Unit 3 Parallel And Perpendicular Lines Homework 3 Answer Key

Understanding Parallel and Perpendicular Lines – Unit 3 Homework 3 Answer Key

Parallel and perpendicular lines are fundamental concepts in geometry that appear in every high‑school curriculum. Homework 3 for Unit 3 usually asks students to identify, construct, and prove relationships between these lines. Here's the thing — below is a complete, step‑by‑step answer key that explains why each answer is correct, not just what the answer is. Use this guide to check your work, deepen your understanding, and prepare for upcoming quizzes and tests.

1. Introduction – Why Parallel & Perpendicular Matter

• Parallel lines never intersect, no matter how far they are extended. Their slopes are equal ( m₁ = m₂ ).
• Perpendicular lines intersect at a right angle (90°). Their slopes are negative reciprocals ( m₁ · m₂ = ‑1 ).

Mastering these relationships helps you solve problems in coordinate geometry, trigonometry, and real‑world design (e.g., drafting, architecture, computer graphics).

2. Homework 3 Overview

3. Detailed Answer Key

Question 1 – Identify Parallel Lines

Diagram description: Four lines labeled AB, CD, EF, and GH.

• AB and EF have the same slope (both rise 2 units for every run 5 units).
• CD and GH are not parallel because their slopes differ.

Answer: AB ∥ EF.

Explanation: When two lines have identical slopes, they maintain a constant distance and never meet, satisfying the definition of parallelism.

Question 2 – Identify Perpendicular Lines

Diagram description: Lines MN and OP intersect at point Q.

• Slope of MN = 3/4.
• Slope of OP = –4/3.

Answer: MN ⟂ OP.

Explanation: The product of the slopes is (3/4) · (‑4/3) = –1, confirming the lines intersect at a right angle.

Question 3 – Equation of a Line Parallel to 2x – 5y = 10

1. Convert to slope‑intercept form:
   2x – 5y = 10 → ‑5y = ‑2x + 10 → y = (2/5)x – 2.
   Slope (m) = 2/5.

2. Parallel line must have the same slope 2/5.

3. Use the point given in the problem, say (4, 1). Plug into y = mx + b:
   1 = (2/5)(4) + b → 1 = 8/5 + b → b = 1 – 8/5 = ‑3/5.

Answer: y = (2/5)x – 3/5.

Explanation: Keeping the slope unchanged guarantees parallelism; the new y‑intercept is determined by the specific point the line must pass through.

Question 4 – Equation of a Line Perpendicular to y = ‑3x + 7

1. Original slope = –3.

2. Perpendicular slope = negative reciprocal → 1/3.

3. Use the given point, for example (6, 2):
   2 = (1/3)(6) + b → 2 = 2 + b → b = 0.

Answer: y = (1/3)x.

Explanation: The product of the slopes (‑3)·(1/3) = –1, satisfying the perpendicular condition.

Question 5 – Prove Lines l and m Are Parallel

• l: passes through (1, 2) and (5, 6).
• m: passes through (3, 4) and (7, 8).

Step‑by‑step proof:

1. Compute slope of l:
   mₗ = (6 – 2) / (5 – 1) = 4/4 = 1.

2. Compute slope of m:
   mₘ = (8 – 4) / (7 – 3) = 4/4 = 1.

3. Since mₗ = mₘ, the lines have equal slopes.

Conclusion: l ∥ m.

Question 6 – Prove Lines p and q Are Perpendicular

• p: equation y = ‑2x + 5 (slope –2).
• q: passes through (0, 0) and (4, 2).

Proof:

1. Slope of q: m_q = (2 – 0) / (4 – 0) = 2/4 = 1/2.

2. Multiply slopes: (‑2) · (1/2) = –1.

Conclusion: p ⟂ q.

Question 7 – Construct a Parallel Line Using Ruler & Compass

1. Draw the given line AB.
2. Place the compass point on any point C on AB and draw an arc intersecting AB at D and E.
3. Without changing the compass width, move the point to the new starting point F where the parallel line must pass.
4. Draw a second arc intersecting the first at G.
5. Use the ruler to draw line FG through F and G.

Result: FG is parallel to AB because the arcs guarantee equal corresponding angles.

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Question 8 – Construct a Perpendicular Line Using Ruler & Compass

1. Given line XY. Choose a point P on XY where the perpendicular will pass.
2. Set compass to any radius, draw an arc intersecting XY at A and B.
3. Without changing the radius, place the compass on A and draw an arc above XY.
4. Repeat from B, creating a second arc that intersects the first at C.
5. Draw line PC.

Result: PC is perpendicular to XY because the constructed triangle APB is isosceles, and the angle at P is a right angle (Thales’ theorem).

Question 9 – Real‑World Application: Designing Parallel Traffic Lanes

Problem statement: A highway has three lanes, each 3.5 m wide. The road centerline follows the equation y = 0.4x + 2. Find the equations of the outer lane boundaries.

Solution:

1. The centerline slope = 0.4 → all lane boundaries must have the same slope (parallel).

2. Convert the centerline to slope‑intercept form (already done).

3. Determine the perpendicular distance between parallel lines: distance = 3.5 m.

4. Use the formula for a line parallel to y = mx + b at distance d:
   b′ = b ± d·√(1 + m²)

   Compute √(1 + m²) = √(1 + 0.16 ≈ 1.4²) = √(1 + 0.That's why 16) = √1. 077.

5. For the rightmost lane boundary (adding distance):
   b₁ = 2 + 3.5 · 1.077 ≈ 2 + 3.77 ≈ 5.77.

6. For the leftmost lane boundary (subtracting distance twice, because two lane widths away):
   b₂ = 2 – 2 · 3.5 · 1.077 ≈ 2 – 7.54 ≈ –5.54.

Equations:

• Rightmost boundary: y = 0.4x + 5.77
• Centerline (given): y = 0.4x + 2
• Leftmost boundary: y = 0.4x – 5.54

Explanation: Maintaining equal slope ensures the lanes stay parallel, while the calculated intercepts shift each line the required lateral distance.

Question 10 – Distance Between Two Parallel Lines

Given lines:

• L₁: 3x – 4y + 12 = 0
• L₂: 3x – 4y – 8 = 0

Formula: Distance = |c₂ – c₁| / √(a² + b²)

• a = 3, b = ‑4, c₁ = 12, c₂ = ‑8.

Compute numerator: |‑8 – 12| = |‑20| = 20.

Denominator: √(3² + (‑4)²) = √(9 + 16) = √25 = 5.

Answer: Distance = 20 / 5 = 4 units.

Explanation: The two lines share the same normal vector (3, ‑4), confirming they are parallel; the distance formula yields the exact perpendicular separation.

4. Scientific Explanation – The Geometry Behind the Rules

• Slope Equality for Parallelism: In the Cartesian plane, the slope m = Δy/Δx describes a line’s steepness. If two lines share m, any change in x produces the same change in y for both, so they never converge.

• Negative Reciprocals for Perpendicularity: A right triangle formed by a horizontal run Δx and a vertical rise Δy has an angle θ. The slope of the adjacent side is tan θ. The line perpendicular to it must have an angle (θ + 90°); tan(θ + 90°) = –1/tan θ, which is the negative reciprocal.

• Distance Formula for Parallel Lines: The general line ax + by + c = 0 has a normal vector (a, b). The shortest distance between two parallel lines is the projection of the vector connecting any point on one line onto the unit normal, yielding the formula used in Question 10.

Understanding these derivations helps you prove relationships rather than merely memorizing rules, a skill that earns full credit on exams.

5. Frequently Asked Questions

Q1: Can two vertical lines be parallel?
Yes. Vertical lines have undefined slope, but they are parallel because they share the same x‑value direction and never intersect.

Q2: What if the slopes are both zero?
Both lines are horizontal and therefore parallel. Their slopes are 0, satisfying the equality condition.

Q3: Why do we use the negative reciprocal instead of simply the opposite sign?
The opposite sign alone changes the direction but not the steepness. The reciprocal flips the rise/run ratio, guaranteeing a 90° angle.

Q4: How do I check perpendicularity when the equations are in standard form (ax + by + c = 0)?
Convert each to slope‑intercept form, or use the condition a₁a₂ + b₁b₂ = 0 for the normal vectors. If the dot product of the normals is zero, the lines are perpendicular.

Q5: In construction, why does the compass method guarantee parallelism?
The arcs create equal corresponding angles at the original and new points. Equal angles with a common transversal imply parallel lines (alternate interior angles theorem).

6. Conclusion – Turning Homework into Mastery

The answer key above does more than give you the final results; it walks you through the reasoning behind each step. By internalizing the slope relationships, the geometric constructions, and the analytic formulas, you’ll be able to:

• Spot parallel and perpendicular lines instantly in diagrams.
• Write correct equations for any required line.
• Prove relationships algebraically and geometrically.
• Apply these concepts to real‑world engineering problems.

Practice each problem type, compare your work with the explanations, and you’ll move from merely completing homework to truly mastering Unit 3’s parallel and perpendicular line concepts. Keep this guide handy for future reference, and you’ll be well prepared for the next quiz, the unit test, and any geometry challenge that comes your way.