7 Straight-line graphs | IGCSE Additional Maths 0606

Home › Learning Hub › IGCSE Add Maths › 7 Straight-line graphs

🎯What you need to be able to do

Find and use the equation of a straight line in any form.
Use the gradient conditions for parallel and perpendicular lines.
Find midpoints and lengths, and the equation of a perpendicular bisector.
Transform relationships such as \( y = Ax^{n} \) and \( y = Ab^{x} \) into straight-line form, and find the constants from the gradient and intercept.
Go the other way: turn a given straight-line graph (for example \( y^{2} \) against \( x^{3} \)) back into an equation connecting \(x\) and \(y\).

📚The mathematics

The toolkit

gradient \( m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} \)

line: \( y - y_{1} = m(x - x_{1}) \)

midpoint \( \left(\dfrac{x_{1} + x_{2}}{2}, \dfrac{y_{1} + y_{2}}{2}\right) \)

length \( \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \)

Parallel lines have equal gradients. Perpendicular lines have gradients whose product is \(-1\): if one is \( m \), the other is \( -\dfrac{1}{m} \). To get the gradient of a line given as \( 2x - 5y = 7 \), rearrange to \( y = \tfrac{2}{5}x - \tfrac{7}{5} \) first.

The perpendicular bisector of \(AB\) passes through the midpoint of \(AB\) at right angles to it. Every point on it is the same distance from \(A\) as from \(B\) — which is why it turns up in circle questions (the centre of a circle through \(A\) and \(B\) lies on it).

The segment from A(minus 1, 5) to B(5, 1) with its midpoint M(2, 3) and the perpendicular bisector y = 1.5x through M and the origin, marked with a right angle. The bisector meets the line y = x + 2 at C(4, 6), forming the triangle ABC. — Worked example 1: the perpendicular bisector of \(AB\) is \( y = \tfrac{3}{2}x \). It meets \( y = x + 2 \) at \( C(4, 6) \), and \(CM\) is the height of triangle \(ABC\).

Straight-line form: linearising a relationship

Experimental data often follow \( y = Ax^{n} \) or \( y = Ab^{x} \). Taking logs turns them into straight lines, and then the gradient and intercept give the constants:

\( y = Ax^{n} \Rightarrow \lg y = n\lg x + \lg A \): plot \( \lg y \) against \( \lg x \); gradient \(n\), intercept \( \lg A \)

\( y = Ab^{x} \Rightarrow \ln y = (\ln b)x + \ln A \): plot \( \ln y \) against \(x\); gradient \( \ln b \), intercept \( \ln A \)

The pattern is always \( Y = mX + c \): decide what \(Y\) and \(X\) are, then match. The same works in reverse: “\( e^{2y} \) plotted against \( x^{2} \) gives a straight line with gradient 4 and intercept 3” means \( e^{2y} = 4x^{2} + 3 \).

A straight line of ln y against x through the points (1, 2.5) and (4, 4.3), extended back to its intercept 1.9 on the vertical axis, with the gradient triangle showing a rise of 1.8 over a run of 3. — Worked example 2: gradient \( 0.6 = \ln b \), intercept \( 1.9 = \ln A \), so \( b = e^{0.6} \) and \( A = e^{1.9} \).

✏️Worked example 1 (no calculator)

The points \(A\) and \(B\) have coordinates \( (-1, 5) \) and \( (5, 1) \). (a) Find the equation of the perpendicular bisector of \(AB\). [4] (b) The perpendicular bisector meets the line \( y = x + 2 \) at \(C\). Find the area of triangle \(ABC\). [4]

(a) Midpoint \( M = (2, 3) \). Gradient of \(AB\) \( = \dfrac{1 - 5}{5 - (-1)} = -\tfrac{2}{3} \), so the perpendicular gradient is \( \tfrac{3}{2} \). The bisector is \( y - 3 = \tfrac{3}{2}(x - 2) \), which simplifies to \( y = \tfrac{3}{2}x \) (or \( 3x - 2y = 0 \)).

(b) \( \tfrac{3}{2}x = x + 2 \Rightarrow x = 4 \), so \( C(4, 6) \). Because \(CM\) is perpendicular to \(AB\), it is the height of the triangle on base \(AB\):

\[ AB = \sqrt{6^{2} + 4^{2}} = \sqrt{52} = 2\sqrt{13}, \qquad CM = \sqrt{2^{2} + 3^{2}} = \sqrt{13} \]

\[ \text{Area} = \tfrac{1}{2} \times 2\sqrt{13} \times \sqrt{13} = 13 \]

Check it. \(C\) should be equidistant from \(A\) and \(B\): \( CA^{2} = 25 + 1 = 26 \) and \( CB^{2} = 1 + 25 = 26 \) ✓.

✏️Worked example 2 (calculator)

The variables \(x\) and \(y\) are related by \( y = Ab^{x} \), where \(A\) and \(b\) are constants. The graph of \( \ln y \) against \(x\) is a straight line passing through \( (1, 2.5) \) and \( (4, 4.3) \). (a) Find \(A\) and \(b\). [4] (b) Find \(y\) when \( x = 2.5 \). [2]

(a) \( \ln y = x\ln b + \ln A \). Gradient \( = \dfrac{4.3 - 2.5}{4 - 1} = 0.6 = \ln b \), so \( b = e^{0.6} = 1.82 \). Intercept: \( 2.5 = 0.6(1) + \ln A \), so \( \ln A = 1.9 \) and \( A = e^{1.9} = 6.69 \).

(b) \( \ln y = 1.9 + 0.6(2.5) = 3.4 \), so \( y = e^{3.4} = 30.0 \).

Check it. Using the unrounded constants, \( y = e^{1.9}\left(e^{0.6}\right)^{2.5} = e^{3.4} \) — the same number. Working in \( \ln y \) avoids rounding \(A\) and \(b\) at all.

Reading the intercept off the wrong axis. The intercept is \( \ln A \), not \(A\): answering \( A = 1.9 \) is the classic slip. And if the horizontal axis does not start at zero, the intercept is not where the line meets the drawn axis — calculate it from the gradient.

📝Practise

Written in the style of the current Paper 1 (no calculator) and Paper 2 (calculator) questions.

1. (No calculator.) Find the equation of the line through \( (3, -2) \) that is perpendicular to \( 2x - 5y = 7 \). Give your answer in the form \( ax + by = c \), where \(a\), \(b\) and \(c\) are integers. [3]

The given line has gradient \( \tfrac{2}{5} \), so the perpendicular gradient is \( -\tfrac{5}{2} \). \( y + 2 = -\tfrac{5}{2}(x - 3) \Rightarrow 2y + 4 = -5x + 15 \Rightarrow 5x + 2y = 11 \).

2. (No calculator.) The points \( A(2, 7) \) and \( B(k, 1) \) are such that \( AB = 10 \). Find the possible values of \(k\). [3]

\( (k - 2)^{2} + 6^{2} = 100 \Rightarrow (k - 2)^{2} = 64 \Rightarrow k - 2 = \pm 8 \), so \( k = 10 \) or \( k = -6 \).

3. (Calculator.) The variables satisfy \( y = Ax^{n} \). The graph of \( \lg y \) against \( \lg x \) is a straight line through \( (0.2, 1.1) \) and \( (0.8, 2.9) \). Find \(n\) and \(A\). [4]

\( \lg y = n\lg x + \lg A \). Gradient \( n = \dfrac{2.9 - 1.1}{0.8 - 0.2} = 3 \). Then \( 1.1 = 3(0.2) + \lg A \Rightarrow \lg A = 0.5 \Rightarrow A = 10^{0.5} = 3.16 \).

4. (No calculator.) When \( y^{2} \) is plotted against \( x^{3} \), a straight line is obtained passing through \( (1, 5) \) and \( (3, 9) \). Find \( y^{2} \) in terms of \(x\). [3]

Gradient \( = \dfrac{9 - 5}{3 - 1} = 2 \). \( y^{2} = 2x^{3} + c \) with \( 5 = 2(1) + c \Rightarrow c = 3 \). So \( y^{2} = 2x^{3} + 3 \).

5. (No calculator.) The graph of \( e^{2y} \) against \( x^{2} \) is a straight line with gradient 4, passing through \( (2, 11) \). Find the exact value of \(y\) when \( x = 1 \). [4]

\( e^{2y} = 4x^{2} + c \), and \( 11 = 4(2) + c \Rightarrow c = 3 \). (The point is \( (x^{2}, e^{2y}) = (2, 11) \), not \( (x, y) \).) At \( x = 1 \): \( e^{2y} = 7 \Rightarrow y = \tfrac{1}{2}\ln 7 \).

6. (Calculator.) The point \( P(6, 1) \) and the line \( L: y = 2x - 3 \). (a) Find the coordinates of the foot of the perpendicular from \(P\) to \(L\). (b) Hence find the shortest distance from \(P\) to \(L\). [5]

(a) Perpendicular gradient \( -\tfrac{1}{2} \): \( y - 1 = -\tfrac{1}{2}(x - 6) \Rightarrow y = 4 - \tfrac{1}{2}x \). Solve with \( y = 2x - 3 \): \( 2.5x = 7 \Rightarrow x = 2.8 \), \( y = 2.6 \). Foot \( (2.8, 2.6) \). (b) \( \sqrt{3.2^{2} + 1.6^{2}} = \sqrt{12.8} = 3.58 \).

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Desmos — plot \( \ln y \) against \(x\) for a small data set and fit a line, to see linearisation in action
Cambridge 0606 examiner reports — the intercept-versus-constant confusion in linearisation is a recurring comment