Solved Exercises on Equations of Parallel and Perpendicular Lines in Integrated Math 1

Master parallel and perpendicular lines: slope relationships, equations, and applications through these 5 detailed exercises with comprehensive solutions.

Solution: Exercises 1 to 3
1 Parallel Line Through a Point
Exercise 1
Find the equation of a line parallel to \(y = 3x - 4\) that passes through the point (2, 5). Write your answer in slope-intercept form.
Definition:

Parallel Lines: Two lines are parallel if and only if they have the same slope but different y-intercepts

Method for finding parallel line through a point:
  1. Identify the slope of the given line
  2. Use the same slope for the parallel line
  3. Use the point-slope form with the given point
  4. Solve for y to get slope-intercept form
Given Line
y = 3x - 4
Slope
m = 3
Parallel Line
y = 3x - 1
Step 1: Identify the slope of the given line

The given line is \(y = 3x - 4\), so the slope is \(m = 3\)

Step 2: Use the same slope for the parallel line

Parallel lines have identical slopes, so the parallel line also has slope \(m = 3\)

Step 3: Use point-slope form with the given point (2, 5)

\(y - y_1 = m(x - x_1)\)

\(y - 5 = 3(x - 2)\)

Step 4: Solve for y to get slope-intercept form

\(y - 5 = 3x - 6\)

\(y = 3x - 6 + 5\)

\(y = 3x - 1\)

Step 5: Verify the line passes through the given point

Check: When \(x = 2\), \(y = 3(2) - 1 = 6 - 1 = 5\) ✓

y = 3x - 1
Final answer:

The equation of the parallel line is \(y = 3x - 1\).

Applied rules:

• Parallel Lines: Same slope, different y-intercept

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

• Slope-Intercept Form: \(y = mx + b\)

2 Perpendicular Line Through a Point
Exercise 2
Find the equation of a line perpendicular to \(y = -\frac{2}{3}x + 5\) that passes through the point (3, -1). Write your answer in slope-intercept form.
Definition:

Perpendicular Lines: Two lines are perpendicular if and only if their slopes are negative reciprocals of each other

Given Line
y = -(2/3)x + 5
Original Slope
m = -2/3
Perp. Slope
m = 3/2
Step 1: Identify the slope of the given line

The given line is \(y = -\frac{2}{3}x + 5\), so the slope is \(m = -\frac{2}{3}\)

Step 2: Find the slope of the perpendicular line

Perpendicular lines have slopes that are negative reciprocals

Negative reciprocal of \(-\frac{2}{3}\) is \(\frac{3}{2}\)

So the perpendicular line has slope \(m = \frac{3}{2}\)

Step 3: Use point-slope form with the given point (3, -1)

\(y - y_1 = m(x - x_1)\)

\(y - (-1) = \frac{3}{2}(x - 3)\)

\(y + 1 = \frac{3}{2}(x - 3)\)

Step 4: Solve for y to get slope-intercept form

\(y + 1 = \frac{3}{2}x - \frac{9}{2}\)

\(y = \frac{3}{2}x - \frac{9}{2} - 1\)

\(y = \frac{3}{2}x - \frac{9}{2} - \frac{2}{2}\)

\(y = \frac{3}{2}x - \frac{11}{2}\)

Step 5: Verify the line passes through the given point

Check: When \(x = 3\), \(y = \frac{3}{2}(3) - \frac{11}{2} = \frac{9}{2} - \frac{11}{2} = -\frac{2}{2} = -1\) ✓

y = (3/2)x - 11/2
Final answer:

The equation of the perpendicular line is \(y = \frac{3}{2}x - \frac{11}{2}\).

Applied rules:

• Perpendicular Lines: Slopes are negative reciprocals (\(m_1 \cdot m_2 = -1\))

• Negative Reciprocal: Flip the fraction and change the sign

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

3 Converting to Standard Form
Exercise 3
Find the equation of a line parallel to \(2x - 3y = 6\) that passes through the point (4, -2). Write your answer in standard form.
Definition:

Standard Form: A linear equation written as \(Ax + By = C\) where A, B, and C are integers and A ≥ 0

Given Line
2x - 3y = 6
Slope
m = 2/3
Parallel Line
2x - 3y = 14
Step 1: Find the slope of the given line by converting to slope-intercept form

Starting with \(2x - 3y = 6\)

\(-3y = -2x + 6\)

\(y = \frac{2}{3}x - 2\)

So the slope is \(m = \frac{2}{3}\)

Step 2: Use the same slope for the parallel line

The parallel line has slope \(m = \frac{2}{3}\)

Step 3: Use point-slope form with the given point (4, -2)

\(y - (-2) = \frac{2}{3}(x - 4)\)

\(y + 2 = \frac{2}{3}(x - 4)\)

Step 4: Convert to standard form

\(y + 2 = \frac{2}{3}x - \frac{8}{3}\)

Multiply everything by 3: \(3y + 6 = 2x - 8\)

Rearrange: \(-2x + 3y = -8 - 6\)

\(-2x + 3y = -14\)

Multiply by -1: \(2x - 3y = 14\)

Step 5: Verify the line passes through the given point

Check: \(2(4) - 3(-2) = 8 + 6 = 14\) ✓

2x - 3y = 14
Final answer:

The equation of the parallel line is \(2x - 3y = 14\).

Applied rules:

• Parallel Lines: Same slope, different y-intercept

• Standard Form Conversion: Eliminate fractions by multiplying by LCD

• Positive Leading Coefficient: Ensure A ≥ 0 by multiplying by -1 if needed

Parallel and Perpendicular Lines Rules and Methods
Parallel: \(m_1 = m_2\), Perpendicular: \(m_1 \cdot m_2 = -1\)
Slope Relationships
Parallel
\(m_1 = m_2\)
Slopes are equal
Perpendicular
\(m_1 \cdot m_2 = -1\)
Negative reciprocals
Negative Reciprocal
\(m_2 = -\frac{1}{m_1}\)
Flip and change sign
Key definitions:

Parallel Lines: Two lines that never intersect and have the same slope

Perpendicular Lines: Two lines that intersect at a right angle (90°)

Negative Reciprocal: For a number \(a\), its negative reciprocal is \(-\frac{1}{a}\)

Complete methodology:
  1. Identify Relationship: Determine if you need a parallel or perpendicular line
  2. Find Slope: Identify the slope of the given line
  3. Apply Rule: For parallel lines, use the same slope; for perpendicular, use the negative reciprocal
  4. Use Point: Apply point-slope form with the given point
  5. Convert Form: Write in required form (slope-intercept, standard, etc.)
Tip 1: For negative reciprocals, flip the fraction and change the sign.
Tip 2: If the original slope is 0 (horizontal line), perpendicular slope is undefined (vertical line).
Tip 3: If the original slope is undefined (vertical line), perpendicular slope is 0 (horizontal line).
Tip 4: Always verify that the new line passes through the given point.
Common errors: Forgetting to take the negative reciprocal for perpendicular lines, not changing the sign, not verifying the line passes through the given point.
Exam preparation: Practice finding negative reciprocals, memorize the relationships, work with different forms of equations.
Formulas to know by heart:

• Parallel Lines: \(m_1 = m_2\)

• Perpendicular Lines: \(m_1 \cdot m_2 = -1\)

• Negative Reciprocal: \(m_2 = -\frac{1}{m_1}\)

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

Solution: Exercises 4 to 5
4 Multiple Relationships
Exercise 4
Given the line passing through points (1, 3) and (4, 9), find the equations of: (a) a parallel line through (0, -2), and (b) a perpendicular line through (0, -2).
Definition:

Slope Formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\) for points \((x_1, y_1)\) and \((x_2, y_2)\)

Given Points
(1,3) and (4,9)
Slope
m = 2
Parallel & Perp.
y=2x-2, y=-½x-2
Step 1: Find the slope of the original line

Using points (1, 3) and (4, 9):

\(m = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2\)

Step 2: Find the parallel line through (0, -2)

Parallel lines have the same slope: \(m = 2\)

Using point-slope form: \(y - (-2) = 2(x - 0)\)

\(y + 2 = 2x\)

\(y = 2x - 2\)

Step 3: Find the perpendicular line through (0, -2)

Perpendicular lines have negative reciprocal slopes: \(m = -\frac{1}{2}\)

Using point-slope form: \(y - (-2) = -\frac{1}{2}(x - 0)\)

\(y + 2 = -\frac{1}{2}x\)

\(y = -\frac{1}{2}x - 2\)

Step 4: Verify both lines pass through (0, -2)

Parallel: When \(x = 0\), \(y = 2(0) - 2 = -2\) ✓

Perpendicular: When \(x = 0\), \(y = -\frac{1}{2}(0) - 2 = -2\) ✓

Step 5: Verify perpendicularity condition

Product of slopes: \(2 \times (-\frac{1}{2}) = -1\) ✓

Parallel: y = 2x - 2, Perpendicular: y = -½x - 2
Final answer:

(a) Parallel line: \(y = 2x - 2\), (b) Perpendicular line: \(y = -\frac{1}{2}x - 2\)

Applied rules:

• Slope Formula: Calculate slope from two points

• Parallel Lines: Same slope, different y-intercept

• Perpendicular Lines: Slopes are negative reciprocals

5 Real-world Application
Exercise 5
A road runs along the line \(y = -\frac{3}{4}x + 6\). A new road needs to be built perpendicular to this road and pass through the town located at (8, 2). Find the equation of the new road. If another parallel road needs to go through (5, -1), find its equation.
Definition:

Linear Modeling: Using linear equations to represent real-world situations

Original Road
y = -(3/4)x + 6
Perp. Road
y = (4/3)x - 26/3
Parallel Road
y = -(3/4)x + 1/4
Step 1: Identify the slope of the original road

The original road has equation \(y = -\frac{3}{4}x + 6\), so slope is \(m = -\frac{3}{4}\)

Step 2: Find the equation of the perpendicular road through (8, 2)

Perpendicular slope: negative reciprocal of \(-\frac{3}{4}\) is \(\frac{4}{3}\)

Using point-slope form: \(y - 2 = \frac{4}{3}(x - 8)\)

\(y - 2 = \frac{4}{3}x - \frac{32}{3}\)

\(y = \frac{4}{3}x - \frac{32}{3} + 2\)

\(y = \frac{4}{3}x - \frac{32}{3} + \frac{6}{3}\)

\(y = \frac{4}{3}x - \frac{26}{3}\)

Step 3: Find the equation of the parallel road through (5, -1)

Parallel slope: same as original, \(m = -\frac{3}{4}\)

Using point-slope form: \(y - (-1) = -\frac{3}{4}(x - 5)\)

\(y + 1 = -\frac{3}{4}x + \frac{15}{4}\)

\(y = -\frac{3}{4}x + \frac{15}{4} - 1\)

\(y = -\frac{3}{4}x + \frac{15}{4} - \frac{4}{4}\)

\(y = -\frac{3}{4}x + \frac{11}{4}\)

Step 4: Verify the perpendicular roads

Product of slopes: \(-\frac{3}{4} \times \frac{4}{3} = -1\) ✓

Step 5: Verify both new roads pass through their respective points

Perpendicular: When \(x = 8\), \(y = \frac{4}{3}(8) - \frac{26}{3} = \frac{32}{3} - \frac{26}{3} = \frac{6}{3} = 2\) ✓

Parallel: When \(x = 5\), \(y = -\frac{3}{4}(5) + \frac{11}{4} = -\frac{15}{4} + \frac{11}{4} = -\frac{4}{4} = -1\) ✓

Perp. Road: y = (4/3)x - 26/3, Parallel Road: y = -(3/4)x + 11/4
Final answer:

The perpendicular road has equation \(y = \frac{4}{3}x - \frac{26}{3}\), and the parallel road has equation \(y = -\frac{3}{4}x + \frac{11}{4}\).

Applied rules:

• Real-world Context: Apply mathematical concepts to practical scenarios

• Parallel Lines: Same slope, different location

• Perpendicular Lines: Negative reciprocal slopes

Parallel and Perpendicular Lines Fundamentals
Parallel: \(m_1 = m_2\), Perpendicular: \(m_1 \cdot m_2 = -1\)
Slope Relationships
Key definitions:

Parallel Lines: Two lines in the same plane that never intersect and have identical slopes

Perpendicular Lines: Two lines that intersect at a right angle (90°), with slopes that are negative reciprocals

Negative Reciprocal: For a number \(a\), its negative reciprocal is \(-\frac{1}{a}\)

Complete methodology:
  1. Identify the Relationship: Determine if you need parallel or perpendicular
  2. Find the Original Slope: Extract from given equation or calculate from points
  3. Apply the Rule: Use same slope for parallel, negative reciprocal for perpendicular
  4. Use the Given Point: Apply point-slope form
  5. Convert to Required Form: Write in slope-intercept, standard, or other required form
  6. Verify Your Answer: Check that the new line passes through the given point
Tip 1: Remember: "Parallel = Same Slope", "Perpendicular = Negative Reciprocal".
Tip 2: To find negative reciprocal: flip the fraction and change the sign.
Tip 3: The product of perpendicular slopes is always -1.
Tip 4: Always check your work by substituting the given point into your equation.
Applications: Architecture (perpendicular walls), city planning (grid streets), engineering (orthogonal components), computer graphics (coordinate systems).
Properties: Parallel lines maintain constant distance; perpendicular lines create 90° angles; slopes have specific mathematical relationships.
Essential formulas:

• Parallel Lines: \(m_1 = m_2\)

• Perpendicular Lines: \(m_1 \cdot m_2 = -1\) or \(m_2 = -\frac{1}{m_1}\)

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

• Slope Formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

Parallel and Perpendicular Lines Visualization
Exercise 6: Relationship Visualization
Show the relationship between these lines:
Original: y = 2x + 1
Parallel: y = 2x - 3
Perpendicular: y = -½x + 2

Analysis: The chart shows how parallel lines maintain the same slope while perpendicular lines have slopes that are negative reciprocals.

  • Parallel lines (same slope = 2) never intersect
  • Perpendicular line has slope -½, which is negative reciprocal of 2
  • Original and perpendicular lines intersect at right angles

Questions & Answers

Question: Why do perpendicular lines have slopes that are negative reciprocals? What's the mathematical reasoning behind this?

Answer: The negative reciprocal relationship comes from the geometric properties of perpendicular lines and the rotation of vectors:

  • Consider two perpendicular lines with slopes m₁ and m₂
  • If we rotate a line with slope m by 90°, we get a line perpendicular to the original
  • A 90° rotation transforms the direction vector (1, m) to (-m, 1)
  • The slope of this new line is 1/(-m) = -1/m

Alternatively, using the dot product: if two lines are perpendicular, their direction vectors are orthogonal (dot product = 0). For lines with slopes m₁ and m₂, direction vectors are (1, m₁) and (1, m₂). Their dot product is 1·1 + m₁·m₂ = 1 + m₁m₂. For orthogonality: 1 + m₁m₂ = 0, so m₁m₂ = -1, meaning m₂ = -1/m₁.

This is why the product of perpendicular slopes is always -1.

Question: What happens when I try to find the perpendicular slope of a horizontal line? And what about a vertical line?

Answer: These are special cases that require careful consideration:

  • Horizontal Line (slope = 0): A horizontal line has slope 0, so its perpendicular line would have slope = -1/0, which is undefined. This means the perpendicular line is vertical.
  • Vertical Line (slope = undefined): A vertical line has undefined slope, so its perpendicular line would have slope 0. This means the perpendicular line is horizontal.

For example, if you have a horizontal line y = 3 and want a perpendicular line through point (2, 5), the perpendicular line would be x = 2 (vertical line).

If you have a vertical line x = 4 and want a perpendicular line through point (4, 7), the perpendicular line would be y = 7 (horizontal line).

These cases show that the negative reciprocal rule applies to finite, non-zero slopes.

Question: Can I determine if two lines are parallel or perpendicular just by looking at their equations in standard form?

Answer: Yes, you can determine the relationship by converting to slope-intercept form or using the coefficients directly:

For lines in standard form:

  • Line 1: \(A_1x + B_1y = C_1\) → slope = \(-\frac{A_1}{B_1}\)
  • Line 2: \(A_2x + B_2y = C_2\) → slope = \(-\frac{A_2}{B_2}\)

Parallel condition: \(-\frac{A_1}{B_1} = -\frac{A_2}{B_2}\), which simplifies to \(\frac{A_1}{B_1} = \frac{A_2}{B_2}\)

Perpendicular condition: \((-\frac{A_1}{B_1}) \cdot (-\frac{A_2}{B_2}) = -1\), which simplifies to \(\frac{A_1 \cdot A_2}{B_1 \cdot B_2} = -1\), or \(A_1 \cdot A_2 = -B_1 \cdot B_2\)

For example, lines \(2x + 3y = 6\) and \(4x + 6y = 12\) are parallel because \(\frac{2}{3} = \frac{4}{6}\). Lines \(2x + 3y = 6\) and \(3x - 2y = 5\) are perpendicular because \(2 \cdot 3 = -1 \cdot (3 \cdot (-2)) = 6 = 6\).

Question: How can I quickly check if my answer is correct when finding parallel or perpendicular lines?

Answer: Here are several quick verification methods:

  • Point Substitution: Plug the given point into your equation to ensure it satisfies the equation
  • Slope Check: For parallel lines, verify slopes are equal; for perpendicular, verify the product is -1
  • Graphical Check: Sketch both lines to visually confirm the relationship

For example, if you found a parallel line with slope 3 that should pass through (1, 4), substitute: when x = 1, y should equal 4 in your equation.

For a perpendicular check, if your original line has slope 2 and you found a perpendicular line with slope -1/2, verify: 2 × (-1/2) = -1 ✓

The fastest check is usually substituting the given point into your final equation to make sure it works.