Mathematics • Year 10 • Unit 2 • Lesson 19
Parallel and Perpendicular Lines, Skill Drill
Build fluency with the two key rules from Lesson 19: parallel lines have equal gradients (m₁ = m₂); perpendicular lines have gradients whose product is −1 (m₁ × m₂ = −1, i.e. negative reciprocals). Then write the equation of a parallel or perpendicular line through a given point.
1. I do, fully worked example
Equation of a line parallel to a given one, through a given point.
Problem. Find the equation of the line parallel to y = 3x + 5 passing through (2, −1).
Step 1, Read off the gradient of the given line.
m₁ = 3 (coefficient of x in y = 3x + 5)
Step 2, Parallel rule: new gradient is the same.
m₂ = m₁ = 3
Step 3, Use y = mx + c with m = 3 and substitute (2, −1).
−1 = 3(2) + c → −1 = 6 + c → c = −7
Step 4, Write the equation.
y = 3x − 7
Answer: y = 3x − 7.
2. We do, fill in the missing steps
Equation of a perpendicular line. Fill in each blank. 5 marks
Problem. Find the equation of the line perpendicular to y = 2x + 1 passing through (4, 3).
Step 1, Original gradient: m₁ = ____.
Step 2, Perpendicular rule: m₁ × m₂ = ____. So m₂ = ____ (the negative reciprocal of ____).
Step 3, Write y = ____ x + c.
Step 4, Substitute (4, 3) to find c:
3 = ____ (4) + c → c = ____
Step 5, Equation: y = ____.
3. You do, independent practice
For each: state the new gradient, find c if needed, then write the equation. Always check the rule (m₁ = m₂ for parallel; m₁ × m₂ = −1 for perpendicular).
Foundation, find the perpendicular / parallel gradient
3.1 The perpendicular gradient to m = 4 is ____. 1 mark
3.2 The perpendicular gradient to m = −2/3 is ____. 1 mark
3.3 The parallel gradient to y = 5x − 9 is ____. 1 mark
3.4 A horizontal line has m = 0. What is the gradient of a line perpendicular to it? 1 mark
Standard, parallel / perpendicular through a point
3.5 Equation of the line parallel to y = 2x + 3 passing through (1, 5). 2 marks
3.6 Equation of the line perpendicular to y = −3x + 1 passing through (6, 2). 2 marks
Extension, classify and apply
3.7 Classify y = 3x + 1 and y = −(1/3)x + 2 as parallel, perpendicular or neither. Show the gradient product test. 2 marks
3.8 Two lines have equations 2x + 3y = 6 and 3x − 2y = 4. Find the gradient of each (rearrange to y = mx + c), then test if they are parallel, perpendicular, or neither. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (perpendicular to y = 2x + 1 through (4, 3))
Step 1: m₁ = 2. Step 2: m₁ × m₂ = −1, so m₂ = −1/2 (negative reciprocal of 2). Step 3: y = −1/2 x + c. Step 4: 3 = −1/2(4) + c → 3 = −2 + c → c = 5. Step 5: y = −(1/2)x + 5.
3.1, perp to m = 4
−1/4.
3.2, perp to m = −2/3
3/2.
3.3, parallel to y = 5x − 9
5.
3.4, perp to a horizontal line
Undefined (vertical line). The product rule m₁ × m₂ = −1 doesn't apply when m₁ = 0; the perpendicular line is vertical, of form x = constant.
3.5, parallel to y = 2x + 3 through (1, 5)
m = 2 (same as original). 5 = 2(1) + c → c = 3. y = 2x + 3. (This happens to be the original line, (1, 5) lies on it.)
3.6, perp to y = −3x + 1 through (6, 2)
m₁ = −3 → m₂ = 1/3. 2 = (1/3)(6) + c → 2 = 2 + c → c = 0. y = (1/3)x.
3.7, classify
m₁ = 3; m₂ = −1/3. m₁ × m₂ = 3 × (−1/3) = −1. The lines are perpendicular.
3.8, classify (general form)
2x + 3y = 6 → 3y = −2x + 6 → y = (−2/3)x + 2. m₁ = −2/3.
3x − 2y = 4 → −2y = −3x + 4 → y = (3/2)x − 2. m₂ = 3/2.
Test: m₁ × m₂ = (−2/3)(3/2) = −1. The lines are perpendicular.