Comprehensive Study Guide: Equations of Lines, Graphing, Parallel, and Perpendicular Lines

Forms of Linear Equations and Slope Concept

• Three Main Forms of Linear Equations:

  • Point-Slope Form: Used when a point on the line $(x_1, y_1)$ and the slope $m$ are known.
  • Slope-Intercept Form: The most common form alongside point-slope form, requiring the slope $m$ and $y$-intercept $b$.
  • General Form: Formatted as $Ax + By + C = 0$ (or $Ax + By = C$).

• Concept and Definition of Slope ($m$):

  • Slope measures the rate of change of $y$-coordinates relative to $x$-coordinates, expressed as "rise over run":     $\text{Slope} = \frac{\text{Change in } y}{\text{Change in } x}$
  • The letter $m$ is the standard abbreviation for slope.
  • Given two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, the formula for slope is:     $m = \frac{y_2 - y_1}{x_2 - x_1}$

• Behavior of Lines Based on Slope Value ($m$):

  • Positive Slope ($m > 0$): As $x$ increases, $y$ increases (the line rises from left to right). As $x$ decreases, $y$ decreases.
  • Negative Slope ($m < 0$): As $x$ increases, $y$ decreases (the line falls from left to right). As $x$ decreases, $y$ increases.
  • Zero Slope ($m = 0$): Represents a horizontal line.
  • Undefined Slope: Represents a vertical line.

Finding Slope, Point-Slope Form, and Slope-Intercept Form

• Example 1: Line Passing Through $(1, 3)$ and $(5, 7)$

  • Slope Calculation:     Designating $(5, 7)$ as $(x_2, y_2)$ and $(1, 3)$ as $(x_1, y_1)$:     $m = \frac{7 - 3}{5 - 1} = \frac{4}{4} = 1$     This represents a vertical change of $1\, \text{unit}$ for every $1\, \text{unit}$ of horizontal movement.
  • Point-Slope Form Formula:$y - y_1 = m(x - x_1)$
  • Point-Slope Representations:
  • Using point $(1, 3)$:       $y - 3 = 1(x - 1)$
  • Using point $(5, 7)$:       $y - 7 = 1(x - 5)$
  • Conversion to Slope-Intercept Form ($y = mx + b$):
  • Solving from point $(1, 3)$:       $y - 3 = x - 1 \implies y = x + 2$
  • Solving from point $(5, 7)$:       $y - 7 = x - 5 \implies y = x + 2$
  • Both points yield the identical slope-intercept equation $y = x + 2$, where slope $m = 1$ and the $y$-intercept is $b = 2$ (occurring at point $(0, 2)$).

• Example 2: Line Passing Through $(-6, 3)$ and $(10, 11)$

  • Slope Calculation:     Designating $(10, 11)$ as $(x_2, y_2)$ and $(-6, 3)$ as $(x_1, y_1)$:     $m = \frac{11 - 3}{10 - (-6)} = \frac{8}{10 + 6} = \frac{8}{16} = \frac{1}{2}$
  • Point-Slope Representations:
  • Using point $(-6, 3)$:       $y - 3 = \frac{1}{2}(x - (-6)) \implies y - 3 = \frac{1}{2}(x + 6)$
  • Using point $(10, 11)$:       $y - 11 = \frac{1}{2}(x - 10)$
  • Conversion to Slope-Intercept Form:
  • From first point-slope form:       $y - 3 = \frac{1}{2}x + 3 \implies y = \frac{1}{2}x + 6$
  • From second point-slope form:       $y - 11 = \frac{1}{2}x - 5 \implies y = \frac{1}{2}x + 6$
  • The $y$-intercept is $b = 6$, corresponding to the point $(0, 6)$.
  • Graphing via Slope Definition:
  • Plot the $y$-intercept at $(0, 6)$.
  • With $m = \frac{1}{2}$, move UP $1\, \text{unit}$ and RIGHT $2\, \text{units}$ to locate $(2, 7)$, or DOWN $1\, \text{unit}$ and LEFT $2\, \text{units}$ to locate $(-2, 5)$.
  • Continuing down $1$ and left $2$ sequentially lands on $(-4, 4)$ and $(-6, 3)$.

• Example 3: Line Given $x$-intercept $\left(-\frac{1}{3}, 0\right)$ and $y$-intercept $(0, 5)$

  • Slope Calculation:$m = \frac{5 - 0}{0 - \left(-\frac{1}{3}\right)} = \frac{5}{\frac{1}{3}} = 5 \times 3 = 15$
  • Point-Slope Representations:
  • Using $\left(-\frac{1}{3}, 0\right)$:       $y - 0 = 15\left(x - \left(-\frac{1}{3}\right)\right) \implies y = 15\left(x + \frac{1}{3}\right)$
  • Using $(0, 5)$:       $y - 5 = 15(x - 0) \implies y - 5 = 15x$
  • Slope-Intercept Form:     Distributing or rearranging either form gives:     $y = 15x + 5$

Converting General Form to Slope-Intercept Form

• General Form Definition: Expressed as $Ax + By + C = 0$ or $Ax + By = C$.

• Conversion Procedure: Isolate the $y$-term on the left side of the equation and divide all terms by coefficient $B$.

• Example 1: Convert $2x + 7y - 14 = 0$

  • Isolate $y$-term:     $7y = -2x + 14$
  • Divide every term by $7$:     $y = -\frac{2}{7}x + 2$
  • Properties: Slope $m = -\frac{2}{7}$, $y$-intercept $b = 2$ (point $(0, 2)$).
  • Graphing Movement: Start at $(0, 2)$, move DOWN $2\, \text{units}$ and RIGHT $7\, \text{units}$.

• Example 2: Convert $8x + 11y - 33 = 0$

  • Isolate $y$-term:     $11y = -8x + 33$
  • Divide every term by $11$:     $y = -\frac{8}{11}x + 3$
  • Properties: Slope $m = -\frac{8}{11}$, $y$-intercept $b = 3$ (point $(0, 3)$).

Graphing Lines in MyLab Math and Special Linear Equations

• Special Cases for Linear Equations:
  • Direct Variation Line $y = \frac{8}{9}x$:
  • Slope $m = \frac{8}{9}$.
  • The constant term is absent, indicating a $y$-intercept of $0$ (passes through the origin $(0, 0)$).
  • Graphing procedure: Plot $(0, 0)$, then move UP $8\, \text{units}$ and RIGHT $9\, \text{units}$ to plot $(9, 8)$.
  • Vertical Line $x = 1$:
  • A line where all points have an $x$-coordinate of $1$.
  • Slope is undefined.
  • Graphing procedure: Select the line tool, plot any two points with $x = 1$ (e.g., $(1, 0)$ and $(1, 3)$).
  • Horizontal Line $f(x) = 3$:
  • Function notation $f(x)$ is equivalent to $y$, giving $y = 3$.
  • Slope $m = 0$
  • Graphing procedure: Select the line tool, plot any two points with $y = 3$ (e.g., $(0, 3)$ and $(2, 3)$).

Parallel and Perpendicular Lines

• Parallel Lines (Symbol: $\parallel$):

  • Have equal slopes ($m_1 = m_2$).
  • Have different $y$-intercepts ($b_1 \neq b_2$).
  • Lines remain equidistant and never intersect.

• Perpendicular Lines (Symbol: $\perp$):

  • Have slopes that are negative inverses / negative reciprocals of each other ($m_2 = -\frac{1}{m_1}$).
  • Intersect at a $90^\circ$ angle, forming four right angles at their intersection.

• Example 1: Lines Parallel and Perpendicular to $y = 3x - 8$ Passing Through $(0, 4)$

  • Given line slope: $m = 3$
  • Parallel Line:
  • Parallel slope: $m_{\parallel} = 3$
  • Point-slope form using $(0, 4)$: $y - 4 = 3(x - 0) \implies y = 3x + 4$
  • Perpendicular Line:
  • Perpendicular slope: $m_{\perp} = -\frac{1}{3}$
  • Point-slope form using $(0, 4)$: $y - 4 = -\frac{1}{3}(x - 0) \implies y = -\frac{1}{3}x + 4$

• Example 2: Lines Parallel and Perpendicular to $y = 2x + 5$ Passing Through $(-2, 7)$

  • Given line slope: $m = 2$
  • Parallel Line Calculation:
  • Slope $m_{\parallel} = 2$
  • Point-slope form: $y - 7 = 2(x - (-2)) \implies y - 7 = 2(x + 2)$
  • Distribute: $y - 7 = 2x + 4$
  • Solve for $y$: $y = 2x + 11$
  • Perpendicular Line Calculation:
  • Slope $m_{\perp} = -\frac{1}{2}$
  • Point-slope form: $y - 7 = -\frac{1}{2}(x - (-2)) \implies y - 7 = -\frac{1}{2}(x + 2)$
  • Distribute: $y - 7 = -\frac{1}{2}x - 1$
  • Solve for $y$: $y = -\frac{1}{2}x + 6$
  • Graphical Verification:
  • Original line $y = 2x + 5$ has $y$-intercept at $(0, 5)$.
  • Parallel line $y = 2x + 11$ has $y$-intercept at $(0, 11)$ and passes through $(-2, 7)$.
  • Perpendicular line $y = -\frac{1}{2}x + 6$ has $y$-intercept at $(0, 6)$ and forms four right angles with $y = 2x + 5$

Questions and Discussion

• Question: Can point-slope form equations be reduced further by distributing the slope into the $x$ terms?

  • Answer: Yes, distributing and simplifying point-slope form produces the slope-intercept form ($y = mx + b$). However, point-slope form specifically requires maintaining the format $y - y_1 = m(x - x_1)$.

• Question: Is it possible or useful to extract general form directly from looking at a graph of a line?

  • Answer: To get general form from a graph, one must reverse-engineer from point-slope or slope-intercept form first. General form is rarely used in practical application or graphing interpretation and serves mainly as an algebraic manipulation exercise.