The shape of the lesson
By the end of the lesson, learners can
- Same gradient, different intercept: In the lab, drag the point that the second line must pass through.
- Parallel to ax + by = c: (Extended) When the given line is in a form like \(2x + 3y = 6\), rearrange to \(y = mx + c\) to read its gradient.
- Using parallel lines: Equal gradients prove that sides are parallel, which identifies parallelograms and trapezia.
- Gradient: write the given line as \(y = mx + c\) and read \(m\).
- Parallel: the new line has the same gradient \(m\).
- Intercept: substitute the given point into \(y = mx + c\) to find the new \(c\).
- Shortcut (Extended): parallel to \(ax + by = c\) is \(ax + by = k\); put the point in to find \(k\).
- Proof: equal gradients show that two sides are parallel.
One picture every idea returns to
Parallel lines have equal gradients. To find the line parallel to \(y = mx + c\) through a point: keep the same \(m\), substitute the point to find the new \(c\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Same slope, never meet.. Parallel lines climb at exactly the same rate, so they have the same gradient. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Same gradient, different intercept | In the lab, drag the point that the second line must pass through. Project tab 4 of the lab (Straight lines lab, parallel lines) and let a student drive it while the class predicts. Model the worked example Spotting parallel lines on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Parallel to ax + by = c | (Extended) When the given line is in a form like \(2x + 3y = 6\), rearrange to \(y = mx + c\) to read its gradient. Model the worked example Rearranging first (Extended) on the board, one line at a time. Your turn: (Extended) Find the equation of the line parallel to \(3x - y = 4\) that passes through \((-2, 5)\). | Pairs |
| 21 to 30 min | Teach and check: Using parallel lines | Equal gradients prove that sides are parallel, which identifies parallelograms and trapezia. Model the worked example Proving a parallelogram on the board, one line at a time. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 17 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Using parallel lines, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Using the point's \(y\)-coordinate as the intercept. | The line through \((1, -3)\) is NOT \(y = 4x - 3\): the point is not on the \(y\)-axis, so you must substitute to find \(c\). |
| Reading the coefficient of \(x\) in \(2x + 3y = 6\) as the gradient. | The gradient is \(-\dfrac{2}{3}\), not 2: rearrange to \(y = \ldots\) before you read it. |
| Dividing by a negative and losing the sign. | From \(-2y = -6x + 5\), every term is divided by \(-2\): \(y = 3x - 2.5\). Check the sign of your gradient by picturing which way the line slopes. |
Parallel lines: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
Which of these lines are parallel to each other?
A: \(y = 2x + 3\) B: \(y = 3 - 2x\) C: \(y = 2x - 7\) D (Extended): \(2y = 4x + 1\)
Find the equation of the line parallel to \(y = -3x + 2\) that passes through \((2, 5)\).
(Extended) Find the equation of the line parallel to \(5x + 2y = 8\) that passes through \((-2, 3)\). Give your answer in the form \(ax + by = c\).
The vertices of quadrilateral \(ABCD\) are \(A(0, 0)\), \(B(6, 2)\), \(C(5, 5)\) and \(D(2, 4)\).
Show that \(ABCD\) is a trapezium.
A line \(L\) drawn on a grid passes through \((0, 4)\) and \((2, 0)\).
Find the equation of the line parallel to \(L\) that passes through \((3, 1)\).
Cyclist A's distance from a town after \(t\) hours is \(d = 15t + 6\) km. Cyclist B leaves the town at \(t = 0\) and rides at the same speed as A.
(a) Write down the equation for B's distance.
(b) Explain, using the graphs, why B never catches A.
(c) Cyclist C leaves the town at \(t = 0\) riding at 18 km/h. After how long does C catch A?
Answer key and marking notes
- Gradients: A is 2, B is \(-2\), C is 2, and D is \(y = 2x + 0.5\), gradient 2.
- A, C and D are parallel.
- Gradient \(-3\): \(5 = -3 \times 2 + c = -6 + c\), so \(c = 11\).
- \(y = -3x + 11\)
- Parallel line: \(5x + 2y = k\).
- Substitute \((-2, 3)\): \(k = 5 \times (-2) + 2 \times 3 = -10 + 6 = -4\).
- \(5x + 2y = -4\)
- Gradient \(AB = \dfrac{2}{6} = \dfrac{1}{3}\); gradient \(DC = \dfrac{5 - 4}{5 - 2} = \dfrac{1}{3}\). So \(AB \parallel DC\).
- Gradient \(AD = \dfrac{4}{2} = 2\); gradient \(BC = \dfrac{5 - 2}{5 - 6} = -3\). So \(AD\) and \(BC\) are not parallel.
- Exactly one pair of parallel sides, so \(ABCD\) is a trapezium.
- Gradient of \(L = \dfrac{0 - 4}{2 - 0} = -2\).
- \(1 = -2 \times 3 + c\), so \(c = 7\).
- \(y = -2x + 7\)
- (a) Same speed means the same gradient 15, starting from 0: \(d = 15t\).
- (b) The lines \(d = 15t + 6\) and \(d = 15t\) have equal gradients, so they are parallel and never meet: B is always 6 km behind A.
- (c) \(18t = 15t + 6\), so \(3t = 6\) and \(t = 2\) hours.
6 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which of these lines is parallel to \(y = 5x - 2\)?
- A\(y = 5x + 3\)
- B\(y = -5x - 2\)
- C\(y = -\dfrac{1}{5}x + 1\)
- D\(y = 2x + 5\)
Find the equation of the line parallel to \(y = 3x + 4\) that passes through \((2, -1)\).
- A\(y = 3x - 7\)
- B\(y = 3x - 1\)
- C\(y = 3x + 5\)
- D\(y = -\dfrac{1}{3}x - \dfrac{1}{3}\)
The diagram shows the line \(L\). Which of these lines is parallel to \(L\)?
- A\(y = \dfrac{1}{2}x + 3\)
- B\(y = 2x - 1\)
- C\(y = -\dfrac{1}{2}x + 3\)
- D\(y = -2x + 3\)
\(A(0, 0)\), \(B(6, 2)\), \(C(5, 5)\) and \(D(2, 4)\) are the vertices of quadrilateral \(ABCD\). Which statement is true?
- A\(AB\) is parallel to \(DC\) only, so \(ABCD\) is a trapezium
- B\(AD\) is parallel to \(BC\) only, so \(ABCD\) is a trapezium
- CBoth pairs of opposite sides are parallel, so \(ABCD\) is a parallelogram
- DNo two sides are parallel
What is the gradient of a line parallel to \(5x - 2y = 6\)?
- A\(\dfrac{5}{2}\)
- B\(-\dfrac{5}{2}\)
- C\(5\)
- D\(-\dfrac{2}{5}\)
Find the equation of the line parallel to \(4x + 3y = 7\) that passes through \((-1, 2)\).
- A\(4x + 3y = 2\)
- B\(4x + 3y = 7\)
- C\(3x + 4y = 5\)
- D\(4x + 3y = 10\)
Quiz key
\(y = 5x + 3\) has gradient 5, the same as \(y = 5x - 2\), so the lines are parallel (only the intercept differs). The tempting \(y = -5x - 2\) has gradient \(-5\): it slopes the other way. \(y = -\dfrac{1}{5}x + 1\) is perpendicular to the line, not parallel, and \(y = 2x + 5\) swaps the two numbers round.
\(-1 = 3 \times 2 + c\), so \(-1 = 6 + c\) and \(c = -7\): \(y = 3x - 7\). Check: \(6 - 7 = -1\). The tempting \(y = 3x - 1\) uses the point's \(y\)-coordinate as the intercept, but \((2, -1)\) is not on the \(y\)-axis. \(y = 3x + 5\) adds 6 instead of subtracting it, and \(y = -\dfrac{1}{3}x - \dfrac{1}{3}\) uses the perpendicular gradient.
From \((0, -1)\) to \((4, 1)\), \(L\) rises 2 for a run of 4, so its gradient is \(\dfrac{2}{4} = \dfrac{1}{2}\). A parallel line has gradient \(\dfrac{1}{2}\): \(y = \dfrac{1}{2}x + 3\). The tempting \(y = 2x - 1\) divides run by rise (and keeps \(L\)'s intercept, which does not matter for parallel lines). \(y = -\dfrac{1}{2}x + 3\) slopes downhill, and \(y = -2x + 3\) is perpendicular to \(L\).
Gradient \(AB = \dfrac{2}{6} = \dfrac{1}{3}\) and gradient \(DC = \dfrac{5 - 4}{5 - 2} = \dfrac{1}{3}\): equal, so \(AB\) is parallel to \(DC\). Gradient \(AD = \dfrac{4}{2} = 2\) but gradient \(BC = \dfrac{5 - 2}{5 - 6} = -3\): not equal. One pair of parallel sides makes a trapezium. The tempting "\(AD\) parallel to \(BC\)" picks the wrong pair: \(C\) is to the LEFT of \(B\), so the run from \(B\) to \(C\) is \(-1\) and that gradient is negative, unlike \(AD\). It is not a parallelogram, because only one pair is parallel.
\(-2y = -5x + 6\), so \(y = \dfrac{5}{2}x - 3\). The gradient is \(\dfrac{5}{2}\), and any parallel line has the same gradient. The tempting \(-\dfrac{5}{2}\) divides by \(+2\) instead of \(-2\) and loses a sign. 5 reads the coefficient of \(x\) without rearranging, and \(-\dfrac{2}{5}\) is the perpendicular gradient.
The parallel line is \(4x + 3y = k\). At \((-1, 2)\): \(k = 4 \times (-1) + 3 \times 2 = -4 + 6 = 2\), so \(4x + 3y = 2\). The tempting \(4x + 3y = 7\) is the original line itself, which does not pass through \((-1, 2)\). \(3x + 4y = 5\) swaps the coefficients (a different gradient), and \(4x + 3y = 10\) uses \(x = 1\) instead of \(-1\).
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.