The shape of the lesson
By the end of the lesson, learners can
- Three forms of one equation: In the lab, drag the two points and watch the gradient and the equation of the line update as you go.
- Parallel and perpendicular: In the lab, drag the points to turn the line and watch its perpendicular turn with it: the product of their gradients stays at \(-1\).
- Lengths, midpoints and intersections: Most coordinate geometry questions combine several small tools.
- Gradient from two points, or by rearranging a given line to \(y = mx + c\).
- Parallel: same gradient. Perpendicular: \(-\dfrac{1}{m}\).
- Equation: \(y - y_{1} = m(x - x_{1})\), then rearrange to the form asked for.
- Length and midpoint from the formulae; intersections by solving simultaneously.
- Sketch the points to check that your answers are sensible.
One picture every idea returns to
The line through \((x_{1}, y_{1})\) with gradient \(m\) is \(y - y_{1} = m(x - x_{1})\). Parallel lines have equal gradients; perpendicular lines have \(m_{1}m_{2} = -1\), so a perpendicular gradient is the negative reciprocal \(-\dfrac{1}{m}\).
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, Two facts fix a line.. A gradient and a point, or two points, are all you need to write the equation of a line. 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: Three forms of one equation | In the lab, drag the two points and watch the gradient and the equation of the line update as you go. Project tab 3 of the lab (Straight lines lab, equation of a line) and let a student drive it while the class predicts. Model the worked example Through two points on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Parallel and perpendicular | In the lab, drag the points to turn the line and watch its perpendicular turn with it: the product of their gradients stays at \(-1\). Project tab 5 of the lab (Straight lines lab, perpendicular lines) and let a student drive it while the class predicts. Model the worked example Perpendicular bisector on the board, one line at a time. Your turn: Find the equation of the line through \((2, -3)\) perpendicular to \(x + 4y = 1\). | Pairs |
| 21 to 30 min | Teach and check: Lengths, midpoints and intersections | Most coordinate geometry questions combine several small tools. Model the worked example Intersection on the board, one line at a time. | Round Table |
| 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, 22 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 Lengths, midpoints and intersections, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| The gradient of \(3x + 4y = 24\) is not 3 (nor \(\frac{4}{3}\)). | Always rearrange to \(y = mx + c\) before reading \(m\). |
| The perpendicular gradient is the NEGATIVE reciprocal. | For \(m = -2\) it is \(+\frac{1}{2}\); \(-\frac{1}{2}\) (reciprocal only) and \(2\) (sign change only) are both wrong. |
| In the length formula, subtract the coordinates before squaring: from \((-1, 3)\) to \((5, -5)\) the changes are 6 and \(-8\), giving \(\sqrt{36 + 64} = 10\). | Adding the coordinates (4 and \(-2\)) is a common slip. |
Equations of straight lines: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Find the gradient and the \(y\)-intercept of the line \(5x - 2y + 8 = 0\).
Find the equation of the line through \((4, -3)\) and \((-2, 6)\), giving your answer in the form \(ax + by + c = 0\) where \(a\), \(b\) and \(c\) are integers.
The points \(A(-2, 1)\) and \(B(4, 5)\) are given.
(a) Find the equation of the perpendicular bisector of \(AB\).
(b) Find the point where the perpendicular bisector crosses the \(x\)-axis.
The points \(A(0, 2)\) and \(B(6, 5)\) are given. The line through \(B\) perpendicular to \(AB\) meets the \(x\)-axis at \(C\).
(a) Find the coordinates of \(C\).
(b) Find the exact area of triangle \(ABC\).
The points \(P(1, 2)\), \(Q(4, 6)\) and \(R(8, 3)\) are given. Show that triangle \(PQR\) is right-angled and isosceles.
Coordinates are in kilometres. A ship sails in a straight line from \(A(2, 1)\) to \(B(14, 10)\). A lighthouse is at \(L(10, 1)\).
(a) Find the equation of the ship's path \(AB\).
(b) Find the point on \(AB\) where the ship is closest to the lighthouse, and the shortest distance.
Answer key and marking notes
- \(2y = 5x + 8 \Rightarrow y = \dfrac{5}{2}x + 4\).
- Gradient \(\dfrac{5}{2}\), \(y\)-intercept \(4\).
- \(m = \dfrac{6 - (-3)}{-2 - 4} = \dfrac{9}{-6} = -\dfrac{3}{2}\).
- \(y + 3 = -\dfrac{3}{2}(x - 4) \Rightarrow 2y + 6 = -3x + 12\).
- \(3x + 2y - 6 = 0\) (check \((-2, 6)\): \(-6 + 12 - 6 = 0\)).
- (a) Midpoint \((1, 3)\). Gradient \(AB = \dfrac{4}{6} = \dfrac{2}{3}\), so the perpendicular gradient is \(-\dfrac{3}{2}\).
- \(y - 3 = -\dfrac{3}{2}(x - 1) \Rightarrow 2y - 6 = -3x + 3 \Rightarrow 3x + 2y - 9 = 0\).
- (b) \(y = 0\): \(3x = 9\), so \((3,\ 0)\).
- (a) Gradient \(AB = \dfrac{1}{2}\); perpendicular gradient \(-2\): \(y = -2x + 17\).
- \(y = 0 \Rightarrow x = \dfrac{17}{2}\), so \(C\left(\dfrac{17}{2}, 0\right)\).
- (b) \(AB = \sqrt{45} = 3\sqrt{5}\), \(BC = \sqrt{\left(\dfrac{5}{2}\right)^{2} + 5^{2}} = \dfrac{5\sqrt{5}}{2}\).
- Angle \(ABC = 90^{\circ}\), so area \(= \dfrac{1}{2} \times 3\sqrt{5} \times \dfrac{5\sqrt{5}}{2} = \dfrac{75}{4}\).
- Gradient \(PQ = \dfrac{4}{3}\), gradient \(QR = \dfrac{-3}{4}\); \(\dfrac{4}{3} \times \left(-\dfrac{3}{4}\right) = -1\), so angle \(PQR = 90^{\circ}\).
- \(PQ = \sqrt{3^{2} + 4^{2}} = 5\) and \(QR = \sqrt{4^{2} + 3^{2}} = 5\).
- Two equal sides meeting at a right angle: the triangle is right-angled and isosceles.
- (a) Gradient \(\dfrac{9}{12} = \dfrac{3}{4}\): \(y - 1 = \dfrac{3}{4}(x - 2) \Rightarrow 3x - 4y - 2 = 0\).
- (b) The closest point is where the perpendicular from \(L\) meets \(AB\). Perpendicular gradient \(-\dfrac{4}{3}\): \(y - 1 = -\dfrac{4}{3}(x - 10) \Rightarrow 4x + 3y - 43 = 0\).
- Solve \(3x - 4y = 2\) and \(4x + 3y = 43\): \(x = \dfrac{178}{25} = 7.12\), \(y = \dfrac{121}{25} = 4.84\).
- Distance \(= \sqrt{(10 - 7.12)^{2} + (1 - 4.84)^{2}} = \sqrt{8.2944 + 14.7456} = \sqrt{23.04}\).
- Shortest distance \(= 4.8\) km, at \((7.12,\ 4.84)\).
8 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.
Find the gradient of the line \(5x - 2y = 8\).
- A\(\dfrac{5}{2}\)
- B\(5\)
- C\(-\dfrac{5}{2}\)
- D\(-\dfrac{2}{5}\)
Find the equation of the line through \((-1,\ 3)\) and \((3,\ -5)\), in the form \(ax + by + c = 0\).
- A\(2x + y - 1 = 0\)
- B\(2x - y + 5 = 0\)
- C\(x + 2y - 5 = 0\)
- D\(2x + y - 5 = 0\)
Find the gradient of a line perpendicular to \(3x + 4y = 7\).
- A\(\dfrac{4}{3}\)
- B\(-\dfrac{4}{3}\)
- C\(\dfrac{3}{4}\)
- D\(-\dfrac{3}{4}\)
Find the distance between the points \((-2,\ 5)\) and \((4,\ -3)\).
- A\(10\)
- B\(2\sqrt{2}\)
- C\(14\)
- D\(100\)
Find the equation of the perpendicular bisector of the points \(A(-3,\ 2)\) and \(B(5,\ 6)\).
- A\(y = -2x + 6\)
- B\(y = \dfrac{1}{2}x + \dfrac{7}{2}\)
- C\(y = -2x - 4\)
- D\(y = 2x + 2\)
Find the point of intersection of the lines \(3x - 2y = 12\) and \(x + 4y = -10\).
- A\((2,\ -3)\)
- B\((-3,\ 2)\)
- C\((2,\ 3)\)
- D\((4,\ 0)\)
The diagram shows the points \(A(1,\ 1)\) and \(B(5,\ 3)\). The line through B perpendicular to AB meets the \(x\)-axis at C. Find the area of triangle ABC.
- A\(7.5\)
- B\(15\)
- C\(4.5\)
- D\(9.75\)
Find the equation of the line through \((2,\ -1)\) that is parallel to \(4x - 6y = 5\).
- A\(2x - 3y - 7 = 0\)
- B\(3x + 2y - 4 = 0\)
- C\(2x - 3y - 1 = 0\)
- D\(4x - 6y - 5 = 0\)
Quiz key
\(2y = 5x - 8\), so \(y = \dfrac{5}{2}x - 4\) and the gradient is \(\dfrac{5}{2}\). The tempting 5 is the coefficient of \(x\) before dividing by 2. \(-\dfrac{5}{2}\) loses track of the sign when moving \(-2y\) across, and \(-\dfrac{2}{5}\) is the gradient of a PERPENDICULAR line.
Gradient \(= \dfrac{-5 - 3}{3 - (-1)} = \dfrac{-8}{4} = -2\). Then \(y - 3 = -2(x + 1)\), so \(y = -2x + 1\), i.e. \(2x + y - 1 = 0\). Check \((3,\ -5)\): \(6 - 5 - 1 = 0\). The tempting \(2x - y + 5 = 0\) uses gradient \(+2\), from subtracting the coordinates in different orders. \(x + 2y - 5 = 0\) uses the reciprocal \(-\frac{1}{2}\), and \(2x + y - 5 = 0\) uses \(x - 1\) instead of \(x + 1\) for the point \((-1,\ 3)\).
\(y = -\dfrac{3}{4}x + \dfrac{7}{4}\), so \(m = -\dfrac{3}{4}\). The perpendicular gradient is the negative reciprocal: \(-1 \div \left(-\dfrac{3}{4}\right) = \dfrac{4}{3}\). The tempting \(-\dfrac{4}{3}\) takes the reciprocal but not the negative. \(\dfrac{3}{4}\) only changes the sign, and \(-\dfrac{3}{4}\) is the gradient of a PARALLEL line.
\(\Delta x = 4 - (-2) = 6\) and \(\Delta y = -3 - 5 = -8\). Distance \(= \sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10\). The tempting \(2\sqrt{2}\) ADDS the coordinates (\(-2 + 4 = 2\), \(5 - 3 = 2\)) and gives \(\sqrt{8}\). 14 adds the two changes without Pythagoras, and 100 forgets the square root.
Midpoint \(M = (1,\ 4)\). Gradient of AB \(= \dfrac{6 - 2}{5 + 3} = \dfrac{1}{2}\), so the bisector has gradient \(-2\). \(y - 4 = -2(x - 1)\), giving \(y = -2x + 6\). The tempting \(y = \frac{1}{2}x + \frac{7}{2}\) goes through M with the gradient of AB itself, so it is the line AB, not perpendicular to it. \(y = -2x - 4\) has the right gradient but passes through A instead of M, and \(y = 2x + 2\) drops the minus sign of the negative reciprocal.
\(2 \times\) first: \(6x - 4y = 24\). Adding \(x + 4y = -10\): \(7x = 14\), so \(x = 2\). Then \(2 + 4y = -10\) gives \(y = -3\). Check in the first: \(6 + 6 = 12\). The tempting \((-3,\ 2)\) swaps the coordinates. \((2,\ 3)\) loses the sign of \(y\), and \((4,\ 0)\) lies on the first line only (\(4 + 0 \ne -10\)).
Gradient of AB \(= \dfrac{2}{4} = \dfrac{1}{2}\), so BC has gradient \(-2\): \(y - 3 = -2(x - 5)\), \(y = -2x + 13\), and C is \((6.5,\ 0)\). \(\text{AB} = \sqrt{16 + 4} = \sqrt{20}\), \(\text{BC} = \sqrt{1.5^{2} + 3^{2}} = \sqrt{11.25}\). Area \(= \frac{1}{2}\sqrt{20 \times 11.25} = \frac{1}{2}\sqrt{225} = 7.5\). The tempting 15 forgets the \(\frac{1}{2}\). 4.5 uses gradient \(+2\) for BC, giving C at \((3.5,\ 0)\), and 9.75 is the area of triangle OBC, \(\frac{1}{2} \times 6.5 \times 3\).
\(6y = 4x - 5\) gives gradient \(\dfrac{2}{3}\). \(y + 1 = \dfrac{2}{3}(x - 2)\), so \(3y + 3 = 2x - 4\), i.e. \(2x - 3y - 7 = 0\). Check: \(4 + 3 - 7 = 0\). The tempting \(3x + 2y - 4 = 0\) uses the perpendicular gradient \(-\frac{3}{2}\). \(2x - 3y - 1 = 0\) passes through \((2,\ 1)\), a sign slip with \(y_{1}\), and \(4x - 6y - 5 = 0\) is the original line, which does not pass through \((2,\ -1)\).
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.