This set has twelve original questions, ordered from easier to harder, covering all five lessons in straight lines and linearisation. Questions 1 to 4 are about intercepts, 5 to 7 are linearisation, 8 to 10 are perpendicular lines, 11 is extrapolation and 12 mixes skills.
Attempt each question on paper before opening the answer, and write the line Y = mX + c explicitly when a graph is involved. The mistake log and retest queue is a good place to record which step went wrong. Use the non-calculator working trainer if fractional gradients slow you down.
Questions and worked answers
1. A line passes through (−4, 0) and (0, 6). Find its equation.
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m = (6 − 0) ÷ (0 − (−4)) = 6/4 = 3/2. The y-intercept is 6.
y = 1.5x + 6. Check: at x = −4, y = −6 + 6 = 0.
2. The line 5x − 2y = 20 meets the axes at P and Q. Find P, Q and the gradient.
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y = 0: 5x = 20, so x = 4 and P = (4, 0). x = 0: −2y = 20, so y = −10 and Q = (0, −10).
Gradient = (−10 − 0) ÷ (0 − 4) = 5/2. Check by rearranging: y = 2.5x − 10.
P = (4, 0), Q = (0, −10), gradient 5/2.
3. A line has gradient −2 and x-intercept 3. Find its equation, its y-intercept and the area of the triangle it forms with the axes.
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y − 0 = −2(x − 3), so y = −2x + 6. The y-intercept is 6.
The triangle has base 3 and height 6, so area = ½ × 3 × 6 = 9 square units. Check: at x = 3, y = −6 + 6 = 0.
4. A line has x-intercept 5 and y-intercept −2. Write its equation in the form ax + by = c with whole-number coefficients.
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x/5 + y/(−2) = 1, so x/5 − y/2 = 1. Multiply by 10: 2x − 5y = 10.
2x − 5y = 10. Check: (5, 0) gives 10, and (0, −2) gives 10.
5. Data satisfy y = ax² + b.
| x | 1 | 2 | 3 |
|---|---|---|---|
| y | 1 | 10 | 25 |
Find a and b.
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X = x² gives 1, 4, 9. Gradients: (10 − 1) ÷ (4 − 1) = 3 and (25 − 10) ÷ (9 − 4) = 3, so the graph is straight and a = 3.
Then 1 = 3(1) + b, so b = −2.
a = 3, b = −2. Check: 3(9) − 2 = 25.
6. Variables satisfy y = a/x + b. A graph of y against 1/x passes through (0.5, 7) and (2, 13). Find a and b.
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Gradient = (13 − 7) ÷ (2 − 0.5) = 6 ÷ 1.5 = 4, so a = 4. Then 7 = 4(0.5) + b, so b = 5.
a = 4, b = 5. Check: at 1/x = 2, y = 8 + 5 = 13.
7. Variables satisfy y = px² + qx. A graph of y/x against x passes through (2, 9) and (6, 17). Find p and q.
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Y = y/x, and y/x = px + q. Gradient = (17 − 9) ÷ (6 − 2) = 2, so p = 2. Then 9 = 2(2) + q, so q = 5.
p = 2, q = 5, so y = 2x² + 5x. Check at x = 6: y = 72 + 30 = 102, and 102 ÷ 6 = 17.
8. Find the equation of the perpendicular bisector of P(1, −1) and Q(5, 5).
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Midpoint = (3, 2). Gradient of PQ = 6 ÷ 4 = 3/2, so the perpendicular gradient is −2/3.
y − 2 = −(2/3)(x − 3), so 3y − 6 = −2x + 6, giving 2x + 3y = 12.
Check: the midpoint gives 6 + 6 = 12. The point (0, 4) is on the line, and its distance² to P is 1 + 25 = 26 and to Q is 25 + 1 = 26.
9. Show that A(1, 2), B(4, 3) and C(5, 0) form a right angle at B, and find the area of the triangle.
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Gradient of AB = (3 − 2) ÷ (4 − 1) = 1/3. Gradient of BC = (0 − 3) ÷ (5 − 4) = −3.
Product = (1/3)(−3) = −1, so AB is perpendicular to BC.
AB² = 9 + 1 = 10 and BC² = 1 + 9 = 10, so both lengths are √10. Area = ½ × √10 × √10 = 5 square units.
Check with coordinates: ½|(1×3 + 4×0 + 5×2) − (2×4 + 3×5 + 0×1)| = ½|13 − 23| = 5.
10. Find k so that y = kx + 1 is perpendicular to 2x + 3y = 5.
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2x + 3y = 5 gives y = −(2/3)x + 5/3, so the gradient is −2/3.
k × (−2/3) = −1, so k = 3/2. Check: (3/2)(−2/3) = −1.
11. A taxi fare is modelled by F = 1.6d + 4, where F is in RM and d is distance in km. The model was fitted using trips from 2 km to 15 km. Find F for d = 10 and d = 120, and comment on each.
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d = 10: F = 16 + 4 = RM20. It is inside the data range, so it is interpolation and reasonable.
d = 120: F = 192 + 4 = RM196. It is far outside the data range, so it is extrapolation. Long trips may be priced differently, so this value is unreliable.
12. Data follow y = ax + b/x.
| x | 1 | 2 | 4 |
|---|---|---|---|
| y | 5 | 5.5 | 8.75 |
Find a and b, then find y when x = 5.
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Multiply by x: xy = ax² + b. Plot xy against x². Values of xy are 5, 11, 35 and x² is 1, 4, 16.
Gradients: (11 − 5) ÷ (4 − 1) = 2 and (35 − 11) ÷ (16 − 4) = 2. So a = 2, and 5 = 2(1) + b gives b = 3.
The model is y = 2x + 3/x. At x = 5: y = 10 + 0.6 = 10.6. Check at x = 4: 8 + 0.75 = 8.75.
If you got these wrong
- Questions 1 to 4: intercepts, signs in the gradient or the area. Go to finding a line from intercept information.
- Questions 5, 6 and 12: deciding what to plot or reading the wrong rows. Go to transforming data into straight-line form.
- Questions 7 and 12 (the final conversion): constants placed in the wrong formula. Go to recovering model constants.
- Questions 8 to 10: gradient of the perpendicular or the midpoint. Go to determining a perpendicular relationship.
- Question 11: missing or vague comments on a prediction. Go to assessing an extrapolated linear model.
When one pattern of error keeps repeating across different questions, it is usually a habit rather than a gap in knowledge. That is the kind of thing we look for in online one-to-one Additional Mathematics tuition.