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MTH1W Linear Relations & Graphing
Grade 9 Unit 4 — practice questions with full solutions

Slope, intercepts, y = mx + b, standard form, parallel and perpendicular lines, tables of values, and reading meaning out of a graph in a real context.

Free MTH1W linear relations & graphing practice for Grade 9 students in Ontario. These 13 questions cover the same material as a typical unit test or exam review on this topic, and every one comes with a complete worked solution — not just an answer key. Useful whether you are preparing for a MTH1W unit test, catching up on a lesson, or reviewing before the final exam.
MTH1W · Grade 9 Strand C — Algebra 11 core 2 challenge
1Core
Find the slope of the line through (2, 5) and (6, 13).
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  1. Use m = (y2 − y1)⁄(x2 − x1).
  2. Substitute: (13 − 5)⁄(6 − 2) = 8⁄4.
  3. Simplify.
Final answerm = 2
2Core
Write the equation of the line with slope −3 and y-intercept 7.
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  1. Use the form y = mx + b.
  2. Here m = −3 and b = 7.
  3. Substitute directly.
Final answery = −3x + 7
3Core
Find the equation of the line passing through (1, 4) and (3, 10).
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  1. First find the slope: m = (10 − 4)⁄(3 − 1) = 6⁄2 = 3.
  2. Substitute one point into y = 3x + b. Using (1, 4): 4 = 3(1) + b.
  3. Solve for b: b = 1.
  4. Check with the other point: 3(3) + 1 = 10. ✔
Final answery = 3x + 1
4Core
-6-5-4-3-2-1123456-5-4-3-2-1123456xy(0, -1)(2, 3)
From the graph above, determine: (a) the slope of the line, (b) its y-intercept, (c) the equation in the form y = mx + b, and (d) the value of y when x = 4.
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  1. (a) Use the two marked points (0, −1) and (2, 3). Slope = rise ÷ run = (3 − (−1)) ÷ (2 − 0) = 4 ÷ 2 = 2.
  2. (b) The line crosses the y-axis at −1, so b = −1.
  3. (c) Substitute into y = mx + b: y = 2x − 1.
  4. (d) Substitute x = 4: y = 2(4) − 1 = 7.
  5. Note that (4, 7) lies beyond the drawn portion — the equation still works because a line continues forever.
Final answer(a) m = 2   (b) −1   (c) y = 2x − 1   (d) y = 7
5Core
Find the x-intercept and y-intercept of 2x + 5y = 20.
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  1. For the x-intercept, set y = 0: 2x = 20, so x = 10.
  2. For the y-intercept, set x = 0: 5y = 20, so y = 4.
  3. Write them as coordinates.
Final answerx-intercept (10, 0);   y-intercept (0, 4)
6Core
Does the point (3, −2) lie on the line y = −2x + 4? Justify your answer.
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  1. Substitute x = 3 into the equation.
  2. y = −2(3) + 4 = −6 + 4 = −2.
  3. The calculated y matches the given y-coordinate, so the point satisfies the equation.
Final answerYes — substituting x = 3 gives y = −2
7Core
A line has equation y = 2⁄3x − 5. State the slope of a line parallel to it, and the slope of a line perpendicular to it.
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  1. Parallel lines have identical slopes, so the parallel slope is 2⁄3.
  2. Perpendicular slopes are negative reciprocals — flip the fraction and change the sign.
  3. Flipping 2⁄3 gives 3⁄2; changing the sign gives −3⁄2.
  4. Quick check: 2⁄3 × (−3⁄2) = −1. ✔
Final answerparallel: 2⁄3    perpendicular: −3⁄2
8Challenge
01234504080120160200ABCTime (hours)Distance (km)
The distance–time graph shows a cyclist's journey in three segments A, B and C. (a) Describe what is happening in each segment. (b) Find the speed during segment A and during segment C. (c) Find the average speed for the whole 5-hour journey.
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  1. (a) Segment A rises steadily — the cyclist is moving away at a constant speed. Segment B is flat — distance is not changing, so the cyclist has stopped. Segment C rises more steeply than A — moving again, and faster than before.
  2. (b) Speed = change in distance ÷ change in time. Segment A: 80 km in 2 h = 40 km/h.
  3. Segment C: from 80 km to 200 km is 120 km, over 2 hours (from t = 3 to t = 5) = 60 km/h.
  4. (c) Average speed uses total distance over total time, not the average of the two speeds.
  5. Total distance 200 km ÷ total time 5 h = 40 km/h.
  6. Note it is not 50 km/h — the rest stop pulls the average down.
Final answer(a) A: constant speed; B: stopped; C: faster constant speed   (b) A = 40 km/h, C = 60 km/h   (c) 40 km/h
9Core
A table of values gives x = 0, 1, 2, 3 and y = 5, 8, 11, 14. Is the relation linear? If so, write its equation.
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  1. Calculate the first differences in y: 8 − 5 = 3, 11 − 8 = 3, 14 − 11 = 3.
  2. The first differences are constant, so the relation is linear.
  3. The constant difference is the slope: m = 3.
  4. The value of y when x = 0 is the y-intercept: b = 5.
Final answerLinear;   y = 3x + 5
10Core
A table gives x = 0, 1, 2, 3 and y = 1, 2, 5, 10. Is this relation linear? Explain.
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  1. Calculate the first differences: 2 − 1 = 1, 5 − 2 = 3, 10 − 5 = 5.
  2. The differences are 1, 3, 5 — they are not constant.
  3. A relation is linear only when the first differences are equal.
Final answerNot linear — first differences (1, 3, 5) are not constant
11Core
A taxi charges a $4.00 flat fee plus $2.25 per kilometre. Write an equation for the cost C after d km, state what the slope and intercept represent, and find the cost of a 12 km trip.
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  1. The flat fee is the fixed starting value, so it is the C-intercept.
  2. The per-kilometre charge is the rate of change, so it is the slope.
  3. Equation: C = 2.25d + 4.
  4. For d = 12: C = 2.25(12) + 4 = 27 + 4 = 31.
Final answerC = 2.25d + 4; slope = cost per km, intercept = flat fee; 12 km costs $31.00
12Core
Find the point of intersection of y = 2x + 1 and y = −x + 7.
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  1. At the intersection both equations give the same y, so set them equal: 2x + 1 = −x + 7.
  2. Add x to both sides: 3x + 1 = 7.
  3. Solve: 3x = 6, so x = 2.
  4. Substitute back: y = 2(2) + 1 = 5.
Final answer(2, 5)
13Challenge
A line has an x-intercept of 4 and passes through (0, −6). Find its equation in the form y = mx + b.
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  1. An x-intercept of 4 means the line passes through (4, 0).
  2. You now have two points: (4, 0) and (0, −6).
  3. Slope: m = (−6 − 0)⁄(0 − 4) = −6⁄−4 = 3⁄2.
  4. The point (0, −6) is already on the y-axis, so b = −6.
Final answery = 3⁄2x − 6

MTH1W Linear Relations & Graphing — common questions

Short answers to the things students ask most about this unit.

How do you tell if a relation is linear from a table?

Calculate the first differences in the y-values. If they are constant, the relation is linear; if they change, it is non-linear.

What do slope and y-intercept mean in a word problem?

The slope is the rate of change — the amount added per unit — and the y-intercept is the starting value or fixed fee before anything changes.

More MTH1W practice — grade 9 math

MTH1W Number SenseMTH1W Algebraic Expressions & Exponent LawsMTH1W Solving Linear EquationsMTH1W Geometry & MeasurementAll MTH1W units
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