Mathematics • Year 10 • Unit 2 • Lesson 19
Parallel and Perpendicular Lines in the Real World
Apply parallel and perpendicular gradients to real geometry — road layouts, garden borders, shelving, triangulation and tile patterns. Then explain why the negative-reciprocal rule fails for horizontal and vertical lines.
1. Word problems
For each: state the gradient you need, find the equation, and check the parallel/perpendicular rule holds.
1.1 — Two roads. Smith Street is modelled by y = (1/2)x + 3 on a town grid (km). Council wants to build a new "Jones Lane" parallel to Smith Street passing through point (4, 1) (where a roundabout exists).
(a) State the gradient of Jones Lane.
(b) Find the equation of Jones Lane in y = mx + c form.
(c) State the y-intercept of Jones Lane and compare to Smith Street. 4 marks
1.2 — Garden border. A fence runs along the line y = −2x + 8. A garden border must be perpendicular to the fence, passing through the corner post at (2, 1).
(a) Find the gradient of the border (perpendicular to the fence).
(b) Find the equation of the border.
(c) Find the point where the border meets the fence (solve the two equations simultaneously). 4 marks
1.3 — Shelving brackets. A wall-mounted shelf lies along the line y = 4 (a horizontal shelf at height 4 m above the floor). Marco wants to install brackets perpendicular to the shelf, anchored to the wall.
(a) What is the gradient of any bracket perpendicular to a horizontal shelf?
(b) Write the equation of a bracket installed at horizontal position x = 3.
(c) Why does the m₁ × m₂ = −1 rule fail to apply here? 3 marks
1.4 — Triangle altitude. A triangle has vertices A(0, 0), B(6, 0), C(2, 4). The altitude from C is perpendicular to AB and passes through C.
(a) What is the gradient of AB? (Note: AB lies along the x-axis.)
(b) What is the gradient of the altitude from C?
(c) Find the equation of the altitude. 3 marks
1.5 — Tile pattern. A diagonal stripe on a kitchen floor lies along y = (4/3)x. A second diagonal stripe is perpendicular to it and passes through (4, 5).
(a) Find the gradient of the second stripe.
(b) Find its equation.
(c) Verify the product of the two gradients equals −1. 3 marks
2. Explain your thinking
Communication, not just numbers. 4 marks
2.1 A classmate states: "The perpendicular gradient to m = 0 is −1/0 = undefined, so the rule m₁ × m₂ = −1 always works, even for horizontal lines." Using the words horizontal, vertical and undefined, explain (i) why m₁ × m₂ = −1 actually breaks down for horizontal/vertical pairs, (ii) what the perpendicular to a horizontal line really is (in terms of orientation, not gradient), and (iii) how we should state the perpendicular rule to handle this edge case properly.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Two roads
(a) m = 1/2 (parallel → same gradient). (b) 1 = (1/2)(4) + c → c = −1. y = (1/2)x − 1. (c) Jones Lane y-intercept = −1; Smith Street y-intercept = 3. They are parallel (same slope) but offset vertically by 4 units.
1.2 — Garden border
(a) Fence m₁ = −2, so border m₂ = 1/2. (b) 1 = (1/2)(2) + c → c = 0. y = (1/2)x. (c) Solve −2x + 8 = (1/2)x → 8 = (5/2)x → x = 16/5 = 3.2, y = (1/2)(3.2) = 1.6. Meeting point (3.2, 1.6).
1.3 — Shelving brackets
(a) Perpendicular to a horizontal line is vertical, gradient undefined. (b) The bracket at x = 3 has equation x = 3. (c) The rule m₁ × m₂ = −1 needs both gradients to be real numbers. Here m₁ = 0 and we'd have to compute −1/0 = undefined, so the rule cannot be applied algebraically — horizontal/vertical is a special case.
1.4 — Triangle altitude
(a) AB along x-axis: m = 0 (horizontal). (b) Perpendicular to horizontal is vertical: m = undefined. (c) The altitude is vertical through C(2, 4): equation x = 2.
1.5 — Tile pattern
(a) Original m₁ = 4/3, so second m₂ = −3/4. (b) 5 = (−3/4)(4) + c → 5 = −3 + c → c = 8. y = −(3/4)x + 8. (c) m₁ × m₂ = (4/3)(−3/4) = −1 ✓.
2.1 — Explain (sample response)
(i) For m₁ × m₂ = −1 to apply, both m₁ and m₂ must be real numbers you can multiply. A horizontal line has m = 0; a vertical line has m undefined. You cannot multiply "undefined" by 0 and get −1 — the multiplication isn't defined. So strictly the formula fails for horizontal-vertical pairs. (ii) The perpendicular to a horizontal line is geometrically a vertical line — picture the floor (horizontal) and the wall (vertical) meeting at 90°. They are perpendicular, but their gradient relationship can't be captured by the product rule. (iii) A cleaner statement is: "Two lines are perpendicular if and only if (a) one is horizontal and the other is vertical, OR (b) both have defined gradients m₁, m₂ with m₁ × m₂ = −1." This handles both cases.
Marking: 1 for explaining the rule breaks at undefined; 1 for "perpendicular to horizontal is vertical"; 1 for using all three required words; 1 for the corrected two-case statement.