - 2$.

• Rules Connecting Non-Standard Variables:

  • Graph with vertical axis $V$ and horizontal axis $t$, passing through $(0, 1)$ and $(4, 3)$:
    • Gradient: $m = \frac{3 - 1}{4 - 0} = \frac{2}{4} = \frac{1}{2}$
    • Vertical intercept: $c = 1$
    • Rule: $V = \frac{1}{2}t + 1$
  • Graph with vertical axis $N$ and horizontal axis $d$, passing through $(0, 2)$ and $(2, 0)$:
    • Gradient: $m = \frac{0 - 2}{2 - 0} = -1$
    • Vertical intercept: $c = 2$
    • Rule: $N = 2 - d$
  • Graph with vertical axis $C$ and horizontal axis $t$, passing through $(0, 3)$ and $(-5, -1)$:
    • Gradient: $m = \frac{3 - (-1)}{0 - (-5)} = \frac{4}{5}$
    • Vertical intercept: $c = 3$
    • Rule: $C = \frac{4}{5}t + 3$