Mathematics • Year 10 • Unit 2 • Lesson 19
Parallel and Perpendicular — Mixed Challenge
Six mixed problems testing the parallel/perpendicular rules across slope-intercept and general form, including classification, finding equations, and using the perpendicular rule to prove right angles in geometry. Spot a "forgot the negative" reciprocal error, then design two lines satisfying given conditions.
1. Mixed problems
Show every step and apply the right rule. 3 marks each
1.1 Find the equation of the line parallel to y = −2x + 4 passing through (3, −1).
1.2 Find the equation of the line perpendicular to y = (3/5)x − 2 passing through (15, 4).
1.3 Classify the pair y = (1/2)x + 7 and 2y − x = 10 as parallel, perpendicular or neither.
1.4 Classify the pair 4x + y = 5 and x − 4y = 3 as parallel, perpendicular or neither.
1.5 Show that triangle ABC with A(1, 1), B(5, 3) and C(3, 7) has a right angle. (Hint: compute the gradients of all three sides and find which pair multiplies to −1.)
1.6 A line through (0, 6) is perpendicular to y = (2/3)x − 1. (a) Find its gradient. (b) Find its equation. (c) Find where it crosses the x-axis.
2. Find the mistake
A Year 10 student finds the line perpendicular to y = (2/3)x + 1 through (6, 4). Their working is below. One line contains a Year-10-realistic reciprocal error. Find it. 3 marks
Student's working:
Line 1: m₁ = 2/3
Line 2: Perpendicular m₂ = 3/2 (just flip the fraction)
Line 3: y = (3/2)x + c
Line 4: (6, 4): 4 = (3/2)(6) + c → 4 = 9 + c → c = −5
Line 5: y = (3/2)x − 5
(a) Which line is wrong?
(b) Explain the missing-sign mistake in one or two sentences.
(c) Redo the working with the correct perpendicular gradient and find the correct equation.
Stuck? Perpendicular is the negative reciprocal: m₂ = −3/2, not 3/2.3. Open-ended challenge — design two lines
Many valid answers. 4 marks
3.1 Design two lines L₁ and L₂ that satisfy all four conditions: (i) L₁ and L₂ are perpendicular, (ii) L₁ passes through (2, 5), (iii) L₂ passes through (2, 5) as well, (iv) the gradient of L₁ is exactly −3/4.
Submit:
(a) Equation of L₁.
(b) Equation of L₂.
(c) Verification that both equations pass through (2, 5).
(d) Product-of-gradients check showing m₁ × m₂ = −1.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — parallel to y = −2x + 4 through (3, −1)
m = −2. −1 = −2(3) + c → c = 5. y = −2x + 5.
1.2 — perp to y = (3/5)x − 2 through (15, 4)
m₂ = −5/3. 4 = (−5/3)(15) + c → 4 = −25 + c → c = 29. y = −(5/3)x + 29.
1.3 — classify y = (1/2)x + 7 and 2y − x = 10
2y − x = 10 → 2y = x + 10 → y = (1/2)x + 5. Both have m = 1/2 and different y-intercepts → parallel.
1.4 — classify 4x + y = 5 and x − 4y = 3
4x + y = 5 → y = −4x + 5, m₁ = −4. x − 4y = 3 → −4y = −x + 3 → y = (1/4)x − 3/4, m₂ = 1/4. m₁ × m₂ = −4 × (1/4) = −1 → perpendicular.
1.5 — Right-angle in triangle ABC
AB: m = (3−1)/(5−1) = 1/2. BC: m = (7−3)/(3−5) = 4/(−2) = −2. AC: m = (7−1)/(3−1) = 3. Test: AB × BC = (1/2)(−2) = −1 → AB ⊥ BC. Therefore right angle at B.
1.6 — perp to y = (2/3)x − 1 through (0, 6)
(a) m₂ = −3/2. (b) Through (0, 6): c = 6. y = −(3/2)x + 6. (c) x-intercept: 0 = −(3/2)x + 6 → x = 4. So (4, 0).
2 — Find the mistake
(a) Line 2.
(b) The student flipped the fraction but forgot the negative — perpendicular gradient is the negative reciprocal, so m₂ = −3/2, not 3/2.
(c) Correct: m₂ = −3/2. y = −(3/2)x + c. (6, 4): 4 = −(3/2)(6) + c → 4 = −9 + c → c = 13. y = −(3/2)x + 13. Check product: (2/3)(−3/2) = −1 ✓.
3 — Open-ended (sample solution)
L₁: m₁ = −3/4. 5 = (−3/4)(2) + c → c = 6.5. L₁: y = −(3/4)x + 6.5.
L₂: m₂ = 4/3 (negative reciprocal of −3/4). 5 = (4/3)(2) + c → c = 5 − 8/3 = 7/3. L₂: y = (4/3)x + 7/3.
Verify (2, 5) on L₁: −(3/4)(2) + 6.5 = −1.5 + 6.5 = 5 ✓. On L₂: (4/3)(2) + 7/3 = 8/3 + 7/3 = 15/3 = 5 ✓.
Product check: (−3/4)(4/3) = −12/12 = −1 ✓.
Marking: 1 for correct L₁ with given gradient; 1 for L₂ with correct perpendicular gradient; 1 for verifying both pass through (2, 5); 1 for the m₁ × m₂ = −1 check.