Mathematics • Year 10 • Unit 2 • Lesson 18
Equations of Lines — Mixed Challenge
Six mixed problems testing all three "find the equation" methods plus general-form conversion. Spot a student who skipped the verification step, then design a line through given constraints.
1. Mixed problems — find the equation
Always check by substituting one point at the end. 3 marks each
1.1 Find the equation of the line with gradient 4 and y-intercept −7. Then write it in general form.
1.2 Find the equation of the line through (−3, 4) with gradient −2/3.
1.3 Find the equation of the line through (1, −4) and (5, 8).
1.4 Convert 3x + 4y − 12 = 0 to y = mx + c form and state m and c.
1.5 Find the equation of the horizontal line through (3, −2) and the vertical line through (5, 1).
1.6 A line has x-intercept 6 and y-intercept −2. (a) Find the gradient. (b) Find the equation in y = mx + c form. (c) Write it in general form.
2. Find the mistake
A Year 10 student finds the equation through (2, 5) and (6, 13). Their working is below. One line contains an error — find it. 3 marks
Student's working:
Line 1: m = (13 − 5)/(6 − 2) = 8/4 = 2
Line 2: y = 2x + c
Line 3: Using (2, 5): 5 = 2(2) + c → c = 5 − 4 = 1
Line 4: Final: y = 2x + 1
Line 5: (No verification done.)
(a) Which line shows the omission?
(b) Explain why verification matters (in one or two sentences).
(c) Add the verification using the OTHER point (6, 13). Show whether the line is correct.
Stuck? Substitute (6, 13) into y = 2x + 1: y = 2(6) + 1 = 13 ✓ — happens to be correct, but the student couldn't know without checking.3. Open-ended challenge — design a line
Many valid answers. 4 marks
3.1 Design a line in general form Ax + By + C = 0 (with integer A, B, C and A > 0) that satisfies all three of: (i) gradient is a non-integer fraction, (ii) passes through (4, 1), (iii) has positive y-intercept.
Submit:
(a) Your line in y = mx + c form first.
(b) Convert to general form.
(c) Verify (4, 1) satisfies both forms.
(d) State the y-intercept and confirm it is positive.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — m = 4, c = −7
y = 4x − 7. General form: 4x − y − 7 = 0.
1.2 — m = −2/3 through (−3, 4)
4 = (−2/3)(−3) + c → 4 = 2 + c → c = 2. y = −(2/3)x + 2.
1.3 — (1, −4) and (5, 8)
m = (8 − (−4))/(5 − 1) = 12/4 = 3. y = 3x + c. (1, −4): −4 = 3 + c → c = −7. y = 3x − 7. Check (5, 8): 15 − 7 = 8 ✓.
1.4 — 3x + 4y − 12 = 0
4y = −3x + 12 → y = (−3/4)x + 3. m = −3/4, c = 3.
1.5 — Horizontal and vertical lines
Horizontal through (3, −2): y = −2 (any x gives y = −2). Vertical through (5, 1): x = 5.
1.6 — x-int 6, y-int −2
(a) Two points (6, 0) and (0, −2): m = (−2 − 0)/(0 − 6) = 1/3. (b) y = (1/3)x − 2. (c) × 3 → 3y = x − 6 → x − 3y − 6 = 0.
2 — Find the mistake
(a) Line 5: the verification step is missing.
(b) Without checking the second point, you can't be sure the gradient or c-value calculation didn't contain an arithmetic slip. Marks are typically given for the verification step in HSC marking.
(c) Verify with (6, 13): y = 2(6) + 1 = 13 ✓. The line y = 2x + 1 is correct.
3 — Open-ended (sample solution)
(a) Sample: y = −(1/4)x + 2 (m = −1/4 is a non-integer fraction ✓).
(b) × 4 → 4y = −x + 8 → x + 4y − 8 = 0 (A = 1 > 0 ✓).
(c) Verify (4, 1): y = −(1/4)(4) + 2 = −1 + 2 = 1 ✓ in slope form. General form: 4 + 4(1) − 8 = 0 ✓.
(d) y-intercept = 2, positive ✓.
Marking: 1 for a non-integer-fraction gradient; 1 for correct general-form conversion with positive A; 1 for verifying (4, 1) in both forms; 1 for positive y-intercept.