MathsICSE Class 10 · Equation of a Line

Slope (Gradient) of a Line: Formula, tan θ and Collinearity (ICSE Class 10)

Q: What is the slope of a line?

The slope m measures steepness: rise divided by run between two points, m = (y₂ − y₁)/(x₂ − x₁). It also equals tan θ, where θ is the angle the line makes with the positive x-axis.

Q: How do I find the slope from an equation?

Rearrange to y = mx + c and read the coefficient of x, or use m = −a/b for ax + by + c = 0.

Q: How do I show three points are collinear?

Show that the slope from the first point to the second equals the slope from the second to the third. Equal slopes through a common point mean one straight line.

Updated 15 October 2026 · Fact-checked

The slope of the line through (x₁, y₁) and (x₂, y₂) is m = (y₂ − y₁)/(x₂ − x₁), the rise divided by the run. It equals tan θ, where θ is the angle the line makes with the positive x-axis. For ax + by + c = 0 the slope is −a/b.

Rise over run between any two points on the line gives m, and it equals tan θ, where θ is the angle with the positive x-axis.

Illustration drawn for QuestionBuddy.

Understand Slope of a Line

The slope (or gradient) $m$ of a line measures its steepness: how much $y$ rises for each 1 unit increase in $x$ .

From two points. For $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ with $x_{1} \neq = x_{2}$ :

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{rise}{run} .$

Keep the same order on top and bottom; swapping both gives the same answer, swapping only one changes the sign.

**Slope as $tan θ$ .** The CISCE syllabus asks for the geometric meaning of $m$ as $tan θ$ , where $θ$ is the angle the line makes with the positive direction of the x-axis (the inclination). So a line inclined at $45°$ has slope $tan 45° = 1$ ; at $60°$ , slope $3$ ; at $30°$ , slope $\frac{1}{3}$ .

Signs and special lines.

$m > 0$ : the line rises from left to right.
$m < 0$ : it falls from left to right.
$m = 0$ : horizontal (parallel to the x-axis), $θ = 0°$ .
Vertical lines (parallel to the y-axis) have $x_{2} - x_{1} = 0$ , so the slope is not defined ( $θ = 90°$ ).

From an equation. Rearrange $a x + b y + c = 0$ as $y = - \frac{a}{b} x - \frac{c}{b}$ ; the slope is the coefficient of $x$ : $m = - \frac{a}{b}$ . For $y = m x + c$ it is read directly.

Collinear points. $A$ , $B$ , $C$ lie on one straight line exactly when slope $A B$ = slope $B C$ (they share point $B$ , so equal slopes mean the same line). In 2024 Q1(vi) this condition gave $y$ in terms of $x$ ; the examiners reported confusion about the condition and errors simplifying.

Reading slopes from a graph. Pick two clear points on the line, count the rise and the run with signs, and divide. The 2025 report on Q3(iii) noted '0 ÷ 4 = 4'; a horizontal segment has slope $\frac{0}{4} = 0$ .

Slope of a Line: formulas to remember

Slope from two points: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$; $x_{1} \neq = x_{2}$ .
Slope and inclination: $m = tan θ$; $θ$ measured from the positive x-axis.
Slope from ax + by + c = 0: $m = - \frac{a}{b}$; $b \neq = 0$ .
Special lines: $horizontal: m = 0; vertical: m not defined$; $y = k$ and $x = k$ .
Collinearity: $m_{A B} = m_{B C} \Rightarrow A, B, C collinear$; Common point $B$ .

How to solve Slope of a Line questions

Label the points, subtract in the same order, simplify.

1Label the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ .
2Work out $y_{2} - y_{1}$ and $x_{2} - x_{1}$ , with brackets around negatives.
3Divide and simplify; give fractions in lowest terms.
4For an equation, rearrange to $y = m x + c$ or use $m = - \frac{a}{b}$ .
5For inclination, match $m$ with $tan 30° = \frac{1}{3}$ , $tan 45° = 1$ , $tan 60° = 3$ .
6For collinearity, show two slopes through a common point are equal and say so.

Slope of a Line in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Slope as steepness

Suppose the edge of a roof in a side-view drawing rises from $(0, 3)$ to $(4, 6)$ , in metres.

Find the slope of the roof edge and say what it means.

Try it first, then show the solutionSolution

$m = \frac{6 - 3}{4 - 0} = \frac{3}{4}$ .
For every 4 m across, the roof rises 3 m; that is 0.75 m up per metre across.

Answer: $\frac{3}{4}$ : it rises 0.75 m for each metre across.

Memory hook: Slope = rise per one step across.

Negative slope as a rate of decrease

Imagine a water tank whose level is 120 cm at time 0 and 90 cm after 15 minutes, falling steadily. Plot time (minutes) on the x-axis and level (cm) on the y-axis.

Find the slope of the line and interpret it.

Try it first, then show the solutionSolution

$m = \frac{90 - 120}{15 - 0} = \frac{- 30}{15} = - 2$ .
The negative sign shows the level falls: 2 cm every minute.

Answer: $- 2$ : the level drops 2 cm per minute.

Memory hook: Downhill left to right = negative slope.

Tricks to remember Slope of a Line

Short memory aids. Say each one aloud once, then use it on a question the same day.

Rise over run
Up-down change on top, left-right change below: $\frac{Δ y}{Δ x}$ .
Use it when: Getting the formula the right way up.
Minus a over b
For $a x + b y + c = 0$ : slope $= - \frac{a}{b}$ (the x-number over the y-number, with a minus).
Use it when: Slopes straight from general equations.
Flat is zero, upright is undefined
A flat road has no climb (0); a wall cannot be walked up (undefined).
Use it when: Horizontal and vertical lines.

Slope of a Line: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Slope from two points · 1 mark · MCQBoard level

The slope of the line through $(- 2, 3)$ and $(4, - 6)$ is:
(a) $\frac{3}{2}$
(b) $- \frac{3}{2}$
(c) $- \frac{2}{3}$
(d) $\frac{1}{2}$

Try it first, then show the solutionSolution

$m = \frac{- 6 - 3}{4 - ( - 2 )} = \frac{- 9}{6} = - \frac{3}{2}$ .

Answer: (b) $- \frac{3}{2}$

Example 2: Slope and inclination · 3 marks · Short answerBoard level

Find the slope of the line through $(1, 2)$ and $(4, 2 + 33)$ , and hence its inclination.

Try it first, then show the solutionSolution

$m = \frac{( 2 + 3 3 ) - 2}{4 - 1} = \frac{3 3}{3} = 3$ .
$tan θ = 3$ , so $θ = 60°$ .

Answer: $m = 3$ ; inclination $60°$ .

Example 3: An unknown coordinate from a slope · 3 marks · Short answerHarder board level

The slope of the line through $A (3, k)$ and $B (k, 5)$ is $- 2$ . Find $k$ .

Try it first, then show the solutionSolution

$\frac{5 - k}{k - 3} = - 2 \Rightarrow 5 - k = - 2 k + 6 \Rightarrow k = 1$ .
Check: $A (3, 1)$ , $B (1, 5)$ : $\frac{5 - 1}{1 - 3} = - 2$ ✓.

Answer: $k = 1$

Example 4: Collinear points and the line through them · 4 marks · Short answerHarder board level

(a) Show that $A (- 1, - 4)$ , $B (1, 2)$ and $C (3, 8)$ are collinear.
(b) A point $P (x, y)$ lies on the same line. Express $y$ in terms of $x$ .

Try it first, then show the solutionSolution

(a) $m_{A B} = \frac{2 + 4}{1 + 1} = 3$ , $m_{B C} = \frac{8 - 2}{3 - 1} = 3$ . Equal slopes with common point $B$ , so collinear.
(b) $m_{B P} = 3$ : $\frac{y - 2}{x - 1} = 3 \Rightarrow y - 2 = 3 x - 3 \Rightarrow y = 3 x - 1$ .

Answer: (a) Both slopes are 3 (b) $y = 3 x - 1$

Example 5: Finding a point for collinearity · 3 marks · Short answerHOTS / competency

Find $a$ if $A (a, 3 a)$ , $B (- 2, 6)$ and $C (3, 1)$ are collinear.

Try it first, then show the solutionSolution

$m_{B C} = \frac{1 - 6}{3 + 2} = - 1$ .
$m_{A B} = \frac{6 - 3 a}{- 2 - a} = - 1 \Rightarrow 6 - 3 a = 2 + a \Rightarrow a = 1$ .
Check: $A (1, 3)$ ; $m_{A C} = \frac{1 - 3}{3 - 1} = - 1$ ✓.

Answer: $a = 1$

Example 6: Case study: a hill road · 4 marks · Case-basedHOTS / competency

In a side view of a hill road (in metres), three marker posts stand at $P (0, 20)$ , $Q (200, 50)$ and $R (500, 95)$ .
(a) Find the slopes of $P Q$ and $QR$ .
(b) Is the road a straight incline from $P$ to $R$ ? Explain.
(c) How high is the road at a horizontal distance of 400 m?

Try it first, then show the solutionSolution

(a) $m_{P Q} = \frac{30}{200} = 0.15$ ; $m_{QR} = \frac{45}{300} = 0.15$ .
(b) Equal slopes through the common point $Q$ : $P$ , $Q$ , $R$ are collinear, so the road is a straight incline.
(c) Height $= 20 + 0.15 \times 400 = 80$ m.

Answer: (a) Both 0.15 (b) Yes, the posts are collinear (c) 80 m

Where students lose marks in Slope of a Line

The slips examiners see most often, and what to write instead.

Mistake: Mixing the order of subtraction
Top and bottom must use the same order of points.
Wrong working
$m = \frac{- 6 - 3}{- 2 - 4} = \frac{3}{2}$
Right working
$m = \frac{- 6 - 3}{4 - ( - 2 )} = - \frac{3}{2}$
Do this: Write '(second − first) over (second − first)'.
Usually costs: The sign, and every later answer.
Mistake: Treating 0/4 as 4, or leaving it as 0/4
Zero divided by a non-zero number is 0.
Wrong working
Slope of $C B = \frac{0}{4} = 4$ .
Right working
$\frac{0}{4} = 0$ , so $C B$ is horizontal.
Do this: Simplify every slope and state what it means.
Usually costs: Reported in 2025 Q3(iii).
Mistake: Not using the collinearity condition correctly
Collinear means equal slopes through a common point.
Wrong working
Setting the product of slopes equal to $- 1$ .
Right working
$m_{A B} = m_{B C}$ .
Do this: Remember: same line, same slope.
Usually costs: Reported in 2024 Q1(vi).

Exam tips

2024 Q1(vi) (MCQ) used equal slopes for collinear points to get $y$ in terms of $x$ ; simplify carefully before choosing an option.
In 2025 Q3(iii) the slopes of $C B$ and $D A$ were both 0, which proved them parallel; state the conclusion in words.
Read points from a graph carefully: a point on the y-axis is $(0, y)$ , not $(y, 0)$ .
Leave slopes as exact fractions or surds, like $- \frac{3}{2}$ or $3$ .

Practice questions on Slope of a Line

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The slope of a line parallel to the x-axis is:
(a) 1
(b) 0
(c) −1
(d) not defined
Hint
No rise.
Answer
(b) 0
Q2 · 2 marks · Short answerBoard level
Find the slope of the line through $(- 3, 5)$ and $(1, - 3)$ .
Hint
Rise $- 8$ , run 4.
Answer
$- 2$
Q3 · 2 marks · Short answerHarder board level
Find the slope and the y-intercept of $3 x + 4 y - 12 = 0$ .
Hint
$y = - \frac{3}{4} x + 3$ .
Answer
Slope $- \frac{3}{4}$ ; y-intercept 3.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The slope of the line joining $(2, 3)$ and $(2, 7)$ is 0.
Reason (R): The slope of a vertical line is not defined.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Both points have $x = 2$ .
Answer
(b). The line is vertical, so its slope is not defined, not 0.
Q5 · 2 marks · Short answerHarder board level
Find the inclination of the line through $(0, 0)$ and $(3, 1)$ .
Hint
$tan θ = \frac{1}{3}$ .
Answer
$30°$
Q6 · 3 marks · Short answerHarder board level
The points $A (x, y)$ , $B (1, 4)$ and $C (3, 0)$ are collinear. Express $y$ in terms of $x$ .
Hint
$m_{A B} = m_{B C} = - 2$ .
Answer
$\frac{y - 4}{x - 1} = - 2$ , so $y = 6 - 2 x$ .
Q7 · 4 marks · Short answerHOTS / competency
Using slopes, show that $A (0, 0)$ , $B (4, 1)$ , $C (5, 5)$ and $D (1, 4)$ are the vertices of a parallelogram.
Hint
Compare $A B$ with $D C$ and $B C$ with $A D$ .
Answer
$m_{A B} = m_{D C} = \frac{1}{4}$ and $m_{B C} = m_{A D} = 4$ : both pairs of opposite sides are parallel, so $A B C D$ is a parallelogram.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the slope formula.Show answer
$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
2.Slope of a line at $45°$ to the x-axis?Show answer
1
3.Slope of $2 x - 5 y + 1 = 0$ ?Show answer
$\frac{2}{5}$
4.Slope of the line through $(1, 1)$ and $(3, 5)$ ?Show answer
2
5.What is the slope of the y-axis?Show answer
Not defined (it is vertical).

Slope of a Line: frequently asked questions

What is the slope of a line?

The slope m measures steepness: rise divided by run between two points, m = (y₂ − y₁)/(x₂ − x₁). It also equals tan θ, where θ is the angle the line makes with the positive x-axis.

How do I find the slope from an equation?

Rearrange to y = mx + c and read the coefficient of x, or use m = −a/b for ax + by + c = 0.

How do I show three points are collinear?

Show that the slope from the first point to the second equals the slope from the second to the third. Equal slopes through a common point mean one straight line.