IGCSE · Practice questions · Gradient of a straight line

Rise, run, repeat.

Six original Cambridge-style questions on the gradient of a straight line, from reading a grid to interpreting a gradient as a speed.

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

Change in \(y\) over change in \(x\).

Grid: draw a right-angled triangle under the line between two exact grid points.
Coordinates (Extended): \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\), same order top and bottom.
Sign: uphill positive, downhill negative.
Special lines: horizontal 0, vertical undefined.
Meaning: the gradient is the change in \(y\) for each 1 unit of \(x\), with units \(y\)-units per \(x\)-unit.

Recall

[2 marks]

State whether the gradient of each line is positive, negative, zero or undefined.

(a) \(y = 5\) (b) \(x = -2\) (c) \(y = 3x\) (d) \(y = 6 - x\)

(a) zero (horizontal) ✓
(b) undefined (vertical) ✓
(c) positive ✓
(d) negative (\(y\) falls as \(x\) increases) ✓

Calculation

[2 marks]

A straight line on a grid passes through the points \((-1, -4)\) and \((3, 8)\). Find its gradient.

Run: from \(-1\) to \(3\) is 4. Rise: from \(-4\) to \(8\) is 12. ✓
Gradient \(= 12 \div 4 = 3\). ✓

Calculation

[2 marks]

(Extended) Calculate the gradient of the line joining \((-4, 6)\) and \((6, 1)\).

\(m = \dfrac{1 - 6}{6 - (-4)} = \dfrac{-5}{10}\) ✓
\(m = -\dfrac{1}{2}\) ✓

Proof

[3 marks]

(Extended) Show that the points \(A(-1, 1)\), \(B(1, 5)\) and \(C(4, 11)\) lie on a straight line.

Gradient \(AB = \dfrac{5 - 1}{1 - (-1)} = \dfrac{4}{2} = 2\). ✓
Gradient \(BC = \dfrac{11 - 5}{4 - 1} = \dfrac{6}{3} = 2\). ✓
The gradients are equal and the two segments share the point \(B\), so \(A\), \(B\) and \(C\) lie on one straight line. ✓

Analysis

[3 marks]

(Extended) The line through \((3, -2)\) and \((t, 10)\) has gradient 4. Find \(t\).

\(\dfrac{10 - (-2)}{t - 3} = 4\), so \(\dfrac{12}{t - 3} = 4\). ✓
\(12 = 4(t - 3) = 4t - 12\), so \(4t = 24\). ✓
\(t = 6\). ✓

Analysis

[4 marks]

A travel graph for a car journey has two straight sections. The first joins \((0, 0)\) to \((2.5, 200)\) and the second joins \((2.5, 200)\) to \((4, 260)\), where \(x\) is the time in hours and \(y\) is the distance in km.

(a) Find the gradient of each section.

(b) What do the gradients represent? In which section was the car faster?

(a) First: \(\dfrac{200 - 0}{2.5 - 0} = 80\). Second: \(\dfrac{260 - 200}{4 - 2.5} = \dfrac{60}{1.5} = 40\). ✓
(b) Each gradient is a speed in km/h (km per hour). ✓
The first section is steeper, so the car was faster there: 80 km/h compared with 40 km/h. ✓

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.