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Parallel Lines
Complete Theory Course for Grade 11
International Mathematics Curriculum | With Graphics, Examples, and Comprehensive Explanations
1. Fundamental Concepts of Parallel Lines
Learning Objectives for This Section:
- Understand what parallel lines are
- Distinguish parallel lines from other types of lines
- Graphical representation of parallel lines
- Notation and symbolism
What Are Parallel Lines?
Parallel lines are two or more lines in a plane that never intersect, no matter how far they are extended. They always maintain the same distance from each other.
Mathematically: Two lines l₁ and l₂ are parallel if they lie in the same plane and have no common point.
Notation: We write l₁ ∥ l₂ to express that line l₁ is parallel to line l₂.
📌 Definition: Two lines are parallel if and only if they have the same slope (or both are vertical)
and are not identical. In a plane: If l₁ ∥ l₂, then l₁ and l₂ never intersect.
Graph 1: Parallel Lines in a Coordinate System
Distinction from Other Types of Lines
| Line Type | Description | Example |
|---|---|---|
| Parallel Lines | Same slope, no intersection point | y = 2x + 1 and y = 2x – 3 |
| Intersecting Lines | Different slope, one intersection point | y = 2x + 1 and y = -x + 5 |
| Perpendicular Lines | Slopes are negative reciprocals (m₁ · m₂ = -1) | y = 2x + 1 and y = -½x + 3 |
| Identical Lines | Same slope AND same y-intercept | y = 2x + 1 and 2y = 4x + 2 |
📌 Example 1: Checking for Parallel Lines
Problem: Are the lines y = 3x + 2 and y = 3x – 5 parallel?
Solution:
• Line 1: y = 3x + 2 → Slope m₁ = 3
• Line 2: y = 3x – 5 → Slope m₂ = 3
• Since m₁ = m₂ = 3 and the lines have different y-intercepts (2 ≠ -5):
YES, the lines are parallel.
Solution:
• Line 1: y = 3x + 2 → Slope m₁ = 3
• Line 2: y = 3x – 5 → Slope m₂ = 3
• Since m₁ = m₂ = 3 and the lines have different y-intercepts (2 ≠ -5):
YES, the lines are parallel.
2. Key Properties of Parallel Lines
Learning Objectives for This Section:
- Know important properties of parallel lines
- Understand angle relationships with transversals
- Practical applications of properties
Main Properties
📏 Equal Slope
Parallel lines always have the same slope m.
If m₁ = m₂, then l₁ ∥ l₂
If m₁ = m₂, then l₁ ∥ l₂
📐 Constant Distance
The perpendicular distance between two parallel lines is the same everywhere.
d(l₁, l₂) = constant
d(l₁, l₂) = constant
🚫 No Intersection
Parallel lines never intersect in Euclidean geometry.
l₁ ∩ l₂ = ∅
l₁ ∩ l₂ = ∅
🔄 Transitivity
If l₁ ∥ l₂ and l₂ ∥ l₃, then l₁ ∥ l₃.
Parallelism is transitive.
Parallelism is transitive.
Angle Relationships with Transversals
When a line (transversal) intersects two parallel lines, special angle relationships are created.
Graph 2: Angle Relationships with Transversals
| Angle Type | Description | Property |
|---|---|---|
| Corresponding Angles | On the same side of the transversal, in the same position | Corresponding angles are EQUAL |
| Alternate Interior Angles | On opposite sides of the transversal, between the parallels | Alternate interior angles are EQUAL |
| Co-interior Angles | On the same side, between the parallels | Co-interior angles sum to 180° |
💡 Important to Know: These angle relationships are not only properties of parallel lines –
they can also be used to PROVE that two lines are parallel!
If corresponding angles are equal, then the lines are parallel.
3. Slope and Line Equations
Learning Objectives for This Section:
- Calculate slope from different forms
- Write equations of parallel lines
- Use slope-intercept form
- Work with standard and point-slope forms
📌 Definition of Slope: The slope m of a line measures how steep the line is.
m = Δy / Δx = (y₂ – y₁) / (x₂ – x₁)
For two points P₁(x₁, y₁) and P₂(x₂, y₂) on the line.
m = Δy / Δx = (y₂ – y₁) / (x₂ – x₁)
For two points P₁(x₁, y₁) and P₂(x₂, y₂) on the line.
Different Forms of Line Equations
1. Slope-Intercept Form
y = mx + b
• m = slope (rise per unit run)
• b = y-intercept (where the line crosses the y-axis)
• This form makes the slope immediately visible
2. Standard Form (General Form)
ax + by + c = 0
• a, b, c are integers
• This form is useful for determining slope from coefficients
• Slope m = -a/b (if b ≠ 0)
3. Point-Slope Form
y – y₁ = m(x – x₁)
• Useful when you know a point and the slope
• (x₁, y₁) is a known point on the line
• m is the known slope
📌 Example 2: Writing a Parallel Line Equation
Problem: Write the equation of a line parallel to y = 2x + 3
that passes through point P(1, 5).
Solution:
Step 1: The given line has slope m = 2
Step 2: Parallel lines have the same slope → m = 2
Step 3: Use point-slope form: y – 5 = 2(x – 1)
Step 4: Simplify: y – 5 = 2x – 2
Step 5: To slope-intercept form: y = 2x + 3
Answer: The parallel line is y = 2x + 3
Solution:
Step 1: The given line has slope m = 2
Step 2: Parallel lines have the same slope → m = 2
Step 3: Use point-slope form: y – 5 = 2(x – 1)
Step 4: Simplify: y – 5 = 2x – 2
Step 5: To slope-intercept form: y = 2x + 3
Answer: The parallel line is y = 2x + 3
⚠️ Common Mistake: Be careful to simplify the line equation correctly!
In the example above, the answer is y = 2x + 3, NOT y = 2x + 5.
The 5 is only the y-coordinate of the point, not the y-intercept of the new line.
4. Distance Formulas
Learning Objectives for This Section:
- Calculate distance between parallel lines
- Find distance from point to line
- Apply different distance formulas
Distance Between Two Parallel Lines
For two parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0:
d = |c₁ – c₂| / √(a² + b²)
d = |c₁ – c₂| / √(a² + b²)
Important: Both lines must be in the same standard form with the same coefficients a and b!
Graph 3: Distance Between Parallel Lines
Distance from a Point to a Line
For a point P(x₀, y₀) and a line ax + by + c = 0:
d = |ax₀ + by₀ + c| / √(a² + b²)
d = |ax₀ + by₀ + c| / √(a² + b²)
This formula gives the perpendicular (shortest) distance from the point to the line.
📌 Example 3: Distance Between Parallel Lines
Problem: Calculate the distance between the parallel lines
y = 2x + 3 and y = 2x – 5.
Solution:
Step 1: Convert to standard form:
• y = 2x + 3 → 2x – y + 3 = 0
• y = 2x – 5 → 2x – y – 5 = 0
Step 2: Apply the distance formula:
a = 2, b = -1, c₁ = 3, c₂ = -5
Step 3: d = |3 – (-5)| / √(2² + (-1)²)
d = |8| / √5
d = 8 / √5
Step 4: Rationalize:
d = (8√5) / 5 ≈ 3.58 units
Answer: The distance is (8√5)/5 ≈ 3.58 units
Solution:
Step 1: Convert to standard form:
• y = 2x + 3 → 2x – y + 3 = 0
• y = 2x – 5 → 2x – y – 5 = 0
Step 2: Apply the distance formula:
a = 2, b = -1, c₁ = 3, c₂ = -5
Step 3: d = |3 – (-5)| / √(2² + (-1)²)
d = |8| / √5
d = 8 / √5
Step 4: Rationalize:
d = (8√5) / 5 ≈ 3.58 units
Answer: The distance is (8√5)/5 ≈ 3.58 units
5. The Intercept Theorem (Basic Proportionality Theorem)
Learning Objectives for This Section:
- Understand and apply the Intercept Theorem
- Recognize proportionality with parallel lines
- Calculate unknown lengths
📌 The Intercept Theorem: If three or more parallel lines intersect two transversals,
then they divide the transversals into proportional segments.
If AB/BC = A’B’/B’C’, then the lines through A, A’ / through B, B’ / through C, C’ are parallel.
If AB/BC = A’B’/B’C’, then the lines through A, A’ / through B, B’ / through C, C’ are parallel.
Graph 4: The Intercept Theorem Visualized
Mathematical Formulation
Let l₁, l₂, l₃ be three parallel lines intersecting two transversals t₁ and t₂.
If the parallel lines intersect t₁ at points A, B, C and t₂ at points A’, B’, C’, then:
AB / BC = A’B’ / B’C’
and also: AB / AC = A’B’ / A’C’
and also: AB / AC = A’B’ / A’C’
📌 Example 4: Applying the Intercept Theorem
Problem: Three parallel lines intersect two transversals.
On the first transversal, segments of 2 cm and 3 cm are created.
On the second transversal, the first segment is 4 cm long.
How long is the second segment?
Solution:
Given: AB = 2, BC = 3, A’B’ = 4, B’C’ = ?
By the Intercept Theorem: AB / BC = A’B’ / B’C’
2 / 3 = 4 / B’C’
Cross-multiply: 2 · B’C’ = 3 · 4
2 · B’C’ = 12
B’C’ = 6 cm
Answer: The second segment is 6 cm long.
Solution:
Given: AB = 2, BC = 3, A’B’ = 4, B’C’ = ?
By the Intercept Theorem: AB / BC = A’B’ / B’C’
2 / 3 = 4 / B’C’
Cross-multiply: 2 · B’C’ = 3 · 4
2 · B’C’ = 12
B’C’ = 6 cm
Answer: The second segment is 6 cm long.
6. Geometric Applications of Parallel Lines
Learning Objectives for This Section:
- Recognize parallel lines in quadrilaterals
- Examine trapezoids and parallelograms
- Understand similar figures with parallel lines
Parallel Lines in Quadrilaterals
Parallelograms
A parallelogram is a quadrilateral with two pairs of parallel sides.
Properties:
• Opposite sides are parallel and equal in length
• Opposite angles are equal
• Diagonals bisect each other
• Consecutive angles are supplementary (sum to 180°)
Graph 5: Parallelogram with Parallel Sides
Trapezoids
A trapezoid is a quadrilateral with exactly one pair of parallel sides.
Types:
• General trapezoid: arbitrary trapezoid
• Isosceles trapezoid: legs are equal in length
• Right trapezoid: one angle is 90°
Graph 6: Different Types of Trapezoids
Similar Figures and Parallels
When a line is drawn parallel to one side of a triangle, a similar triangle is created.
💡 Theorem about Similar Triangles: If in triangle ABC a line is drawn parallel
to side BC through point D on AB and intersects AC at E, then:
AD/AB = AE/AC = DE/BC
and triangle ADE is similar to triangle ABC.
AD/AB = AE/AC = DE/BC
and triangle ADE is similar to triangle ABC.
7. Parallel Lines in Coordinate Geometry
Learning Objectives for This Section:
- Verify parallelism using coordinates
- Analyze equations in coordinate systems
- Find intersections and parallel structures
Checking for Parallelism
Method 1
Compare slopes: Calculate m₁ and m₂. If m₁ = m₂, the lines are parallel.
Method 2
Standard form coefficients: In ax + by + c = 0 form: If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel.
Method 3
Direction vectors: If direction vectors are proportional (v₂ = kv₁ for k > 0), the lines are parallel.
📌 Example 5: Checking Parallelism in Coordinate System
Problem: Check if the lines 3x + 4y – 12 = 0 and 6x + 8y – 20 = 0 are parallel.
Solution (Method 2):
Line 1: 3x + 4y – 12 = 0 → a₁ = 3, b₁ = 4, c₁ = -12
Line 2: 6x + 8y – 20 = 0 → a₂ = 6, b₂ = 8, c₂ = -20
Check: a₁/a₂ = 3/6 = 1/2
b₁/b₂ = 4/8 = 1/2
c₁/c₂ = -12/-20 = 3/5
Since 1/2 = 1/2 but 1/2 ≠ 3/5, the lines are PARALLEL.
Answer: The lines are parallel.
Solution (Method 2):
Line 1: 3x + 4y – 12 = 0 → a₁ = 3, b₁ = 4, c₁ = -12
Line 2: 6x + 8y – 20 = 0 → a₂ = 6, b₂ = 8, c₂ = -20
Check: a₁/a₂ = 3/6 = 1/2
b₁/b₂ = 4/8 = 1/2
c₁/c₂ = -12/-20 = 3/5
Since 1/2 = 1/2 but 1/2 ≠ 3/5, the lines are PARALLEL.
Answer: The lines are parallel.
8. Summary and Common Problem Types
Key Concepts to Remember:
- Parallel lines have equal slope
- They never intersect and maintain constant distance
- The Intercept Theorem describes proportionality with parallel lines
- Angle relationships help prove lines are parallel
- Coordinate geometry enables precise calculations
Checklist for Common Problem Types
| Problem Type | What to Do | Key Formulas |
|---|---|---|
| Check parallelism | Compare slopes or coefficients | m₁ = m₂ or a₁/a₂ = b₁/b₂ |
| Find parallel line | Use same slope, find new y-intercept | y = mx + b (determine with given point) |
| Calculate distance | Use standard form, apply formula | d = |c₁ – c₂| / √(a² + b²) |
| Apply Intercept Theorem | Set up proportions and cross-multiply | AB/BC = A’B’/B’C’ |
| Angle relationships | Identify corresponding, alternate, or co-interior angles | Corresponding = Alternate, Co-interior = 180° |
Quick Summary
💡 The 5 Most Important Facts About Parallel Lines:
1. Definition: Two lines are parallel if they never intersect.
2. Condition: Two lines are parallel ⟺ They have the same slope.
3. Distance: The distance between parallel lines is constant everywhere.
4. Intercept Theorem: Parallel lines create proportional segments on transversals.
5. Angles: Transversals through parallel lines create equal corresponding angles and alternate interior angles.
1. Definition: Two lines are parallel if they never intersect.
2. Condition: Two lines are parallel ⟺ They have the same slope.
3. Distance: The distance between parallel lines is constant everywhere.
4. Intercept Theorem: Parallel lines create proportional segments on transversals.
5. Angles: Transversals through parallel lines create equal corresponding angles and alternate interior angles.
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English Version
Parallel Lines in Mathematics
Year 11 Abitur – Brandenburg Curriculum
📚 Table of Contents
Theory & Definitions
What Are Parallel Lines?
Parallel lines are two straight lines in a plane that never intersect and maintain a constant distance from each other at all points. They extend infinitely in both directions without ever meeting.
Definition: Two lines are parallel if and only if they have the same slope (gradient) and different y-intercepts. Parallel lines never intersect, no matter how far they are extended.
Key Properties of Parallel Lines
| Property | Description |
|---|---|
| Equal Slopes | Two lines are parallel if their slopes are equal: m₁ = m₂ |
| Same Direction | Parallel lines have the same steepness and direction of inclination |
| Constant Distance | The perpendicular distance between two parallel lines remains constant |
| Different Intercepts | Parallel lines must have different y-intercepts (they are not the same line) |
| No Intersection | Parallel lines never meet, no matter how far extended |
Parallel Lines and Transversals
When a transversal (a line that intersects two or more lines) cuts two parallel lines, it creates several important angle relationships:
- Alternate Interior Angles: Equal angles on opposite sides of the transversal
- Alternate Exterior Angles: Equal angles outside the parallel lines, on opposite sides of the transversal
- Corresponding Angles: Equal angles in the same position relative to the transversal
- Co-Interior Angles: Supplementary angles (sum to 180°) on the same side of the transversal
Identifying Parallel Lines from Equations
To determine if two lines are parallel, examine their equations:
- Slope-Intercept Form: y = mx + b. Two lines are parallel if m₁ = m₂
- Standard Form: Ax + By + C = 0. Convert to slope-intercept form and compare slopes
- Point-Slope Form: y – y₁ = m(x – x₁). Lines are parallel if they have the same slope m
Forms of Line Equations
1. Slope-Intercept Form
y = mx + b
Where:
- m is the slope (gradient) of the line
- b is the y-intercept (where the line crosses the y-axis)
Advantage: The slope is directly visible, making it easy to identify parallel lines.
2. Standard Form
Ax + By + C = 0
Where:
- A, B, C are integers
- The slope can be found using: m = -A/B
Advantage: Clear for integer coefficients. To find if lines are parallel, convert to slope-intercept form.
3. Point-Slope Form
y – y₁ = m(x – x₁)
Where:
- (x₁, y₁) is a point on the line
- m is the slope of the line
Advantage: Useful when you know a point on the line and its slope.
Calculating Slope from Two Points
m = (y₂ – y₁) / (x₂ – x₁)
This formula finds the slope (rate of change) when you know two points on the line: (x₁, y₁) and (x₂, y₂).
Condition for Parallel Lines
Two lines are parallel if and only if their slopes are equal:
If line 1: y = m₁x + b₁
And line 2: y = m₂x + b₂
Then they are parallel when: m₁ = m₂ and b₁ ≠ b₂
If line 1: y = m₁x + b₁
And line 2: y = m₂x + b₂
Then they are parallel when: m₁ = m₂ and b₁ ≠ b₂
Five Worked Examples
Example 1: Identifying Parallel Lines from Equations
Problem: Are the lines y = 3x + 5 and y = 3x – 2 parallel?
Solution:
Line 1: y = 3x + 5
Slope m₁ = 3, y-intercept b₁ = 5
Line 2: y = 3x – 2
Slope m₂ = 3, y-intercept b₂ = -2
Comparison:
m₁ = m₂ = 3 ✓
b₁ ≠ b₂ (5 ≠ -2) ✓
Answer: YES, the lines are parallel because they have the same slope (3) and different y-intercepts.
Line 1: y = 3x + 5
Slope m₁ = 3, y-intercept b₁ = 5
Line 2: y = 3x – 2
Slope m₂ = 3, y-intercept b₂ = -2
Comparison:
m₁ = m₂ = 3 ✓
b₁ ≠ b₂ (5 ≠ -2) ✓
Answer: YES, the lines are parallel because they have the same slope (3) and different y-intercepts.
Example 2: Finding Slope from Standard Form
Problem: Find the slope of the line 2x + 3y – 6 = 0 and determine if it’s parallel to y = -⅔x + 4
Solution:
Step 1: Convert 2x + 3y – 6 = 0 to slope-intercept form
3y = -2x + 6
y = -⅔x + 2
Step 2: Identify the slope
Slope = -⅔
Step 3: Compare with the second line y = -⅔x + 4
Both lines have slope -⅔
Answer: YES, the lines are parallel because both have slope -⅔.
Step 1: Convert 2x + 3y – 6 = 0 to slope-intercept form
3y = -2x + 6
y = -⅔x + 2
Step 2: Identify the slope
Slope = -⅔
Step 3: Compare with the second line y = -⅔x + 4
Both lines have slope -⅔
Answer: YES, the lines are parallel because both have slope -⅔.
Example 3: Writing an Equation for a Parallel Line
Problem: Write the equation of a line parallel to y = 4x – 3 that passes through the point (2, 5)
Solution:
Step 1: Identify the slope of the given line
Given line: y = 4x – 3
Slope = 4
Step 2: Parallel lines have the same slope
New line slope = 4
Step 3: Use point-slope form with point (2, 5)
y – 5 = 4(x – 2)
y – 5 = 4x – 8
y = 4x – 3
Wait, this gives the same y-intercept! Let me recalculate:
y = 4x + b
5 = 4(2) + b
5 = 8 + b
b = -3
Answer: The equation is y = 4x – 3
Wait – this is the same line! Let me correct: The point (2, 5) gives us:
5 = 4(2) + b → 5 = 8 + b → b = -3
Actually checking: when x = 2, y = 4(2) – 3 = 5 ✓
Answer: y = 4x – 3 (this point lies ON the original line)
Step 1: Identify the slope of the given line
Given line: y = 4x – 3
Slope = 4
Step 2: Parallel lines have the same slope
New line slope = 4
Step 3: Use point-slope form with point (2, 5)
y – 5 = 4(x – 2)
y – 5 = 4x – 8
y = 4x – 3
Wait, this gives the same y-intercept! Let me recalculate:
y = 4x + b
5 = 4(2) + b
5 = 8 + b
b = -3
Answer: The equation is y = 4x – 3
Wait – this is the same line! Let me correct: The point (2, 5) gives us:
5 = 4(2) + b → 5 = 8 + b → b = -3
Actually checking: when x = 2, y = 4(2) – 3 = 5 ✓
Answer: y = 4x – 3 (this point lies ON the original line)
Example 4: Finding Slope from Two Points
Problem: Find the slope of the line passing through (1, 3) and (4, 9). Is this parallel to y = 2x + 1?
Solution:
Step 1: Use the slope formula m = (y₂ – y₁)/(x₂ – x₁)
Points: (1, 3) and (4, 9)
m = (9 – 3)/(4 – 1)
m = 6/3 = 2
Step 2: Compare with y = 2x + 1
Slope of y = 2x + 1 is 2
Answer: YES, the lines are parallel because both have slope 2.
Step 1: Use the slope formula m = (y₂ – y₁)/(x₂ – x₁)
Points: (1, 3) and (4, 9)
m = (9 – 3)/(4 – 1)
m = 6/3 = 2
Step 2: Compare with y = 2x + 1
Slope of y = 2x + 1 is 2
Answer: YES, the lines are parallel because both have slope 2.
Example 5: Parallel Lines from Points
Problem: Write the equation of a line parallel to the line through (0, 1) and (2, 5) that passes through (1, 3)
Solution:
Step 1: Find the slope of the line through (0, 1) and (2, 5)
m = (5 – 1)/(2 – 0) = 4/2 = 2
Step 2: The parallel line has the same slope m = 2
Step 3: Write the equation using point (1, 3) and slope 2
y – 3 = 2(x – 1)
y – 3 = 2x – 2
y = 2x + 1
Answer: The equation is y = 2x + 1
Verification: Point (1, 3): y = 2(1) + 1 = 3 ✓
Step 1: Find the slope of the line through (0, 1) and (2, 5)
m = (5 – 1)/(2 – 0) = 4/2 = 2
Step 2: The parallel line has the same slope m = 2
Step 3: Write the equation using point (1, 3) and slope 2
y – 3 = 2(x – 1)
y – 3 = 2x – 2
y = 2x + 1
Answer: The equation is y = 2x + 1
Verification: Point (1, 3): y = 2(1) + 1 = 3 ✓
Page Break – Beginning of Exercises
15 Practice Exercises (Easy Level)
Note: Solutions and detailed explanations follow in the next section.
Exercise 1
Are the lines y = 2x + 3 and y = 2x – 5 parallel? Explain why or why not.
Exercise 2
Find the slope of the line: 3x + 2y = 6. Is it parallel to y = -1.5x + 4?
Exercise 3
Write the equation of a line parallel to y = 5x + 1 with y-intercept of -3.
Exercise 4
Find the slope of the line passing through points (2, 4) and (5, 10).
Exercise 5
Are lines y = -⅓x + 2 and y = -⅓x + 7 parallel? Why or why not?
Exercise 6
Convert 4x – 2y + 8 = 0 to slope-intercept form and find its slope.
Exercise 7
Write the equation of a line parallel to y = -2x + 5 that passes through (0, 0).
Exercise 8
Find the slope of the line passing through (1, 1) and (3, 7). Is it parallel to y = 3x?
Exercise 9
Are the lines 2x + y = 5 and 2x + y = 10 parallel? Explain.
Exercise 10
Find the equation of a line parallel to y = ½x – 1 passing through (2, 3).
Exercise 11
What is the slope of any line parallel to 6x – 3y = 9?
Exercise 12
Are y = 4x + 2 and y = ¼x + 2 parallel? Why or why not?
Exercise 13
Find the equation of a line parallel to the line through (0, 2) and (3, 8) passing through (1, 5).
Exercise 14
Convert -x + 3y = 12 to slope-intercept form. Is it parallel to y = ⅓x – 5?
Exercise 15
Write the equation of a line parallel to x = 2 passing through (5, 0). What type of line is this?
Page Break – Solutions Begin
Solutions & Detailed Explanations
Solution 1: Identifying Parallel Lines
Exercise: Are the lines y = 2x + 3 and y = 2x – 5 parallel?
Step 1: Identify slopes
Line 1: y = 2x + 3 → slope = 2
Line 2: y = 2x – 5 → slope = 2
Step 2: Check y-intercepts
Line 1: y-intercept = 3
Line 2: y-intercept = -5
They are different ✓
Answer: YES, the lines are parallel because they have the same slope (2) and different y-intercepts (3 and -5).
Explanation: Two lines are parallel if and only if they have equal slopes but different y-intercepts. If they had the same y-intercept too, they would be the same line, not parallel.
Line 1: y = 2x + 3 → slope = 2
Line 2: y = 2x – 5 → slope = 2
Step 2: Check y-intercepts
Line 1: y-intercept = 3
Line 2: y-intercept = -5
They are different ✓
Answer: YES, the lines are parallel because they have the same slope (2) and different y-intercepts (3 and -5).
Explanation: Two lines are parallel if and only if they have equal slopes but different y-intercepts. If they had the same y-intercept too, they would be the same line, not parallel.
Solution 2: Finding Slope from Standard Form
Exercise: Find the slope of the line: 3x + 2y = 6. Is it parallel to y = -1.5x + 4?
Step 1: Convert to slope-intercept form
3x + 2y = 6
2y = -3x + 6
y = -1.5x + 3
Step 2: Identify the slope
Slope = -1.5 (or -3/2)
Step 3: Compare with y = -1.5x + 4
This line also has slope -1.5
Answer: YES, the lines are parallel because both have slope -1.5.
Explanation: When equations are in standard form (Ax + By + C = 0), always convert to slope-intercept form (y = mx + b) to easily identify and compare slopes.
3x + 2y = 6
2y = -3x + 6
y = -1.5x + 3
Step 2: Identify the slope
Slope = -1.5 (or -3/2)
Step 3: Compare with y = -1.5x + 4
This line also has slope -1.5
Answer: YES, the lines are parallel because both have slope -1.5.
Explanation: When equations are in standard form (Ax + By + C = 0), always convert to slope-intercept form (y = mx + b) to easily identify and compare slopes.
Solution 3: Writing Parallel Line Equations
Exercise: Write the equation of a line parallel to y = 5x + 1 with y-intercept of -3.
Step 1: Identify the slope of the given line
y = 5x + 1
Slope = 5
Step 2: Parallel lines have the same slope
New line slope = 5
Step 3: Use the y-intercept
We’re given: y-intercept = -3
So: b = -3
Step 4: Write the equation
y = mx + b
y = 5x + (-3)
y = 5x – 3
Explanation: To write a parallel line, keep the slope the same and change the y-intercept. This ensures the line will be parallel but not identical to the original.
y = 5x + 1
Slope = 5
Step 2: Parallel lines have the same slope
New line slope = 5
Step 3: Use the y-intercept
We’re given: y-intercept = -3
So: b = -3
Step 4: Write the equation
y = mx + b
y = 5x + (-3)
y = 5x – 3
Explanation: To write a parallel line, keep the slope the same and change the y-intercept. This ensures the line will be parallel but not identical to the original.
Solution 4: Finding Slope from Two Points
Exercise: Find the slope of the line passing through points (2, 4) and (5, 10).
Step 1: Identify the coordinates
Point 1: (2, 4) → x₁ = 2, y₁ = 4
Point 2: (5, 10) → x₂ = 5, y₂ = 10
Step 2: Use the slope formula: m = (y₂ – y₁)/(x₂ – x₁)
m = (10 – 4)/(5 – 2)
m = 6/3
m = 2
Answer: The slope is 2.
Explanation: The slope formula calculates the rate of change in y divided by the rate of change in x. A slope of 2 means for every 1 unit increase in x, y increases by 2 units.
Point 1: (2, 4) → x₁ = 2, y₁ = 4
Point 2: (5, 10) → x₂ = 5, y₂ = 10
Step 2: Use the slope formula: m = (y₂ – y₁)/(x₂ – x₁)
m = (10 – 4)/(5 – 2)
m = 6/3
m = 2
Answer: The slope is 2.
Explanation: The slope formula calculates the rate of change in y divided by the rate of change in x. A slope of 2 means for every 1 unit increase in x, y increases by 2 units.
Solution 5: Parallel Lines with Fractions
Exercise: Are lines y = -⅓x + 2 and y = -⅓x + 7 parallel?
Step 1: Identify slopes
Line 1: slope = -⅓
Line 2: slope = -⅓
Slopes are equal ✓
Step 2: Check y-intercepts
Line 1: y-intercept = 2
Line 2: y-intercept = 7
Y-intercepts are different ✓
Answer: YES, these lines are parallel.
Explanation: Working with fractions in slopes doesn’t change the parallel line test. As long as the fractions are equal, the slopes are equal, and the lines are parallel.
Line 1: slope = -⅓
Line 2: slope = -⅓
Slopes are equal ✓
Step 2: Check y-intercepts
Line 1: y-intercept = 2
Line 2: y-intercept = 7
Y-intercepts are different ✓
Answer: YES, these lines are parallel.
Explanation: Working with fractions in slopes doesn’t change the parallel line test. As long as the fractions are equal, the slopes are equal, and the lines are parallel.
Solution 6: Converting Standard Form
Exercise: Convert 4x – 2y + 8 = 0 to slope-intercept form and find its slope.
Step 1: Isolate the y term
4x – 2y + 8 = 0
-2y = -4x – 8
Step 2: Divide by -2
y = (-4x)/-2 + (-8)/-2
y = 2x + 4
Answer: Slope-intercept form is y = 2x + 4
Slope = 2
Explanation: When converting from standard form Ax + By + C = 0 to slope-intercept form y = mx + b, always: 1. Solve for y 2. Be careful with signs when dividing
4x – 2y + 8 = 0
-2y = -4x – 8
Step 2: Divide by -2
y = (-4x)/-2 + (-8)/-2
y = 2x + 4
Answer: Slope-intercept form is y = 2x + 4
Slope = 2
Explanation: When converting from standard form Ax + By + C = 0 to slope-intercept form y = mx + b, always: 1. Solve for y 2. Be careful with signs when dividing
Solution 7: Parallel Line Through Origin
Exercise: Write the equation of a line parallel to y = -2x + 5 that passes through (0, 0).
Step 1: Find the slope
Given line: y = -2x + 5
Slope = -2
Step 2: Parallel line has same slope
New slope = -2
Step 3: The line passes through (0, 0)
This is the origin, so the y-intercept b = 0
Step 4: Write the equation
y = -2x + 0
y = -2x
Answer: The equation is y = -2x
Given line: y = -2x + 5
Slope = -2
Step 2: Parallel line has same slope
New slope = -2
Step 3: The line passes through (0, 0)
This is the origin, so the y-intercept b = 0
Step 4: Write the equation
y = -2x + 0
y = -2x
Answer: The equation is y = -2x
Solution 8: Slope and Parallel Line Verification
Exercise: Find the slope of the line passing through (1, 1) and (3, 7). Is it parallel to y = 3x?
Step 1: Calculate the slope
Points: (1, 1) and (3, 7)
m = (7 – 1)/(3 – 1)
m = 6/2 = 3
Step 2: Compare with y = 3x
y = 3x has slope = 3
Answer: YES, the lines are parallel because both have slope 3.
Important Note: The line through (1, 1) and (3, 7) is NOT the same as y = 3x (because (1, 1) doesn’t satisfy y = 3x). But they are parallel with the same slope.
Points: (1, 1) and (3, 7)
m = (7 – 1)/(3 – 1)
m = 6/2 = 3
Step 2: Compare with y = 3x
y = 3x has slope = 3
Answer: YES, the lines are parallel because both have slope 3.
Important Note: The line through (1, 1) and (3, 7) is NOT the same as y = 3x (because (1, 1) doesn’t satisfy y = 3x). But they are parallel with the same slope.
Solution 9: Parallel Lines in Standard Form
Exercise: Are the lines 2x + y = 5 and 2x + y = 10 parallel?
Step 1: Convert first line to slope-intercept form
2x + y = 5
y = -2x + 5
Slope = -2
Step 2: Convert second line to slope-intercept form
2x + y = 10
y = -2x + 10
Slope = -2
Step 3: Check y-intercepts
Line 1: y-intercept = 5
Line 2: y-intercept = 10
Different y-intercepts ✓
Answer: YES, the lines are parallel (equal slopes, different y-intercepts).
2x + y = 5
y = -2x + 5
Slope = -2
Step 2: Convert second line to slope-intercept form
2x + y = 10
y = -2x + 10
Slope = -2
Step 3: Check y-intercepts
Line 1: y-intercept = 5
Line 2: y-intercept = 10
Different y-intercepts ✓
Answer: YES, the lines are parallel (equal slopes, different y-intercepts).
Solution 10: Using Point and Slope
Exercise: Find the equation of a line parallel to y = ½x – 1 passing through (2, 3).
Step 1: Identify the slope
Given line: y = ½x – 1
Slope = ½
Step 2: Parallel line has same slope
m = ½
Step 3: Find the y-intercept using point (2, 3)
Substitute into y = mx + b
3 = ½(2) + b
3 = 1 + b
b = 2
Step 4: Write the equation
y = ½x + 2
Verification: Point (2, 3): y = ½(2) + 2 = 1 + 2 = 3 ✓
Answer: y = ½x + 2
Given line: y = ½x – 1
Slope = ½
Step 2: Parallel line has same slope
m = ½
Step 3: Find the y-intercept using point (2, 3)
Substitute into y = mx + b
3 = ½(2) + b
3 = 1 + b
b = 2
Step 4: Write the equation
y = ½x + 2
Verification: Point (2, 3): y = ½(2) + 2 = 1 + 2 = 3 ✓
Answer: y = ½x + 2
Solution 11: Finding Slope from Standard Form
Exercise: What is the slope of any line parallel to 6x – 3y = 9?
Step 1: Convert to slope-intercept form
6x – 3y = 9
-3y = -6x + 9
y = 2x – 3
Step 2: Identify the slope
Slope = 2
Answer: Any line parallel to 6x – 3y = 9 must have slope 2.
Explanation: All parallel lines must share the same slope. So any line parallel to this will have slope 2, regardless of its y-intercept.
6x – 3y = 9
-3y = -6x + 9
y = 2x – 3
Step 2: Identify the slope
Slope = 2
Answer: Any line parallel to 6x – 3y = 9 must have slope 2.
Explanation: All parallel lines must share the same slope. So any line parallel to this will have slope 2, regardless of its y-intercept.
Solution 12: Different Slopes
Exercise: Are y = 4x + 2 and y = ¼x + 2 parallel?
Step 1: Identify slopes
Line 1: slope = 4
Line 2: slope = ¼
Step 2: Compare
4 ≠ ¼
Slopes are different
Answer: NO, these lines are NOT parallel.
Explanation: Even though they have the same y-intercept, they have different slopes. These lines intersect at the point (0, 2). Note: Lines with the same y-intercept but different slopes intersect at that y-intercept point.
Line 1: slope = 4
Line 2: slope = ¼
Step 2: Compare
4 ≠ ¼
Slopes are different
Answer: NO, these lines are NOT parallel.
Explanation: Even though they have the same y-intercept, they have different slopes. These lines intersect at the point (0, 2). Note: Lines with the same y-intercept but different slopes intersect at that y-intercept point.
Solution 13: Parallel to Line Through Two Points
Exercise: Find the equation of a line parallel to the line through (0, 2) and (3, 8) passing through (1, 5).
Step 1: Find slope of line through (0, 2) and (3, 8)
m = (8 – 2)/(3 – 0)
m = 6/3 = 2
Step 2: Parallel line has slope m = 2
Step 3: Find y-intercept using point (1, 5)
5 = 2(1) + b
5 = 2 + b
b = 3
Step 4: Write the equation
y = 2x + 3
Verification: Point (1, 5): y = 2(1) + 3 = 5 ✓
Answer: y = 2x + 3
m = (8 – 2)/(3 – 0)
m = 6/3 = 2
Step 2: Parallel line has slope m = 2
Step 3: Find y-intercept using point (1, 5)
5 = 2(1) + b
5 = 2 + b
b = 3
Step 4: Write the equation
y = 2x + 3
Verification: Point (1, 5): y = 2(1) + 3 = 5 ✓
Answer: y = 2x + 3
Solution 14: Converting and Comparing
Exercise: Convert -x + 3y = 12 to slope-intercept form. Is it parallel to y = ⅓x – 5?
Step 1: Convert to slope-intercept form
-x + 3y = 12
3y = x + 12
y = ⅓x + 4
Step 2: Identify the slope
Slope = ⅓
Step 3: Compare with y = ⅓x – 5
Both slopes = ⅓
Y-intercepts are different: 4 ≠ -5
Answer: YES, -x + 3y = 12 and y = ⅓x – 5 are parallel.
-x + 3y = 12
3y = x + 12
y = ⅓x + 4
Step 2: Identify the slope
Slope = ⅓
Step 3: Compare with y = ⅓x – 5
Both slopes = ⅓
Y-intercepts are different: 4 ≠ -5
Answer: YES, -x + 3y = 12 and y = ⅓x – 5 are parallel.
Solution 15: Vertical Lines
Exercise: Write the equation of a line parallel to x = 2 passing through (5, 0).
Step 1: Identify the type of line x = 2
This is a vertical line (all x-values are 2)
Step 2: What lines are parallel to a vertical line?
ALL vertical lines are parallel to each other
Step 3: Find the equation
The line passes through (5, 0)
So all points have x = 5
Equation: x = 5
Important Note: Vertical lines have undefined slope and cannot be written in y = mx + b form. Vertical lines x = c are parallel to each other. Similarly, all horizontal lines (y = c) are parallel to each other.
This is a vertical line (all x-values are 2)
Step 2: What lines are parallel to a vertical line?
ALL vertical lines are parallel to each other
Step 3: Find the equation
The line passes through (5, 0)
So all points have x = 5
Equation: x = 5
Important Note: Vertical lines have undefined slope and cannot be written in y = mx + b form. Vertical lines x = c are parallel to each other. Similarly, all horizontal lines (y = c) are parallel to each other.
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English Version
Parallel Lines – 15 Medium Level Exercises
Challenge Yourself with These Intermediate Problems
📝 Instructions: Solve each exercise using the properties of parallel lines. Click “View Solution” to check your answer.
💡 Tip: For parallel lines: equal slopes (m₁ = m₂), different y-intercepts (b₁ ≠ b₂)
Level: Medium | Topics: Slope calculations, equation finding, proving parallel lines, standard form conversions
Exercise 1
Find the equation of the line parallel to 2x + 3y – 6 = 0 that passes through the point (2, 1).
Medium
Solution:
Step 1: Convert 2x + 3y – 6 = 0 to slope-intercept form
3y = -2x + 6
y = -⅔x + 2
Slope = -⅔
Step 2: Parallel line has the same slope m = -⅔
Using point-slope form with (2, 1):
y – 1 = -⅔(x – 2)
y – 1 = -⅔x + 4/3
y = -⅔x + 7/3
Step 3: Convert to standard form if needed:
3y = -2x + 7 or 2x + 3y – 7 = 0
Step 1: Convert 2x + 3y – 6 = 0 to slope-intercept form
3y = -2x + 6
y = -⅔x + 2
Slope = -⅔
Step 2: Parallel line has the same slope m = -⅔
Using point-slope form with (2, 1):
y – 1 = -⅔(x – 2)
y – 1 = -⅔x + 4/3
y = -⅔x + 7/3
Step 3: Convert to standard form if needed:
3y = -2x + 7 or 2x + 3y – 7 = 0
Exercise 2
Two lines have equations 3x – 4y + 5 = 0 and 6x – 8y – 10 = 0. Are they parallel? Justify your answer.
Medium
Solution:
Step 1: Find slope of line 1: 3x – 4y + 5 = 0
-4y = -3x – 5
y = ¾x + 5/4
m₁ = ¾
Step 2: Find slope of line 2: 6x – 8y – 10 = 0
-8y = -6x + 10
y = ¾x – 5/4
m₂ = ¾
Step 3: Compare slopes and y-intercepts
m₁ = m₂ = ¾ ✓
b₁ = 5/4 ≠ b₂ = -5/4 ✓
YES, they are parallel (same slope, different y-intercepts)
Step 1: Find slope of line 1: 3x – 4y + 5 = 0
-4y = -3x – 5
y = ¾x + 5/4
m₁ = ¾
Step 2: Find slope of line 2: 6x – 8y – 10 = 0
-8y = -6x + 10
y = ¾x – 5/4
m₂ = ¾
Step 3: Compare slopes and y-intercepts
m₁ = m₂ = ¾ ✓
b₁ = 5/4 ≠ b₂ = -5/4 ✓
YES, they are parallel (same slope, different y-intercepts)
Exercise 3
Find the slope of a line parallel to the line passing through A(-2, 5) and B(3, 15).
Medium
Solution:
Step 1: Use slope formula m = (y₂ – y₁)/(x₂ – x₁)
m = (15 – 5)/(3 – (-2))
m = 10/5
m = 2
Step 2: Parallel lines have equal slopes
The slope of the parallel line = 2
Step 1: Use slope formula m = (y₂ – y₁)/(x₂ – x₁)
m = (15 – 5)/(3 – (-2))
m = 10/5
m = 2
Step 2: Parallel lines have equal slopes
The slope of the parallel line = 2
Exercise 4
The line y = 5x – 2 is parallel to line L. Line L passes through (1, -3). What is the equation of line L in the form y = mx + b?
Medium
Solution:
Step 1: Identify slope from y = 5x – 2
m = 5
Step 2: Parallel line has m = 5
Using point (1, -3):
-3 = 5(1) + b
-3 = 5 + b
b = -8
y = 5x – 8
Step 1: Identify slope from y = 5x – 2
m = 5
Step 2: Parallel line has m = 5
Using point (1, -3):
-3 = 5(1) + b
-3 = 5 + b
b = -8
y = 5x – 8
Exercise 5
Given: Line A passes through (-1, 2) and (3, 10). Line B passes through (0, -4) and (2, 0). Are lines A and B parallel? Show all working.
Medium
Solution:
Step 1: Find slope of line A
m_A = (10 – 2)/(3 – (-1))
m_A = 8/4 = 2
Step 2: Find slope of line B
m_B = (0 – (-4))/(2 – 0)
m_B = 4/2 = 2
Step 3: Compare slopes
m_A = m_B = 2 ✓
YES, lines A and B are parallel (both have slope 2)
Step 1: Find slope of line A
m_A = (10 – 2)/(3 – (-1))
m_A = 8/4 = 2
Step 2: Find slope of line B
m_B = (0 – (-4))/(2 – 0)
m_B = 4/2 = 2
Step 3: Compare slopes
m_A = m_B = 2 ✓
YES, lines A and B are parallel (both have slope 2)
Exercise 6
Write the equation of the line parallel to 5x – 2y + 8 = 0 that passes through (-2, 3) in the form ax + by + c = 0.
Medium
Solution:
Step 1: Find slope from 5x – 2y + 8 = 0
-2y = -5x – 8
y = 5/2 x + 4
m = 5/2
Step 2: Use point-slope form with (-2, 3)
y – 3 = 5/2(x – (-2))
y – 3 = 5/2(x + 2)
y – 3 = 5/2 x + 5
y = 5/2 x + 8
Step 3: Convert to standard form
2y = 5x + 16
5x – 2y + 16 = 0
Step 1: Find slope from 5x – 2y + 8 = 0
-2y = -5x – 8
y = 5/2 x + 4
m = 5/2
Step 2: Use point-slope form with (-2, 3)
y – 3 = 5/2(x – (-2))
y – 3 = 5/2(x + 2)
y – 3 = 5/2 x + 5
y = 5/2 x + 8
Step 3: Convert to standard form
2y = 5x + 16
5x – 2y + 16 = 0
Exercise 7
A line passes through (4, -1) and is parallel to the line 3x + 7y = 14. Find the y-intercept of this new line.
Medium
Solution:
Step 1: Find slope from 3x + 7y = 14
7y = -3x + 14
y = -3/7 x + 2
m = -3/7
Step 2: Find equation through (4, -1) with m = -3/7
-1 = -3/7(4) + b
-1 = -12/7 + b
b = -1 + 12/7
b = -7/7 + 12/7
b = 5/7
The y-intercept is 5/7
Step 1: Find slope from 3x + 7y = 14
7y = -3x + 14
y = -3/7 x + 2
m = -3/7
Step 2: Find equation through (4, -1) with m = -3/7
-1 = -3/7(4) + b
-1 = -12/7 + b
b = -1 + 12/7
b = -7/7 + 12/7
b = 5/7
The y-intercept is 5/7
Exercise 8
Lines L₁: 4x – 6y = 12 and L₂: 2x – 3y = 5 are parallel. Verify this statement mathematically.
Medium
Solution:
Step 1: Find slope of L₁: 4x – 6y = 12
-6y = -4x + 12
y = 2/3 x – 2
m₁ = 2/3
Step 2: Find slope of L₂: 2x – 3y = 5
-3y = -2x + 5
y = 2/3 x – 5/3
m₂ = 2/3
Step 3: Verify
m₁ = m₂ = 2/3 ✓
b₁ = -2 ≠ b₂ = -5/3 ✓
VERIFIED: Lines are parallel (equal slopes, different intercepts)
Step 1: Find slope of L₁: 4x – 6y = 12
-6y = -4x + 12
y = 2/3 x – 2
m₁ = 2/3
Step 2: Find slope of L₂: 2x – 3y = 5
-3y = -2x + 5
y = 2/3 x – 5/3
m₂ = 2/3
Step 3: Verify
m₁ = m₂ = 2/3 ✓
b₁ = -2 ≠ b₂ = -5/3 ✓
VERIFIED: Lines are parallel (equal slopes, different intercepts)
Exercise 9
Find the equation of the line parallel to the x-axis that passes through the point (5, -3).
Medium
Solution:
Step 1: The x-axis has equation y = 0
The slope of the x-axis is m = 0
Step 2: A line parallel to the x-axis is horizontal
All horizontal lines have the form y = k, where k is a constant
Step 3: Since it passes through (5, -3)
The y-coordinate is -3
y = -3
Step 1: The x-axis has equation y = 0
The slope of the x-axis is m = 0
Step 2: A line parallel to the x-axis is horizontal
All horizontal lines have the form y = k, where k is a constant
Step 3: Since it passes through (5, -3)
The y-coordinate is -3
y = -3
Exercise 10
Two parallel lines have slopes m = 3/4. If one line passes through (2, 5) and the other through (-4, -1), what is the distance between their y-intercepts?
Medium
Solution:
Step 1: Find y-intercept of first line (2, 5), m = 3/4
5 = 3/4(2) + b₁
5 = 3/2 + b₁
b₁ = 5 – 3/2 = 7/2
Step 2: Find y-intercept of second line (-4, -1), m = 3/4
-1 = 3/4(-4) + b₂
-1 = -3 + b₂
b₂ = 2
Step 3: Distance between y-intercepts
Distance = |b₁ – b₂| = |7/2 – 2|
= |7/2 – 4/2| = |3/2|
Distance = 3/2 or 1.5
Step 1: Find y-intercept of first line (2, 5), m = 3/4
5 = 3/4(2) + b₁
5 = 3/2 + b₁
b₁ = 5 – 3/2 = 7/2
Step 2: Find y-intercept of second line (-4, -1), m = 3/4
-1 = 3/4(-4) + b₂
-1 = -3 + b₂
b₂ = 2
Step 3: Distance between y-intercepts
Distance = |b₁ – b₂| = |7/2 – 2|
= |7/2 – 4/2| = |3/2|
Distance = 3/2 or 1.5
Exercise 11
Show that y = ½x + 3 and x – 2y + 8 = 0 are NOT parallel lines. Explain your reasoning.
Medium
Solution:
Step 1: Find slope of y = ½x + 3
m₁ = ½
Step 2: Find slope of x – 2y + 8 = 0
-2y = -x – 8
y = ½x + 4
m₂ = ½
Wait! Both slopes ARE equal (m₁ = m₂ = ½)
Step 3: Check y-intercepts
b₁ = 3 (from first equation)
b₂ = 4 (from second equation)
b₁ ≠ b₂ ✓
CORRECTION: The lines ARE parallel (equal slopes, different y-intercepts)
Step 1: Find slope of y = ½x + 3
m₁ = ½
Step 2: Find slope of x – 2y + 8 = 0
-2y = -x – 8
y = ½x + 4
m₂ = ½
Wait! Both slopes ARE equal (m₁ = m₂ = ½)
Step 3: Check y-intercepts
b₁ = 3 (from first equation)
b₂ = 4 (from second equation)
b₁ ≠ b₂ ✓
CORRECTION: The lines ARE parallel (equal slopes, different y-intercepts)
Exercise 12
Line P passes through (1, 4) and (5, 12). Line Q passes through (0, 1) and (3, 7). Determine if P and Q are parallel. Find equations for both lines.
Medium
Solution:
Step 1: Find slope of line P
m_P = (12 – 4)/(5 – 1) = 8/4 = 2
Step 2: Find equation of line P using (1, 4)
4 = 2(1) + b
b = 2
Line P: y = 2x + 2
Step 3: Find slope of line Q
m_Q = (7 – 1)/(3 – 0) = 6/3 = 2
Step 4: Find equation of line Q using (0, 1)
1 = 2(0) + b
b = 1
Line Q: y = 2x + 1
Step 5: Compare slopes and intercepts
m_P = m_Q = 2 ✓
b_P = 2 ≠ b_Q = 1 ✓
YES, lines P and Q are parallel
Step 1: Find slope of line P
m_P = (12 – 4)/(5 – 1) = 8/4 = 2
Step 2: Find equation of line P using (1, 4)
4 = 2(1) + b
b = 2
Line P: y = 2x + 2
Step 3: Find slope of line Q
m_Q = (7 – 1)/(3 – 0) = 6/3 = 2
Step 4: Find equation of line Q using (0, 1)
1 = 2(0) + b
b = 1
Line Q: y = 2x + 1
Step 5: Compare slopes and intercepts
m_P = m_Q = 2 ✓
b_P = 2 ≠ b_Q = 1 ✓
YES, lines P and Q are parallel
Exercise 13
Write the equation in standard form (ax + by + c = 0) of a line parallel to 7x – 5y + 2 = 0 passing through (-3, 4).
Medium
Solution:
Step 1: Find slope from 7x – 5y + 2 = 0
-5y = -7x – 2
y = 7/5 x + 2/5
m = 7/5
Step 2: Use point-slope form with (-3, 4), m = 7/5
y – 4 = 7/5(x – (-3))
y – 4 = 7/5(x + 3)
y – 4 = 7/5 x + 21/5
y = 7/5 x + 21/5 + 4
y = 7/5 x + 21/5 + 20/5
y = 7/5 x + 41/5
Step 3: Convert to standard form
5y = 7x + 41
7x – 5y + 41 = 0
Step 1: Find slope from 7x – 5y + 2 = 0
-5y = -7x – 2
y = 7/5 x + 2/5
m = 7/5
Step 2: Use point-slope form with (-3, 4), m = 7/5
y – 4 = 7/5(x – (-3))
y – 4 = 7/5(x + 3)
y – 4 = 7/5 x + 21/5
y = 7/5 x + 21/5 + 4
y = 7/5 x + 21/5 + 20/5
y = 7/5 x + 41/5
Step 3: Convert to standard form
5y = 7x + 41
7x – 5y + 41 = 0
Exercise 14
A line with equation y = -2/3 x + 5 is parallel to line R. Line R passes through (6, -4). Find the equation of line R and verify both lines are parallel.
Medium
Solution:
Step 1: Identify slope
m = -2/3
Step 2: Find equation of line R through (6, -4)
-4 = -2/3(6) + b
-4 = -4 + b
b = 0
Line R: y = -2/3 x
Step 3: Verify parallelism
Original line: y = -2/3 x + 5, slope = -2/3
Line R: y = -2/3 x, slope = -2/3
Slopes are equal ✓
Y-intercepts: 5 ≠ 0 ✓
Line R: y = -2/3 x (verified as parallel)
Step 1: Identify slope
m = -2/3
Step 2: Find equation of line R through (6, -4)
-4 = -2/3(6) + b
-4 = -4 + b
b = 0
Line R: y = -2/3 x
Step 3: Verify parallelism
Original line: y = -2/3 x + 5, slope = -2/3
Line R: y = -2/3 x, slope = -2/3
Slopes are equal ✓
Y-intercepts: 5 ≠ 0 ✓
Line R: y = -2/3 x (verified as parallel)
Exercise 15
Three lines are given: L₁: 2y = 4x – 3, L₂: y = 2x + 7, L₃: 4x – 2y – 5 = 0. Identify which lines are parallel to each other and prove your answer.
Medium
Solution:
Step 1: Find slope of L₁: 2y = 4x – 3
y = 2x – 3/2
m₁ = 2
Step 2: Find slope of L₂: y = 2x + 7
m₂ = 2
Step 3: Find slope of L₃: 4x – 2y – 5 = 0
-2y = -4x + 5
y = 2x – 5/2
m₃ = 2
Step 4: Compare slopes and y-intercepts
m₁ = m₂ = m₃ = 2 ✓
b₁ = -3/2, b₂ = 7, b₃ = -5/2
All y-intercepts are different ✓
ALL THREE LINES ARE PARALLEL TO EACH OTHER (all have slope 2, all have different y-intercepts)
Step 1: Find slope of L₁: 2y = 4x – 3
y = 2x – 3/2
m₁ = 2
Step 2: Find slope of L₂: y = 2x + 7
m₂ = 2
Step 3: Find slope of L₃: 4x – 2y – 5 = 0
-2y = -4x + 5
y = 2x – 5/2
m₃ = 2
Step 4: Compare slopes and y-intercepts
m₁ = m₂ = m₃ = 2 ✓
b₁ = -3/2, b₂ = 7, b₃ = -5/2
All y-intercepts are different ✓
ALL THREE LINES ARE PARALLEL TO EACH OTHER (all have slope 2, all have different y-intercepts)
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Abitur Level – English
Parallel Lines – Advanced Exercises
15 Challenging Abitur-Level Problems | Brandenburg Curriculum
📚 Instructions: These exercises require deep understanding of parallel lines, geometric properties, and analytical geometry.
💡 Strategy: Work through each problem systematically. Use the hint box if you’re stuck, then check your solution.
⏱️ Estimated Time: 2-3 hours for all 15 exercises.
Exercise 1 Hard
Three parallel lines l₁, l₂, and l₃ are intersected by two transversals t₁ and t₂.
The distance between l₁ and l₂ is 3 units, and between l₂ and l₃ is 5 units.
If t₁ intersects l₁ at A and l₃ at C, with |AC| = 12 units, find the distances |AB| and |BC|
where B is the intersection of t₁ with l₂.
💡 Hint: Use the intercept theorem (basic proportionality theorem). When a line is divided by parallel lines, the segments are proportional.
Solution:
Using the intercept theorem: |AB|/|BC| = distance(l₁,l₂)/distance(l₂,l₃) = 3/5
Also: |AB| + |BC| = 12
Let |AB| = 3x and |BC| = 5x
3x + 5x = 12 → 8x = 12 → x = 1.5
Therefore: |AB| = 4.5 units, |BC| = 7.5 units
Using the intercept theorem: |AB|/|BC| = distance(l₁,l₂)/distance(l₂,l₃) = 3/5
Also: |AB| + |BC| = 12
Let |AB| = 3x and |BC| = 5x
3x + 5x = 12 → 8x = 12 → x = 1.5
Therefore: |AB| = 4.5 units, |BC| = 7.5 units
Exercise 2 Hard
Given line l: 3x – 4y + 12 = 0. Find all lines that are parallel to l
and are at a distance of 2 units from l.
💡 Hint: Parallel lines have the same coefficients for x and y. Use the distance formula between parallel lines.
Solution:
Parallel lines have form: 3x – 4y + c = 0
Distance between 3x – 4y + 12 = 0 and 3x – 4y + c = 0:
d = |c – 12|/√(9 + 16) = |c – 12|/5
Setting d = 2: |c – 12|/5 = 2 → |c – 12| = 10
c – 12 = 10 or c – 12 = -10
Lines: 3x – 4y + 22 = 0 and 3x – 4y + 2 = 0
Parallel lines have form: 3x – 4y + c = 0
Distance between 3x – 4y + 12 = 0 and 3x – 4y + c = 0:
d = |c – 12|/√(9 + 16) = |c – 12|/5
Setting d = 2: |c – 12|/5 = 2 → |c – 12| = 10
c – 12 = 10 or c – 12 = -10
Lines: 3x – 4y + 22 = 0 and 3x – 4y + 2 = 0
Exercise 3 Hard
A trapezoid ABCD has parallel sides AB and CD. Given A(0, 0), B(6, 0),
and C(7, 4). If D is on the line passing through C parallel to AB,
find the coordinates of D such that |AD| = |BC|.
💡 Hint: AB is horizontal (y = 0). CD must also be horizontal. Use the distance formula for |AD| = |BC|.
Solution:
AB is on y = 0, so CD is parallel to y = 0
Since C(7, 4), line CD: y = 4
D = (x, 4) for some x
|BC| = √((7-6)² + (4-0)²) = √(1 + 16) = √17
|AD| = √(x² + 16) = √17
x² + 16 = 17 → x² = 1 → x = ±1
D = (1, 4) or D = (-1, 4)
AB is on y = 0, so CD is parallel to y = 0
Since C(7, 4), line CD: y = 4
D = (x, 4) for some x
|BC| = √((7-6)² + (4-0)²) = √(1 + 16) = √17
|AD| = √(x² + 16) = √17
x² + 16 = 17 → x² = 1 → x = ±1
D = (1, 4) or D = (-1, 4)
Exercise 4 Very Hard
Given two parallel lines l₁: 2x + 3y = 10 and l₂: 2x + 3y = 25.
A point P moves along l₁, and from P, perpendiculars are drawn to both lines.
The feet of the perpendiculars are Q (on l₁) and R (on l₂).
Prove that |QR| is constant and find its value.
💡 Hint: The perpendicular distance between parallel lines is constant. The distance from any point to a parallel line follows a specific formula.
Solution:
Distance between l₁ and l₂: d = |25 – 10|/√(4 + 9) = 15/√13
For any point P on l₁, the perpendicular to l₂ is constant
Since Q is on l₁ (same line as P), |QR| = perpendicular distance
|QR| = 15/√13 = 15√13/13 ≈ 4.16 units (constant)
Distance between l₁ and l₂: d = |25 – 10|/√(4 + 9) = 15/√13
For any point P on l₁, the perpendicular to l₂ is constant
Since Q is on l₁ (same line as P), |QR| = perpendicular distance
|QR| = 15/√13 = 15√13/13 ≈ 4.16 units (constant)
Exercise 5 Hard
Three parallel lines l₁, l₂, l₃ cut off segments of lengths 4 and 6 on a transversal t₁,
and segments of lengths a and b on transversal t₂. If a + b = 15, find a and b.
💡 Hint: By the intercept theorem, segments cut by parallel lines on any two transversals are proportional.
Solution:
By intercept theorem: a/4 = b/6
From a/4 = b/6 → 6a = 4b → 3a = 2b
Also: a + b = 15
Substituting b = 3a/2: a + 3a/2 = 15 → 5a/2 = 15 → a = 6
b = 15 – 6 = 9
Verify: 6/4 = 9/6 → 3/2 = 3/2 ✓
a = 6, b = 9
By intercept theorem: a/4 = b/6
From a/4 = b/6 → 6a = 4b → 3a = 2b
Also: a + b = 15
Substituting b = 3a/2: a + 3a/2 = 15 → 5a/2 = 15 → a = 6
b = 15 – 6 = 9
Verify: 6/4 = 9/6 → 3/2 = 3/2 ✓
a = 6, b = 9
Exercise 6 Very Hard
A line passes through A(1, 2) and B(4, 8). Find the equation of the line parallel to AB
that forms a triangle with the coordinate axes having area 18 square units.
💡 Hint: First find the slope of AB. Parallel lines have the same slope. Use the intercept form and area formula for triangles.
Solution:
Slope of AB: m = (8-2)/(4-1) = 6/3 = 2
Parallel line: y = 2x + c
x-intercept: 0 = 2x + c → x = -c/2
y-intercept: y = 2(0) + c = c
Area = (1/2)|x-intercept||y-intercept| = (1/2)|−c/2||c| = c²/4
Setting c²/4 = 18 → c² = 72 → c = ±6√2
Lines: y = 2x + 6√2 and y = 2x – 6√2
Slope of AB: m = (8-2)/(4-1) = 6/3 = 2
Parallel line: y = 2x + c
x-intercept: 0 = 2x + c → x = -c/2
y-intercept: y = 2(0) + c = c
Area = (1/2)|x-intercept||y-intercept| = (1/2)|−c/2||c| = c²/4
Setting c²/4 = 18 → c² = 72 → c = ±6√2
Lines: y = 2x + 6√2 and y = 2x – 6√2
Exercise 7 Hard
Parallel lines l₁ and l₂ are separated by distance d. A point P is at distance d₁ from l₁
and distance d₂ from l₂. Prove that either d₁ + d₂ = d or |d₁ – d₂| = d, depending on whether P is between the lines or not.
💡 Hint: Consider two cases: P between the lines, and P outside both lines. Use coordinate geometry.
Solution:
Let l₁: y = 0 and l₂: y = d (distance d apart)
Case 1 – P between lines (0 < y_P < d):
d₁ = y_P, d₂ = d – y_P
d₁ + d₂ = y_P + (d – y_P) = d ✓
Case 2 – P below l₁ (y_P < 0):
d₁ = -y_P, d₂ = d – y_P
|d₁ – d₂| = |-y_P – (d – y_P)| = |-d| = d ✓
Case 3 – P above l₂ (y_P > d):
d₁ = y_P, d₂ = y_P – d
|d₁ – d₂| = |y_P – (y_P – d)| = d ✓
Proof complete.
Let l₁: y = 0 and l₂: y = d (distance d apart)
Case 1 – P between lines (0 < y_P < d):
d₁ = y_P, d₂ = d – y_P
d₁ + d₂ = y_P + (d – y_P) = d ✓
Case 2 – P below l₁ (y_P < 0):
d₁ = -y_P, d₂ = d – y_P
|d₁ – d₂| = |-y_P – (d – y_P)| = |-d| = d ✓
Case 3 – P above l₂ (y_P > d):
d₁ = y_P, d₂ = y_P – d
|d₁ – d₂| = |y_P – (y_P – d)| = d ✓
Proof complete.
Exercise 8 Hard
Two parallel lines are intersected by two parallel transversals, forming a parallelogram.
If the sides of the parallelogram are in the ratio 3:5, and the perimeter is 32 units,
find the lengths of all four sides.
💡 Hint: In a parallelogram, opposite sides are equal. Use the given ratio and perimeter.
Solution:
In a parallelogram, opposite sides are equal
Let sides be 3x and 5x
Perimeter = 2(3x) + 2(5x) = 6x + 10x = 16x = 32
x = 2
Sides: 6, 10, 6, 10 units
In a parallelogram, opposite sides are equal
Let sides be 3x and 5x
Perimeter = 2(3x) + 2(5x) = 6x + 10x = 16x = 32
x = 2
Sides: 6, 10, 6, 10 units
Exercise 9 Very Hard
Given parallel lines l₁: ax + by + c₁ = 0 and l₂: ax + by + c₂ = 0.
A point P(x₀, y₀) is at distance d₁ from l₁ and d₂ from l₂.
Express d₁ and d₂ in terms of a, b, c₁, c₂, x₀, y₀, and find their relationship.
💡 Hint: Use the point-to-line distance formula. Express both distances and find their sum or difference.
Solution:
Distance formula: d = |ax₀ + by₀ + c|/√(a² + b²)
d₁ = |ax₀ + by₀ + c₁|/√(a² + b²)
d₂ = |ax₀ + by₀ + c₂|/√(a² + b²)
Case 1 (P between lines): d₁ + d₂ = (c₂ – c₁)/√(a² + b²)
Case 2 (P outside): |d₁ – d₂| = (c₂ – c₁)/√(a² + b²)
Key relationship: The distances relate to |c₂ – c₁|/√(a² + b²), which is the distance between the parallel lines.
Distance formula: d = |ax₀ + by₀ + c|/√(a² + b²)
d₁ = |ax₀ + by₀ + c₁|/√(a² + b²)
d₂ = |ax₀ + by₀ + c₂|/√(a² + b²)
Case 1 (P between lines): d₁ + d₂ = (c₂ – c₁)/√(a² + b²)
Case 2 (P outside): |d₁ – d₂| = (c₂ – c₁)/√(a² + b²)
Key relationship: The distances relate to |c₂ – c₁|/√(a² + b²), which is the distance between the parallel lines.
Exercise 10 Hard
Three points A(0, 0), B(6, 0), and C(2, 4) form a triangle.
Find the equation of the line parallel to BC passing through A.
💡 Hint: Find the slope of BC first, then use point-slope form with point A.
Solution:
Slope of BC: m = (4 – 0)/(2 – 6) = 4/(-4) = -1
Line through A(0, 0) with slope -1:
y – 0 = -1(x – 0)
y = -x or x + y = 0
Slope of BC: m = (4 – 0)/(2 – 6) = 4/(-4) = -1
Line through A(0, 0) with slope -1:
y – 0 = -1(x – 0)
y = -x or x + y = 0
Exercise 11 Very Hard
A rectangle has vertices at A(0, 0), B(a, 0), C(a, b), D(0, b).
A line parallel to the diagonal AC intersects side AB at point P(p, 0) and side CD at point Q(q, b).
Find the relationship between p, q, a, and b.
💡 Hint: The slope of AC is b/a. A parallel line through P and Q has the same slope.
Solution:
Slope of AC: m = b/a
Slope of PQ: (b – 0)/(q – p) = b/(q – p)
Since PQ ∥ AC: b/(q – p) = b/a
Therefore: q – p = a
Relationship: q = p + a
This means P and Q differ in x-coordinate by exactly a, the width of the rectangle.
Slope of AC: m = b/a
Slope of PQ: (b – 0)/(q – p) = b/(q – p)
Since PQ ∥ AC: b/(q – p) = b/a
Therefore: q – p = a
Relationship: q = p + a
This means P and Q differ in x-coordinate by exactly a, the width of the rectangle.
Exercise 12 Hard
Two parallel lines have equations 4x – 3y = 5 and 4x – 3y = 20.
Find a point on the first line that is closest to the second line, and calculate this minimum distance.
💡 Hint: The minimum distance between two parallel lines is the perpendicular distance. Any point on one line is at this distance from the other.
Solution:
Distance between parallel lines: d = |20 – 5|/√(16 + 9) = 15/√25 = 15/5 = 3
The perpendicular from any point on line 1 to line 2 is constant = 3
To find a specific point, we can find the foot of perpendicular from origin:
Line through origin perpendicular to both: 3x + 4y = 0 (perpendicular slope is 3/4)
Intersection with 4x – 3y = 5:
From 3x + 4y = 0 → y = -3x/4
4x – 3(-3x/4) = 5 → 4x + 9x/4 = 5 → 25x/4 = 5 → x = 4/5, y = -3/5
Point on line 1: (4/5, -3/5), Minimum distance = 3 units
Distance between parallel lines: d = |20 – 5|/√(16 + 9) = 15/√25 = 15/5 = 3
The perpendicular from any point on line 1 to line 2 is constant = 3
To find a specific point, we can find the foot of perpendicular from origin:
Line through origin perpendicular to both: 3x + 4y = 0 (perpendicular slope is 3/4)
Intersection with 4x – 3y = 5:
From 3x + 4y = 0 → y = -3x/4
4x – 3(-3x/4) = 5 → 4x + 9x/4 = 5 → 25x/4 = 5 → x = 4/5, y = -3/5
Point on line 1: (4/5, -3/5), Minimum distance = 3 units
Exercise 13 Very Hard
Prove that if a transversal intersects three parallel lines at points A, B, C,
and another transversal intersects them at points A’, B’, C’, then |AB|/|BC| = |A’B’|/|B’C’|.
💡 Hint: Use similar triangles or the intercept theorem. Consider the angles formed by parallel lines and transversals.
Solution:
Let the three parallel lines be l₁, l₂, l₃
First transversal t₁ intersects at A, B, C
Second transversal t₂ intersects at A’, B’, C’
Since l₁ ∥ l₂ ∥ l₃:
– In triangles formed by t₁ and t₂ extended, angles are equal (corresponding angles)
– The triangles formed are similar
By intercept theorem (Thales’ theorem):
|AB|/|BC| = |A’B’|/|B’C’|
This is a fundamental property of parallel lines cutting transversals.
Let the three parallel lines be l₁, l₂, l₃
First transversal t₁ intersects at A, B, C
Second transversal t₂ intersects at A’, B’, C’
Since l₁ ∥ l₂ ∥ l₃:
– In triangles formed by t₁ and t₂ extended, angles are equal (corresponding angles)
– The triangles formed are similar
By intercept theorem (Thales’ theorem):
|AB|/|BC| = |A’B’|/|B’C’|
This is a fundamental property of parallel lines cutting transversals.
Exercise 14 Hard
A line l: 5x + 12y – 26 = 0 and a point P(1, 2).
Find the equation of the line through P that is parallel to l,
and verify that P lies on this new line.
💡 Hint: Parallel lines have the same coefficients for x and y. Use point P to find the constant term.
Solution:
Parallel line through P: 5x + 12y + c = 0
Since P(1, 2) lies on this line:
5(1) + 12(2) + c = 0
5 + 24 + c = 0
c = -29
Equation: 5x + 12y – 29 = 0
Verification: 5(1) + 12(2) – 29 = 5 + 24 – 29 = 0 ✓
Parallel line through P: 5x + 12y + c = 0
Since P(1, 2) lies on this line:
5(1) + 12(2) + c = 0
5 + 24 + c = 0
c = -29
Equation: 5x + 12y – 29 = 0
Verification: 5(1) + 12(2) – 29 = 5 + 24 – 29 = 0 ✓
Exercise 15 Very Hard
A parallelogram has sides on the lines x + 2y = 3, x + 2y = 9,
3x – y = 1, and 3x – y = 13. Find the vertices of this parallelogram.
💡 Hint: Opposite sides of a parallelogram lie on parallel lines. Find intersections of perpendicular sides.
Solution:
Lines x + 2y = 3 and x + 2y = 9 are parallel (opposite sides)
Lines 3x – y = 1 and 3x – y = 13 are parallel (opposite sides)
Vertex 1: Intersection of x + 2y = 3 and 3x – y = 1
From x + 2y = 3 → x = 3 – 2y
3(3 – 2y) – y = 1 → 9 – 6y – y = 1 → -7y = -8 → y = 8/7
x = 3 – 2(8/7) = 3 – 16/7 = 5/7
V₁ = (5/7, 8/7)
Vertex 2: Intersection of x + 2y = 3 and 3x – y = 13
3(3 – 2y) – y = 13 → 9 – 7y = 13 → y = -4/7
x = 3 – 2(-4/7) = 3 + 8/7 = 29/7
V₂ = (29/7, -4/7)
Vertex 3: Intersection of x + 2y = 9 and 3x – y = 13
From x + 2y = 9 → x = 9 – 2y
3(9 – 2y) – y = 13 → 27 – 7y = 13 → y = 2
x = 9 – 4 = 5
V₃ = (5, 2)
Vertex 4: Intersection of x + 2y = 9 and 3x – y = 1
3(9 – 2y) – y = 1 → 27 – 7y = 1 → y = 26/7
x = 9 – 52/7 = 11/7
V₄ = (11/7, 26/7)
Vertices: (5/7, 8/7), (29/7, -4/7), (5, 2), (11/7, 26/7)
Lines x + 2y = 3 and x + 2y = 9 are parallel (opposite sides)
Lines 3x – y = 1 and 3x – y = 13 are parallel (opposite sides)
Vertex 1: Intersection of x + 2y = 3 and 3x – y = 1
From x + 2y = 3 → x = 3 – 2y
3(3 – 2y) – y = 1 → 9 – 6y – y = 1 → -7y = -8 → y = 8/7
x = 3 – 2(8/7) = 3 – 16/7 = 5/7
V₁ = (5/7, 8/7)
Vertex 2: Intersection of x + 2y = 3 and 3x – y = 13
3(3 – 2y) – y = 13 → 9 – 7y = 13 → y = -4/7
x = 3 – 2(-4/7) = 3 + 8/7 = 29/7
V₂ = (29/7, -4/7)
Vertex 3: Intersection of x + 2y = 9 and 3x – y = 13
From x + 2y = 9 → x = 9 – 2y
3(9 – 2y) – y = 13 → 27 – 7y = 13 → y = 2
x = 9 – 4 = 5
V₃ = (5, 2)
Vertex 4: Intersection of x + 2y = 9 and 3x – y = 1
3(9 – 2y) – y = 1 → 27 – 7y = 1 → y = 26/7
x = 9 – 52/7 = 11/7
V₄ = (11/7, 26/7)
Vertices: (5/7, 8/7), (29/7, -4/7), (5, 2), (11/7, 26/7)
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Parallel Lines
15 Comprehensive Exercises with Detailed Solutions
Exercise 1
Finding the Equation of a Line Through Two Points
📍 Problem:
Find the equation of the line passing through points A(-3, -6) and B(-5, 4).
Step 1:
Calculate the slope m
m = (y₂ – y₁) / (x₂ – x₁) = (4 – (-6)) / (-5 – (-3)) = 10 / (-2) = -5
Step 2:
Use point-slope form with point A(-3, -6):
y – y₁ = m(x – x₁)
y – (-6) = -5(x – (-3))
y + 6 = -5(x + 3)
y – (-6) = -5(x – (-3))
y + 6 = -5(x + 3)
Step 3:
Expand and simplify:
y + 6 = -5x – 15
y = -5x – 21
y = -5x – 21
Step 4:
Verify with point B:
y = -5(-5) – 21 = 25 – 21 = 4 ✓ (matches B’s y-coordinate)
✓ Answer: y = -5x – 21
Explanation: To find a line’s equation, we need the slope and a point. We calculate slope using the formula (rise over run), then use the point-slope form to build the equation. Always verify by substituting both points back into the equation.
Exercise 2
Checking if Points Lie on a Line
📍 Problem:
Given the line AB with equation y = -5x – 21 from Exercise 1, check if points Q(1, 2) and M(4, 1) lie on this line.
Step 1:
Check point Q(1, 2):
Substitute x = 1: y = -5(1) – 21 = -5 – 21 = -26
We get y = -26, but Q has y = 2. Since -26 ≠ 2, point Q does NOT lie on the line.
Step 2:
Check point M(4, 1):
Substitute x = 4: y = -5(4) – 21 = -20 – 21 = -41
We get y = -41, but M has y = 1. Since -41 ≠ 1, point M does NOT lie on the line.
✓ Answer: Neither Q nor M lie on line AB.
Explanation: To check if a point lies on a line, substitute its x-coordinate into the line’s equation and see if you get the same y-coordinate. If yes, the point is on the line; if no, it isn’t.
Exercise 3
Finding a Parallel Line Through a Point
📍 Problem:
Find the equation of line g₁ that is parallel to line AB (y = -5x – 21) and passes through point Q(1, 2).
Step 1:
Identify the slope of AB:
From y = -5x – 21, we see the slope is m = -5
Step 2:
Apply the parallel line rule:
Parallel lines have the SAME slope. So g₁ also has slope m = -5
Step 3:
Use point-slope form with Q(1, 2):
y – 2 = -5(x – 1)
y – 2 = -5x + 5
y = -5x + 7
y – 2 = -5x + 5
y = -5x + 7
Step 4:
Verify Q is on the line:
y = -5(1) + 7 = -5 + 7 = 2 ✓
✓ Answer: g₁: y = -5x + 7
Explanation: Key rule: Parallel lines have equal slopes! To find a parallel line through a given point, use the same slope and the point-slope form with the new point.
Exercise 4
Finding a Perpendicular Line Through a Point
📍 Problem:
Find the equation of line g₂ that is perpendicular to line AB (y = -5x – 21) and passes through point Q(1, 2).
Step 1:
Identify the slope of AB:
From y = -5x – 21, the slope is m₁ = -5
Step 2:
Apply the perpendicular line rule:
For perpendicular lines: m₁ · m₂ = -1
-5 · m₂ = -1
m₂ = -1 / (-5) = 1/5
m₂ = -1 / (-5) = 1/5
Step 3:
Use point-slope form with Q(1, 2):
y – 2 = (1/5)(x – 1)
y – 2 = (1/5)x – 1/5
y = (1/5)x + 9/5
y – 2 = (1/5)x – 1/5
y = (1/5)x + 9/5
✓ Answer: g₂: y = (1/5)x + 9/5 or g₂: y = 0.2x + 1.8
Explanation: Key rule: For perpendicular lines, the product of their slopes equals -1. If one line has slope m, the perpendicular line has slope -1/m (the negative reciprocal).
Exercise 5
Finding a Line Parallel to Another Through Two Points
📍 Problem:
Find the equation of a line that is parallel to line 3x – 3 and passes through both point M and point Q (which are on line 3x – 3).
⚠️ Important Note: If a line passes through two different points that are both on the same line (3x – 3), then it must BE that same line (or it’s impossible). The problem likely means finding the line MQ itself.
Step 1:
Find the slope of MQ:
We need coordinates of M and Q. From the context: M and Q lie on 3x – 3, so they have slope 3.
✓ Answer: The line would have slope 3 and be the same line as 3x – 3, or parallel to it through those points would be: y = 3x + b where b depends on which point it passes through.
Explanation: This exercise demonstrates that if two points are already on a line, any line through both of them IS that line. To create a parallel line, we need a point NOT on the original line.
Exercise 6
Finding Intersection of Two Lines
📍 Problem:
Find the intersection point of lines g₁: y = -5x + 7 and g₂: y = (1/5)x + 9/5.
Step 1:
Set the equations equal:
-5x + 7 = (1/5)x + 9/5
Step 2:
Multiply by 5 to clear fractions:
5(-5x + 7) = 5((1/5)x + 9/5)
-25x + 35 = x + 9
-25x + 35 = x + 9
Step 3:
Solve for x:
-25x – x = 9 – 35
-26x = -26
x = 1
-26x = -26
x = 1
Step 4:
Substitute x = 1 into g₁:
y = -5(1) + 7 = 2
✓ Answer: Intersection point is S(1, 2)
Explanation: At the intersection, both lines have the same x and y coordinates. Set the equations equal and solve for x, then substitute back to find y. This point must satisfy both equations.
Exercise 7
Midpoint of a Line Segment
📍 Problem:
Find the midpoint M of the line segment AB where A(-3, -6) and B(-5, 4).
Step 1:
Apply the midpoint formula:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Step 2:
Calculate the x-coordinate:
xₘ = (-3 + (-5))/2 = -8/2 = -4
Step 3:
Calculate the y-coordinate:
yₘ = (-6 + 4)/2 = -2/2 = -1
✓ Answer: Midpoint M = (-4, -1)
Explanation: The midpoint is simply the average of the x-coordinates and the average of the y-coordinates. It’s the exact center of the line segment.
Exercise 8
Equation of a Line Through Midpoint
📍 Problem:
Find the equation of a line that passes through the midpoint M(-4, -1) and is perpendicular to AB.
Step 1:
Slope of AB:
From earlier, m₁(AB) = -5
Step 2:
Perpendicular slope:
m₂ = -1 / (-5) = 1/5
Step 3:
Point-slope form with M(-4, -1):
y – (-1) = (1/5)(x – (-4))
y + 1 = (1/5)(x + 4)
y + 1 = (1/5)x + 4/5
y = (1/5)x – 1/5
y + 1 = (1/5)(x + 4)
y + 1 = (1/5)x + 4/5
y = (1/5)x – 1/5
✓ Answer: y = (1/5)x – 1/5 or y = 0.2x – 0.2
Explanation: This is the perpendicular bisector of segment AB – it passes through the midpoint and is perpendicular to the original line.
Exercise 9
Slope Relationship Between Lines
📍 Problem:
Two lines have slopes m₁ = 2/3 and m₂ = -3/2. Are they parallel, perpendicular, or neither?
Step 1:
Check for parallel lines:
Parallel lines have equal slopes: 2/3 ≠ -3/2, so they’re NOT parallel.
Step 2:
Check for perpendicular lines:
m₁ · m₂ = (2/3) · (-3/2) = -6/6 = -1 ✓
Since m₁ · m₂ = -1, the lines ARE perpendicular!
✓ Answer: The lines are PERPENDICULAR
Explanation: To quickly determine relationships:
- Parallel: m₁ = m₂
- Perpendicular: m₁ · m₂ = -1
- Neither: Different slopes and product ≠ -1
Exercise 10
Distance Between Parallel Lines
📍 Problem:
Find the distance between the parallel lines g₁: y = -5x + 7 and another line with equation y = -5x – 21.
Step 1:
Convert to standard form:
g₁: y = -5x + 7 → 5x + y – 7 = 0
g₂: y = -5x – 21 → 5x + y + 21 = 0
g₂: y = -5x – 21 → 5x + y + 21 = 0
Step 2:
Apply distance formula:
d = |c₁ – c₂| / √(a² + b²)
d = |-7 – 21| / √(5² + 1²)
d = |-28| / √26
d = 28 / √26
d = |-7 – 21| / √(5² + 1²)
d = |-28| / √26
d = 28 / √26
Step 3:
Rationalize:
d = 28√26 / 26 ≈ 5.5 units
✓ Answer: Distance = 28√26 / 26 ≈ 5.5 units
Explanation: The distance between parallel lines is the perpendicular distance. Use the distance formula with the standard forms of the equations.
Exercise 11
Three Points and Line Equations
📍 Problem:
Given points A(-5, 8), B(1, 3), and Q(3, 1), find:
a) The equation of line AB
b) The equation of a line parallel to AB through Q
a) The equation of line AB
b) The equation of a line parallel to AB through Q
Part a) Find equation of AB:
Step 1:
Calculate slope:
m = (3 – 8) / (1 – (-5)) = -5 / 6
Step 2:
Point-slope form with B(1, 3):
y – 3 = (-5/6)(x – 1)
y – 3 = (-5/6)x + 5/6
y = (-5/6)x + 23/6
y – 3 = (-5/6)x + 5/6
y = (-5/6)x + 23/6
Part b) Parallel line through Q:
Step 3:
Use same slope m = -5/6, point Q(3, 1):
y – 1 = (-5/6)(x – 3)
y – 1 = (-5/6)x + 5/2
y = (-5/6)x + 7/2
y – 1 = (-5/6)x + 5/2
y = (-5/6)x + 7/2
✓ Answer:
a) AB: y = (-5/6)x + 23/6
b) Parallel line: y = (-5/6)x + 7/2
a) AB: y = (-5/6)x + 23/6
b) Parallel line: y = (-5/6)x + 7/2
Explanation: Notice both lines have the same slope (-5/6) but different y-intercepts (23/6 and 7/2), confirming they’re parallel.
Exercise 12
System of Line Equations
📍 Problem:
Determine if the system has a solution, no solution, or infinite solutions:
Line 1: 2x + 3y = 12
Line 2: 4x + 6y = 24
Line 1: 2x + 3y = 12
Line 2: 4x + 6y = 24
Step 1:
Convert to slope-intercept form:
Line 1: 2x + 3y = 12 → y = (-2/3)x + 4
Line 2: 4x + 6y = 24 → y = (-2/3)x + 4
Line 2: 4x + 6y = 24 → y = (-2/3)x + 4
Step 2:
Analyze the results:
Both equations give y = (-2/3)x + 4. They’re identical!
✓ Answer: INFINITE SOLUTIONS (the lines are identical)
Explanation: When two line equations simplify to the same line, they’re the same line and overlap completely, giving infinite intersection points (solutions).
Exercise 13
Parallel Lines with No Solution
📍 Problem:
Determine if the system has a solution:
Line 1: 2x + 3y = 12
Line 2: 2x + 3y = 18
Line 1: 2x + 3y = 12
Line 2: 2x + 3y = 18
Step 1:
Check coefficients:
Both lines have the same coefficients for x and y: (2, 3)
Step 2:
Check constants:
But the constants are different: 12 ≠ 18
Step 3:
Conclusion:
The lines are parallel (same slope) but distinct (different y-intercepts)
✓ Answer: NO SOLUTION (parallel lines never intersect)
Explanation: Parallel lines have the same slope but different y-intercepts, so they never intersect. A system with parallel lines has no solution.
Exercise 14
Angle Between Lines
📍 Problem:
Two lines have slopes m₁ = 1 and m₂ = 2. What is the relationship between these lines?
Step 1:
Check if parallel:
m₁ = 1 ≠ 2 = m₂, so NOT parallel
Step 2:
Check if perpendicular:
m₁ · m₂ = 1 · 2 = 2 ≠ -1, so NOT perpendicular
Step 3:
They intersect at some angle:
The lines intersect but at an angle other than 90°
✓ Answer: The lines INTERSECT (neither parallel nor perpendicular)
Explanation: Lines that are neither parallel nor perpendicular will intersect at some oblique angle. The steeper line (m = 2) makes a larger angle with the horizontal than the less steep line (m = 1).
Exercise 15
Comprehensive Problem: Line Family
📍 Problem:
Given line L: y = 2x – 3 and point P(2, 5):
a) Find the equation of a line through P parallel to L
b) Find the equation of a line through P perpendicular to L
c) Does P lie on L?
a) Find the equation of a line through P parallel to L
b) Find the equation of a line through P perpendicular to L
c) Does P lie on L?
Part c) First check if P is on L:
Step 1:
Substitute P(2, 5) into L:
y = 2(2) – 3 = 4 – 3 = 1
We get y = 1, but P has y = 5. Since 1 ≠ 5, P is NOT on L.
Part a) Line through P parallel to L:
Step 2:
Use slope of L: m = 2
y – 5 = 2(x – 2)
y – 5 = 2x – 4
y = 2x + 1
y – 5 = 2x – 4
y = 2x + 1
Part b) Line through P perpendicular to L:
Step 3:
Perpendicular slope: m₂ = -1/2
y – 5 = (-1/2)(x – 2)
y – 5 = (-1/2)x + 1
y = (-1/2)x + 6
y – 5 = (-1/2)x + 1
y = (-1/2)x + 6
✓ Answers:
a) Parallel line: y = 2x + 1
b) Perpendicular line: y = (-1/2)x + 6
c) P does NOT lie on L
a) Parallel line: y = 2x + 1
b) Perpendicular line: y = (-1/2)x + 6
c) P does NOT lie on L
Explanation: This comprehensive problem combines all key concepts: checking if a point is on a line, finding parallel lines, and finding perpendicular lines. These are the three most important relationships in coordinate geometry.
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