IGCSE · Practice questions · Reducing to straight-line form

Straighten, read, convert.

Six original Cambridge-style questions on reducing relationships to straight-line form: choosing \(X\) and \(Y\), finding \(A\), \(n\) and \(b\) from two points on a transformed line, and turning a given line back into an equation to find unknown values.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Rearrange to \(Y = mX + c\), read \(m\) and \(c\), then undo.

Rearrange: take logs (for \(Ax^{n}\) or \(Ab^{x}\)) or multiply/divide (for sums of terms) until you have \(Y = mX + c\).
Match: \(X\) and \(Y\) are made only from \(x\) and \(y\); the unknown constants are in \(m\) and \(c\).
Gradient and intercept from two points: \(m = \dfrac{Y_2 - Y_1}{X_2 - X_1}\), \(c = Y_1 - mX_1\).
Undo the logs in the same base: \(A = 10^{c}\) or \(A = e^{c}\); keep exact values until the final 3 s.f. answer.
Back to the equation: put the plotted quantities back into \(Y = mX + c\), then substitute.

Recall

[2 marks]

The variables \(x\) and \(y\) are related by \(y = Ae^{kx}\), where \(A\) and \(k\) are constants. State what should be plotted to obtain a straight line, and state the gradient and the intercept of that line in terms of \(A\) and \(k\).

Take ln: \(\ln y = \ln A + \ln e^{kx} = kx + \ln A\). ✓
Plot \(\ln y\) against \(x\): the gradient is \(k\) and the intercept is \(\ln A\). ✓

Calculation

[3 marks]

The variables \(x\) and \(y\) are related by \(y = Ax^{n}\). The graph of \(\lg y\) against \(\lg x\) is a straight line passing through the points \((0.5,\ 2.3)\) and \((2,\ 5.3)\). Find the value of \(n\) and the value of \(A\).

\(\lg y = n\lg x + \lg A\). ✓
\(n = \dfrac{5.3 - 2.3}{2 - 0.5} = \dfrac{3}{1.5} = 2\). ✓
\(\lg A = 2.3 - 2 \times 0.5 = 1.3\), so \(A = 10^{1.3} = 20.0\) (3 s.f.). ✓

Calculation

[5 marks]

The variables \(x\) and \(y\) are related by \(y = Ab^{x}\). The graph of \(\lg y\) against \(x\) is a straight line passing through \((1,\ 1.1)\) and \((4,\ 2.3)\).

(a) Find the value of \(A\) and the value of \(b\).

(b) Find the value of \(y\) when \(x = 3\).

(a) Take lg: \(\lg y = (\lg b)x + \lg A\). ✓
Gradient: \(\lg b = \dfrac{2.3 - 1.1}{4 - 1} = 0.4\), so \(b = 10^{0.4} = 2.51\) (3 s.f.). ✓
Intercept: \(\lg A = 1.1 - 0.4 \times 1 = 0.7\), so \(A = 10^{0.7} = 5.01\) (3 s.f.). ✓
(b) \(\lg y = 0.4 \times 3 + 0.7 = 1.9\), so \(y = 10^{1.9} = 79.4\) (3 s.f.). ✓

Calculation

[4 marks]

When \(e^{y}\) is plotted against \(\sqrt{x}\), a straight line is obtained passing through \((1,\ 5)\) and \((4,\ 11)\).

(a) Express \(e^{y}\) in terms of \(x\).

(b) Find the exact value of \(y\) when \(x = 9\).

(a) Gradient: \(m = \dfrac{11 - 5}{4 - 1} = 2\). Intercept: \(c = 5 - 2 \times 1 = 3\). ✓
So \(e^{y} = 2\sqrt{x} + 3\). ✓
(b) When \(x = 9\): \(e^{y} = 2 \times 3 + 3 = 9\). ✓
\(y = \ln 9 = 2\ln 3\) (\(\approx 2.20\)). ✓

Analysis

[5 marks]

The variables \(x\) and \(y\) are related by \(y = ax^{2} + bx\), where \(a\) and \(b\) are constants.

(a) Show that plotting \(\dfrac{y}{x}\) against \(x\) gives a straight line.

(b) This line passes through \((1,\ 7)\) and \((3,\ 11)\). Find \(a\) and \(b\).

(a) Divide every term by \(x\): \(\dfrac{y}{x} = ax + b\), which is \(Y = mX + c\) with \(Y = \dfrac{y}{x}\), \(X = x\). ✓
(b) Gradient \(a = \dfrac{11 - 7}{3 - 1} = 2\); intercept \(b = 7 - 2 \times 1 = 5\). ✓
(c) \(y = 2x^{2} + 5x = 12\), so \(2x^{2} + 5x - 12 = 0\). ✓
\((2x - 3)(x + 4) = 0\), so \(x = 1.5\) (rejecting \(x = -4\)). ✓

Analysis

[6 marks]

The period \(T\) seconds of a swinging object is thought to be related to a length \(L\) metres by \(T = kL^{n}\), where \(k\) and \(n\) are constants. The graph of \(\lg T\) against \(\lg L\) is a straight line passing through \((-1,\ -0.2)\) and \((0.4,\ 0.5)\).

(a) Find the value of \(n\) and the value of \(k\).

(b) Find \(T\) when \(L = 0.8\).

(a) \(\lg T = n\lg L + \lg k\). Gradient: \(n = \dfrac{0.5 - (-0.2)}{0.4 - (-1)} = \dfrac{0.7}{1.4} = 0.5\). ✓
Intercept: \(\lg k = 0.5 - 0.5 \times 0.4 = 0.3\), so \(k = 10^{0.3} = 2.00\) (3 s.f.). ✓
(b) \(\lg T = 0.5\lg 0.8 + 0.3 = 0.2515\ldots\), so \(T = 1.78\) (3 s.f.). ✓
(c) \(\lg 3 = 0.5\lg L + 0.3\), so \(\lg L = \dfrac{\lg 3 - 0.3}{0.5} = 0.3542\ldots\). ✓
\(L = 10^{0.3542\ldots} = 2.26\) (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.