Honors Geometry Unit 1: Parallel and Perpendicular Lines Study Guide

Unit 1: Parallel and Perpendicular Lines Overview

• Learning Objectives:

  • Use slope and equations of lines to determine where two lines intersect.

  • Use slope and equations of lines to determine whether two lines are parallel, perpendicular, or neither.

  • Set up and solve equations using the distance and midpoint formulas.

  • Use slope, equations of lines, and the distance formula to find distances.

1.1 Slopes and Equations of Lines

• Slope Formula:

  • Slope ($m$) describes the steepness and direction of a line.

  • $m = \frac{\text{rise}}{\text{run}} = \frac{\text{change in y}}{\text{change in x}} = \frac{y_2 - y_1}{x_2 - x_1}$

  • Example: For points $A(1, 5)$ and $B(4, 4)$, the slope is calculated as $m = \frac{4 - 5}{4 - 1} = -\frac{1}{3}$.

• Forms of Linear Equations:

  • Slope-Intercept Form: $y = mx + b$

    • $m$: slope.

    • $b$: y-intercept.

  • Point-Slope Form: $y - y_1 = m(x - x_1)$

    • $m$: slope.

    • $(x_1, y_1)$: any specific point on the line.

  • Standard Form: $Ax + By = C$

    • $A$, $B$, and $C$ are constants (typically whole numbers, where $A$ is non-negative).

• Converting Between Forms (Example: Points A(1, 5) and B(4, 4)):

  • To Slope-Intercept: Start with $y = mx + b$. Plug in the slope $-\frac{1}{3}$ and coordinates for point B (4, 4) to solve for $b$.

    • $4 = -\frac{1}{3}(4) + b$

    • $4 = -\frac{4}{3} + b$

    • $\frac{12}{3} = -\frac{4}{3} + b \rightarrow b = \frac{16}{3}$

    • Equation: $y = -\frac{1}{3}x + \frac{16}{3}$

  • To Point-Slope: Use the slope and either point A or B.

    • Equation (using B): $y - 4 = -\frac{1}{3}(x - 4)$

  • To Standard Form: Start with Point-Slope form, multiply by the denominator to eliminate fractions, and rearrange terms.

    • $3(y - 4) = -1(x - 4)$

    • $3y - 12 = -x + 4$

    • $x + 3y = 16$

• Equation Construction Exercises:

  • Slope: $-\frac{2}{3}$, Point: (0, 3)

    • Slope-Intercept: $y = -\frac{2}{3}x + 3$

    • Point-Slope: $y - 3 = -\frac{2}{3}(x - 0)$

    • Standard Form: $2x + 3y = 9$

  • Slope: $-3$, Point: (-1, -3)

    • Slope-Intercept: $y = -3x - 6$

    • Point-Slope: $y + 3 = -3(x + 1)$

    • Standard Form: $3x + y = -6$

  • Point (3, 0) and Point (5, -5)

    • Slope ($m$): $\frac{-5 - 0}{5 - 3} = -\frac{5}{2}$

    • Point-Slope: $y - 0 = -\frac{5}{2}(x - 3)$

    • Slope-Intercept: $y = -\frac{5}{2}x + \frac{15}{2}$

    • Standard Form: $5x + 2y = 15$

Systems of Equations

• Definition: A system of equations consists of two or more equations. The solution is the point index $(x, y)$ where the lines intersect.

• Graphing Method: Graph both equations using a straight edge to find the shared point.

  • Example: $x + y = 2$ and $y = \frac{3}{2}x - 3$. Intersection: $(2, 0)$.

  • Example: $5x + 2y = 8$ and $y = -1$. Intersection: $(2, -1)$.

• Substitution Method: Replace a variable in one equation with its equivalent expression from the other equation to solve for the intersection point.

  • Example: $y = 3$ and $2x + 3y = 15$

    • $2x + 3(3) = 15 \rightarrow 2x = 6 \rightarrow x = 3$. Solution: $(3, 3)$.

  • Example: $y = 2x - 7$ and $y = x - 1$

    • $x - 1 = 2x - 7 \rightarrow 7 - 1 = 2x - x \rightarrow x = 6$

    • $y = 6 - 1 = 5$. Solution: $(6, 5)$.

  • Example: $y = \frac{3}{4}x + 1$ and $y = -\frac{3}{2}x - 8$

    • Solution: $(-4, -2)$.

  • Example: $y = -3x + 6$ and $y = 2x - 5$

    • Solution: $(2.2, -0.6)$ (approximate based on graphing; calculated solution is $(11/5, -3/5)$).

1.2 Parallel and Perpendicular Lines

• Slopes of Parallel Lines: Parallel lines have equal slopes ($m_1 = m_2$).

• Slopes of Perpendicular Lines: Perpendicular lines have opposite reciprocal slopes. This involves a sign change (positive to negative or vice versa) and flipping the fraction ($m_1 = -\frac{1}{m_2}$).

• Determining Relationships:

  • Example 1: $A(-6, -9); B(8, 19)$ and $C(0, -4); D(2, 0)$

    • $m_{AB} = \frac{19 - (-9)}{8 - (-6)} = \frac{28}{14} = 2$

    • $m_{CD} = \frac{0 - (-4)}{2 - 0} = \frac{4}{2} = 2$

    • Result: Parallel (same slope).

  • Example 2: $A(4, 2); B(-3, 1)$ and $C(6, 0); D(-10, 8)$

    • $m_{AB} = \frac{1 - 2}{-3 - 4} = \frac{-1}{-7} = \frac{1}{7}$

    • $m_{CD} = \frac{8 - 0}{-10 - 6} = \frac{8}{-16} = -\frac{1}{2}$

    • Result: Neither.

  • Example 3: $A(8, -2); B(4, -1)$ and $C(3, 11); D(-2, -9)$

    • $m_{AB} = \frac{-1 - (-2)}{4 - 8} = -\frac{1}{4}$

    • $m_{CD} = \frac{-9 - 11}{-2 - 3} = \frac{-20}{-5} = 4$

    • Result: Perpendicular (negative reciprocals).

• Analysis from Equations:

  • $y = 7x - 12$ and $y - 3 = 7(x + 2)$: Slopes are both 7, so the lines are Parallel.

  • $y = 2x + 4$ and $-2x + y = -10$: Rearranging the second equation yields $y = 2x - 10$. Both slopes are 2. Lines are Parallel.

  • $y - 4 = 3(x + 5)$ and $y + 3 = -\frac{1}{3}(x + 1)$: Slopes are 3 and $-\frac{1}{3}$. Lines are Perpendicular.

  • $3x + 5y = 10$ and $y - 3 = \frac{5}{3}(x - 4)$: Rearranging the first equation yields $y = -\frac{3}{5}x + 2$. Slopes are $-\frac{3}{5}$ and $\frac{5}{3}$. Lines are Perpendicular.

• Writing Equations for Specific Conditions:

  • Point (-7, -4), perpendicular to $y = \frac{1}{2}x + 9$.

    1. Current slope is $\frac{1}{2}$, so perpendicular slope is $-2$.

    2. Use point-slope: $y - (-4) = -2(x - (-7))$

    3. Solve for $y$: $y + 4 = -2x - 14 \Rightarrow y = -2x - 18$.

  • Point (-1, -10), parallel to $y = 5x - 9$.

    • Slope: 5. Equation: $y + 10 = 5(x + 1) \Rightarrow y = 5x - 5$.

• Solving for Missing Variables:

  • Perpendicular Condition: Line through (5, 3) and ($x$, -5) is perpendicular to line through (4, 6) and (-2, 9).

    • Slope 2: $\frac{9 - 6}{-2 - 4} = \frac{3}{-6} = -\frac{1}{2}$.

    • Slope 1 must be 2.

    • $2 = \frac{-5 - 3}{x - 5} \rightarrow 2 = \frac{-8}{x - 5} \rightarrow 2(x - 5) = -8 \rightarrow x - 5 = -4 \rightarrow x = 1$.

  • Parallel Condition: Same line conditions, but parallel.

    • Slope 1 must be $-\frac{1}{2}$.

    • $-\frac{1}{2} = \frac{-8}{x - 5} \rightarrow -1(x - 5) = -16 \rightarrow x - 5 = 16 \rightarrow x = 21$.

1.3 Midpoint and Distance Formula

• Midpoint Formula:

  • A midpoint bisects a segment into two equal halves.

  • $M = \begin{pmatrix} \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \end{pmatrix}$

  • Example: Find midpoint of (2, 6) and (6, 1).

    • $x_M = \frac{2 + 6}{2} = 4$; $y_M = \frac{6 + 1}{2} = 3.5$. Midpoint: (4, 3.5).

• Finding an Endpoint:

  • If point $B(3, 1)$ is the midpoint of segment $AC$, and $A$ is (0, -1):

    • $3 = \frac{0 + x_C}{2} \rightarrow x_C = 6$

    • $1 = \frac{-1 + y_C}{2} \rightarrow y_C = 3$. Coordinate $C$: (6, 3).

• Distance Formula:

  • This calculates the measure of the shortest possible distance between two points.

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

  • Example: Distance between (2, 6) and (6, 1).

    • $d = \sqrt{(6 - 2)^2 + (1 - 6)^2} = \sqrt{(4)^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41} \approx 6.40$.

• Real-World Distance Problems:

  • Pedestrian Pathway: From (-3, 8) to (9, -1) where 1 unit = 10 meters.

    • $d = \sqrt{(9 + 3)^2 + (-1 - 8)^2} = \sqrt{12^2 + (-9)^2} = \sqrt{144 + 81} = \sqrt{225} = 15$

    • Total distance: $15 \times 10 = 150 \, \text{meters}$.

  • Boat Dock: Dock at (50, 300) to snack shop at (550, 50).

    • $d = \sqrt{(550 - 50)^2 + (50 - 300)^2} = \sqrt{500^2 + (-250)^2} = \sqrt{250000 + 62500} = \sqrt{312500}$

    • $d \approx 559.02$ meters.

1.4A Distance between a Point and a Line

• Steps to Find Distance from Point P to Line l:

  • Step 1: Write the equation for line $l$.

  • Step 2: Write the equation for the line perpendicular to $l$ that passes through point $P$.

  • Step 3: Find the Point of Perpendicularity (PoP) by solving the system of equations created by the two lines.

  • Step 4: Use the Distance Formula to find the distance between point $P$ and the PoP.

• Detailed Example:

  • Line $l$ contains (0, -6) and (5, 4). Point $P$ is (4, 12).

  • Step 1: Slope of $l$ is $\frac{4 - (-6)}{5 - 0} = 2$. Equation: $y = 2x - 6$.

  • Step 2: Perpendicular slope is $-\frac{1}{2}$. Equation through (4, 12): $y - 12 = -\frac{1}{2}(x - 4) \Rightarrow y = -\frac{1}{2}x + 14$.

  • Step 3: Solve $2x - 6 = -\frac{1}{2}x + 14$.

    • $2.5x = 20 \rightarrow x = 8$

    • $y = 2(8) - 6 = 10$. PoP is (8, 10).

  • Step 4: Distance between (4, 12) and (8, 10).

    • $d = \sqrt{(8 - 4)^2 + (10 - 12)^2} = \sqrt{16 + 4} = \sqrt{20} \approx 4.47$.

1.4B Distance between Parallel Lines

• Steps to Find Distance between Parallel Lines:

  • Step 1: Identify point $P$ by choosing the y-intercept of the first equation.

  • Step 2: Write the equation for a line perpendicular to both lines passing through $P$.

  • Step 3: Find the Point of Perpendicularity (PoP) on the second line by solving the system.

  • Step 4: Use the distance formula between $P$ and PoP.

• Detailed Example:

  • Parallel lines: $y = -3x + 6$ and $y = -3x - 4$.

  • Step 1: Choose y-intercept of $y = -3x + 6$ as $P(0, 6)$.

  • Step 2: Perpendicular slope is $\frac{1}{3}$. Perpendicular line through $P$: $y = \frac{1}{3}x + 6$.

  • Step 3: Solve intersection of $y = \frac{1}{3}x + 6$ and $y = -3x - 4$.

    • $\frac{1}{3}x + 6 = -3x - 4$

    • Multiply by 3: $x + 18 = -9x - 12$

    • $10x = -30 \rightarrow x = -3$

    • $y = -3(-3) - 4 = 5$. PoP is (-3, 5).

  • Step 4: Distance between (0, 6) and (-3, 5).

    • $d = \sqrt{(0 - (-3))^2 + (6 - 5)^2} = \sqrt{3^2 + 1^2} = \sqrt{10} \approx 3.16$.

Questions & Discussion

• Question: Why can different points be chosen for solving the distance between parallel lines?

• Response: Both distances will come out the same regardless of which line you start from or which point you pick on the line. The math to get there may differ in the intermediate steps, but the final result is constant.

• Problem Solver Feedback: A self-assessment scale suggests that students consider themselves mastering the material if they can successfully calculate intersection points. Scoring a "4" indicates readiness for testing, while scoring a "0" indicates a need for help.