Try each problem on paper first, then open the answer to check your working. For problems with two points, the link under the answer shows the line in the Slope Finder.
Finding slope
1. Find the slope of the line through (1, 3) and (4, 9).
2. Find the slope of the line through (−2, 5) and (2, −3).
3. Find the slope of the line through (−5, −1) and (3, 5).
4. Find the slope of the line through (2, 7) and (2, −1). What is its equation?
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The run is 2 − 2 = 0, so the slope is undefined. The line is vertical, with equation x = 2.
5. Find the slope of the line through (−4, 6) and (5, 6). What is its equation?
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The rise is 6 − 6 = 0, so the slope is 0. The line is horizontal, with equation y = 6.
Writing equations
6. Write the equation of the line with slope 5 and y-intercept −2.
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Substitute m = 5 and b = −2 into y = mx + b: y = 5x − 2
7. Find the equation of the line through (3, 1) and (6, 3).
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m = (3 − 1) ÷ (6 − 3) = 2/3. Then 1 = ⅔ × 3 + b, so 1 = 2 + b, and b = −1.
y = ⅔x − 1. Check: ⅔ × 6 − 1 = 3. ✓
8. Find the equation of the line through (−1, 4) and (2, −5).
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m = (−5 − 4) ÷ (2 − (−1)) = −9 ÷ 3 = −3. Then 4 = −3 × (−1) + b, so 4 = 3 + b, and b = 1.
y = −3x + 1. Check: −3 × 2 + 1 = −5. ✓
9. Write the point-slope form of the line with slope ½ through (−2, 3).
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y − 3 = ½(x + 2). The point has x = −2, so x − (−2) becomes x + 2.
Changing form
10. Find the slope and y-intercept of 4x − 2y = 10.
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Rearrange: −2y = −4x + 10, then divide by −2: y = 2x − 5.
Slope 2, y-intercept −5.
11. Write y = −⅗x + 2 in standard form.
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Multiply by 5: 5y = −3x + 10. Move the x term: 3x + 5y = 10.
Parallel, perpendicular and tables
12. Find the line parallel to y = −4x + 7 through (2, 1).
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Same slope, −4. Then 1 = −4 × 2 + b, so 1 = −8 + b, and b = 9.
y = −4x + 9
13. Find the line perpendicular to y = 3x − 1 through (6, 2).
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The negative reciprocal of 3 is −1/3. Then 2 = −⅓ × 6 + b, so 2 = −2 + b, and b = 4.
y = −⅓x + 4
14. The table shows points on a line: when x = 1, 3, 5, y = −1, 5, 11. Find the equation.
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Each time x goes up by 2, y goes up by 6, so m = 6 ÷ 2 = 3. Then −1 = 3 × 1 + b, so b = −4.
y = 3x − 4. Check: 3 × 5 − 4 = 11. ✓
Systems of equations
15. Solve y = 3x − 2 and y = −x + 6.
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Set them equal: 3x − 2 = −x + 6, so 4x = 8 and x = 2. Then y = 3 × 2 − 2 = 4.
(2, 4). Check: −2 + 6 = 4 ✓.
16. Solve x + y = 8 and x − y = 2.
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Add the equations: 2x = 10, so x = 5. Then 5 + y = 8, so y = 3.
(5, 3). Check: 5 − 3 = 2 ✓.
17. Solve 2x + 3y = 7 and 4x + 6y = 10.
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Multiply the first equation by 2: 4x + 6y = 14. Subtract the second: 0 = 4, which is false.
No solution. The lines are parallel, both with slope −2/3.
18. Solve 3x + 2y = 12 and x − 2y = −4.
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The y terms are +2y and −2y, so add: 4x = 8 and x = 2. Then 3 × 2 + 2y = 12, so 2y = 6 and y = 3.
(2, 3). Check: 2 − 2 × 3 = −4 ✓.
Quadratics by intersecting lines
19. Solve x² − 2x − 8 = 0 by the intersecting lines method.
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Rearrange to x² = 2x + 8 and draw the line y = 2x + 8. It crosses y = x² at (−2, 4) and (4, 16).
x = −2 or x = 4. Check: 4 + 4 − 8 = 0 ✓ and 16 − 8 − 8 = 0 ✓.
20. Which line would you draw to solve x² + 3x − 4 = 0? What are the solutions?
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Rearrange to x² = −3x + 4, so draw y = −3x + 4. It crosses the parabola at (−4, 16) and (1, 1).
x = −4 or x = 1. Check: 16 − 12 − 4 = 0 ✓ and 1 + 3 − 4 = 0 ✓.
21. Solve 2x² − 6x + 4 = 0 by the intersecting lines method.
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Divide by 2 first: x² − 3x + 2 = 0. Rearrange: x² = 3x − 2, so draw y = 3x − 2. It crosses the parabola at (1, 1) and (2, 4).
x = 1 or x = 2. Check: 2 − 6 + 4 = 0 ✓ and 8 − 12 + 4 = 0 ✓.
22. How many solutions does x² − 6x + 9 = 0 have? Use the intersecting lines method to explain.
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Rearrange to x² = 6x − 9. The line y = 6x − 9 just touches the parabola at (3, 9): it is a tangent.
One repeated solution, x = 3. The discriminant agrees: 36 − 36 = 0.
Factorising and the quadratic formula
23. Solve (x + 4)(x − 1) = 0.
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One of the brackets must be zero: x + 4 = 0 or x − 1 = 0.
x = −4 or x = 1.
24. Solve x² + 7x + 12 = 0 by factorising.
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Two numbers that multiply to 12 and add to 7: 3 and 4. So (x + 3)(x + 4) = 0.
x = −3 or x = −4. Check: 9 − 21 + 12 = 0 ✓.
25. Solve 3x² + 5x − 2 = 0 by factorising.
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a × c = −6. Two numbers that multiply to −6 and add to 5: 6 and −1.
Split the middle term: 3x² + 6x − x − 2 = 3x(x + 2) − 1(x + 2) = (3x − 1)(x + 2) = 0.
x = ⅓ or x = −2.
26. Use the quadratic formula to solve x² + 4x + 1 = 0. Give exact answers.
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a = 1, b = 4, c = 1, so b² − 4ac = 16 − 4 = 12.
x = (−4 ± √12) ÷ 2. Since √12 = 2√3, x = −2 ± √3, about −3.732 and −0.268.
Scatter graphs and lines of best fit
27. A scatter graph has correlation coefficient r = −0.85. Describe the correlation.
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r is negative and close to −1, so there is a strong negative correlation: as one variable increases, the other tends to decrease, and the points lie fairly close to a straight line.
28. The line of best fit for revision hours (x) and test score (y) is y = 4.96x + 37.3, from data covering 1 to 10 hours. Predict the score for 7 hours of revision. Would a prediction for 15 hours be reliable?
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y = 4.96 × 7 + 37.3 = 72.02, so about 72 marks. This is interpolation, inside the data, so it is fairly reliable.
15 hours gives 4.96 × 15 + 37.3 ≈ 112 marks, more than the maximum of 100. 15 hours is outside the data (extrapolation), so the prediction is not reliable.
29. Find the least squares regression line for the points (1, 3), (2, 5), (3, 6) and (4, 8).
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Means: x̄ = 10 ÷ 4 = 2.5 and ȳ = 22 ÷ 4 = 5.5.
x − x̄: −1.5, −0.5, 0.5, 1.5. y − ȳ: −2.5, −0.5, 0.5, 2.5.
Sxy = 3.75 + 0.25 + 0.25 + 3.75 = 8 and Sxx = 2.25 + 0.25 + 0.25 + 2.25 = 5.
m = 8 ÷ 5 = 1.6 and c = 5.5 − 1.6 × 2.5 = 1.5, so y = 1.6x + 1.5.
30. For used cars, the line of best fit for value in £1000s (y) against age in years (x) is y = −1.53x + 19.1. What does the slope tell you?
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The slope is −1.53, and the units are £1000s per year. On average, a car’s value falls by about £1,530 for each year of age.
Want more help with any of these? Each topic has its own lesson on the Lessons page.