Ch 1 · 7 Straight-line Graphs · O Level Additional Mathematics 4037
Topic 1 of 2 in Coordinate Geometry — O Level Additional Mathematics 4037 — 46 lessons.
Introduction
Scientists love straight lines: two points fix one, and its gradient and intercept are easy to read. So when an experiment gives a curve, they change the variables until the graph becomes straight, then read off the constants. This topic reviews line equations, midpoints, perpendicular bisectors, and then that transformation trick. (Syllabus topic 7.)
Key ideas — Equations of a line
Gradient m = y_2 - y_1/x_2 - x_1. The line through (x_1, y_1) with gradient m is y - y_1 = m(x - x_1), or y = mx + c (intercept c), or in general form ax + by + c = 0. Points of intersection come from solving the equations simultaneously.
🧑🏫 Teacher's explanation
Think of a straight line as a constant 'rate of change'. If you are walking up a hill at a steady pace, every step forward (x) results in the same amount of height gained (y). The gradient, m, is simply that ratio of 'rise over run'. If the hill is steep, m is large; if it is flat, m is zero; if it goes downhill, m is negative.
📖 Definitions
🔍 Worked example — Easy
Find the equation of the line passing through the points A(2, 5) and B(4, 9) in the form y = mx + c.
⚠️ Common errors
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🎯 Exam question
Key ideas — Parallel and perpendicular
🧑🏫 Teacher's explanation
Imagine you are walking up a hill. If your friend is walking up a different hill that has the exact same steepness, you are both moving in a parallel direction. In coordinate geometry, 'steepness' is the gradient, m. If two lines are parallel, they never meet because they rise and run at the exact same rate, meaning m_1 = m_2.
📖 Definitions
🔍 Worked example — Easy
Line L_1 passes through (2, 5) and (4, 9). Find the gradient of a line L_2 that is parallel to L_1.
🔍 Worked example — Standard
A line has the equation y = -3x + 7. Find the equation of the line that is perpendicular to this line and passes through the point (4, 2).
🔍 Worked example — Exam-style
Find the equation of the perpendicular bisector of the line segment joining A(1, 4) and B(5, 10).
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Midpoint, length and perpendicular bisector
M = (x_1 + x_2/2,\ y_1 + y_2/2 ) length = √((x_2 - x_1)^2 + (y_2 - y_1)^2)
🧑🏫 Teacher's explanation
Imagine you are standing exactly halfway between two friends, Alice and Bob. If you want to walk in a direction that keeps you exactly the same distance from both of them at all times, you must walk along a line that cuts the path between them perfectly in half at a 90^° angle. This is the 'perpendicular bisector'.
📖 Definitions
🔍 Worked example — Easy
Find the midpoint and the length of the line segment joining A(2, 3) and B(6, 11).
🔍 Worked example — Standard
Find the equation of the perpendicular bisector of the line segment joining P(-1, 4) and Q(3, 2).
🔍 Worked example — Exam-style
The line L_1 passes through (2, 5) and (4, 9). Find the equation of line L_2 which is the perpendicular bisector of L_1.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Transforming to straight-line form
Take logs (or change variables) so the equation looks like Y = mX + c:
🧑🏫 Teacher's explanation
Sometimes, mathematical relationships don't look like a straight line when you plot them on a standard graph. For example, y = Ax^n is a curve. However, if we 'repackage' our variables—by taking logarithms or squaring them—we can force that curve to behave like a straight line. Think of it like looking at a curved road through a special lens that makes it appear straight; once we see the straight line, we can easily find the 'slope' and 'starting point' of the original curve.
📖 Definitions
🔍 Worked example — Easy
Show that the equation y = Ax^n can be written in the form lg y = mlg x + c and identify m and c.
🔍 Worked example — Standard
A curve has the equation y = Ab^x. A graph of lg y against x is plotted. If the gradient is 0.3 and the y-intercept is 1.5, find the values of A and b.
🔍 Worked example — Exam-style
The relationship between y and x is given by y^2 = Ax^3 + B. A graph of y^2 against x^3 is plotted. The line passes through (8, 100) and (27, 250). Find the values of A and B.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Worked examples
A(-2, 5), B(4, 1). Midpoint (1, 3); gradient AB = 1 - 5/4 + 2 = -tfrac23, so the bisector has gradient tfrac32: y - 3 = tfrac32(x - 1), i.e. 3x - 2y + 3 = 0. Length AB = √(36 + 16) = 2√(13).
🔍 Worked example — Easy
Find the equation of the perpendicular bisector of the line segment joining A(-2, 5) and B(4, 1).
🔍 Worked example — Standard
A graph of lg y against lg x is a straight line passing through (1, 2.5) and (3, 6.5). Find the relationship between y and x in the form y = Ax^n.
🔍 Worked example — Exam-style
A graph of y^2 against x^3 is a straight line passing through (2, 11) and (5, 26). Find the equation of the line in the form y^2 = Ax^3 + B.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) gradient of the given line tfrac25, so -tfrac52: y + 2 = -tfrac52(x - 3), i.e. 5x + 2y - 11 = 0. (2) midpoint (4, 3), gradient PQ = -tfrac43, bisector y - 3 = tfrac34(x - 4), i.e. 3x - 4y = 0; PQ = 10. (3) ln b = 0.5 ⇒ b = e^0.5 = 1.65; ln A = 1.2 ⇒ A = e^1.2 = 3.32. (4) e^2y = 3x^2 + 2, so y = tfrac12ln(3x^2 + 2).