MathsICSE Class 10 · Equation of a Line

Equation of a Line in Different Forms: Slope-Intercept, Point-Slope and Two-Point (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

A straight line can be written as y = mx + c (slope m, y-intercept c), as y − y₁ = m(x − x₁) through a known point with slope m, or through two points by first finding m. ICSE answers are usually given in the form ax + by + c = 0 with integer coefficients.

Slope m is rise divided by run between any two points, and c is where the line cuts the y-axis; finish with integer coefficients in ax + by + c = 0.

Illustration drawn for QuestionBuddy.

Understand Equation of a Line in Different Forms

Every non-vertical straight line has an equation that every point on it satisfies, and no other point does.

Slope-intercept form. $y = m x + c$ . Here $m$ is the slope and $c$ is the y-intercept: the line meets the y-axis at $(0, c)$ , the point on the line where $x = 0$ . The CISCE syllabus asks for this geometric meaning of $c$ as well as of $m$ .

Point-slope form. Through $(x_{1}, y_{1})$ with slope $m$ :

$y - y_{1} = m (x - x_{1}) .$

Two-point form. Through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ : find $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ , then use the point-slope form with either point.

General form. Any line can be written $a x + b y + c = 0$ . Its slope is $- \frac{a}{b}$ and its y-intercept is $- \frac{c}{b}$ . Examiners expect final answers with integer coefficients: $y = - \frac{5}{3} x$ should become $5 x + 3 y = 0$ (the 2025 Q7(ii) report lists answers left with fractions).

Special lines.

Horizontal through $(h, k)$ : $y = k$ . Vertical: $x = h$ (no slope-intercept form).
Through the origin: $c = 0$ , so $y = m x$ or $a x + b y = 0$ with no constant term. The 2025 Q1(iv) report notes candidates who missed this.

Intercepts. To find where a line meets the axes, put $y = 0$ (x-axis) and $x = 0$ (y-axis). If a line meets the x-axis at $(p, 0)$ and the y-axis at $(0, q)$ , its slope is $\frac{q - 0}{0 - p} = - \frac{q}{p}$ .

Checking. Substitute each given point into your final equation; both sides must agree. This catches sign slips at once.

Lines in context. A straight-line relationship $y = m x + c$ has a fixed part $c$ (the value when $x = 0$ ) and a rate $m$ (the change in $y$ for each unit of $x$ ).

Equation of a Line in Different Forms: formulas to remember

Slope-intercept form: $y = m x + c$; $c$ is the y-intercept: the line passes through $(0, c)$ .
Point-slope form: $y - y_{1} = m (x - x_{1})$; Through $(x_{1}, y_{1})$ with slope $m$ .
Two-point form: $y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$; Find $m$ first.
General form: $a x + b y + c = 0 : m = - \frac{a}{b}, y-intercept = - \frac{c}{b}$; Give answers with integer coefficients.
Line through the origin: $y = m x (a x + b y = 0)$; No constant term.

How to solve Equation of a Line in Different Forms questions

Find the slope, choose a point, write the form, then tidy to integer coefficients.

1Decide what you know: slope and intercept, slope and a point, or two points.
2If needed, find $m$ from two points.
3Write $y - y_{1} = m (x - x_{1})$ (or $y = m x + c$ ).
4Clear fractions by multiplying through by the denominator.
5Rearrange to $a x + b y + c = 0$ with $a$ positive and integer coefficients.
6Check by substituting the given point(s).

Equation of a Line in Different Forms 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.

y = mx + c as fixed charge plus rate

Suppose a mobile plan costs ₹199 a month plus ₹2 for each extra GB used.

Write the monthly bill $y$ for $x$ extra GB as a line, and find the bill for 10 extra GB.

Try it first, then show the solutionSolution

Fixed part $c = 199$ , rate $m = 2$ : $y = 2 x + 199$ .
$x = 10$ : $y = 20 + 199 = 219$ .

Answer: $y = 2 x + 199$ ; ₹219.

Memory hook: c is where you start; m is how fast you climb.

Equation from two points

Imagine a tank being filled at a steady rate: the water is 20 cm deep at the start and 45 cm deep after 5 minutes.

Find the depth $h$ after $t$ minutes, and when it reaches 100 cm.

Try it first, then show the solutionSolution

Points $(0, 20)$ and $(5, 45)$ : $m = \frac{25}{5} = 5$ ; $c = 20$ . So $h = 5 t + 20$ .
$5 t + 20 = 100 \Rightarrow t = 16$ .

Answer: $h = 5 t + 20$ ; after 16 minutes.

Memory hook: Two readings fix a straight-line rule.

Tricks to remember Equation of a Line in Different Forms

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

Point-slope: subtract the point
In $y - y_{1} = m (x - x_{1})$ , the point's coordinates appear with minus signs; a point $(- 2, 5)$ gives $y - 5 = m (x + 2)$ .
Use it when: Avoiding sign errors.
c lives on the y-axis
Put $x = 0$ : whatever is left is $c$ .
Use it when: Reading and finding the y-intercept.
No fractions in the final answer
Multiply by the denominator, then move everything to one side.
Use it when: Presenting the answer as examiners expect.

Equation of a Line in Different Forms: solved examples

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

Example 1: Slope-intercept form · 1 mark · MCQBoard level

The equation of the line with slope 2 and y-intercept $- 3$ is:
(a) $y = - 3 x + 2$
(b) $y = 2 x + 3$
(c) $y = 2 x - 3$
(d) $2 y = x - 3$

Try it first, then show the solutionSolution

$y = m x + c$ with $m = 2$ , $c = - 3$ .

Answer: (c) $y = 2 x - 3$

Example 2: Point and slope · 3 marks · Short answerBoard level

Find the equation of the line through $(2, - 1)$ with slope $- \frac{3}{2}$ . Give it with integer coefficients.

Try it first, then show the solutionSolution

$y - (- 1) = - \frac{3}{2} (x - 2)$ .
Multiply by 2: $2 y + 2 = - 3 x + 6$ .
$3 x + 2 y - 4 = 0$ . Check $(2, - 1)$ : $6 - 2 - 4 = 0$ ✓.

Answer: $3 x + 2 y - 4 = 0$

Example 3: Through two points · 3 marks · Short answerHarder board level

Find the equation of the line through $(- 2, 5)$ and $(4, - 1)$ .

Try it first, then show the solutionSolution

$m = \frac{- 1 - 5}{4 + 2} = - 1$ .
$y - 5 = - 1 (x + 2) \Rightarrow y - 5 = - x - 2$ .
$x + y - 3 = 0$ . Check $(4, - 1)$ : $4 - 1 - 3 = 0$ ✓.

Answer: $x + y - 3 = 0$

Example 4: Reading a general equation · 4 marks · Short answerHarder board level

For the line $3 x - 4 y + 12 = 0$ , find (a) its slope (b) its y-intercept (c) where it meets the x-axis (d) the point on it with x-coordinate 4.

Try it first, then show the solutionSolution

Rearrange: $4 y = 3 x + 12$ , so $y = \frac{3}{4} x + 3$ .
(a) $m = \frac{3}{4}$ (b) $c = 3$ .
(c) $y = 0$ : $3 x + 12 = 0$ , $x = - 4$ , the point $(- 4, 0)$ .
(d) $x = 4$ : $12 - 4 y + 12 = 0$ , $y = 6$ , the point $(4, 6)$ .

Answer: (a) $\frac{3}{4}$ (b) 3 (c) $(- 4, 0)$ (d) $(4, 6)$

Example 5: Line, intercept and section ratio together · 4 marks · Short answerHOTS / competency

$A$ is $(- 2, 1)$ and $B$ is $(4, 7)$ .
(a) Find the equation of $A B$ .
(b) Find the point $P$ where $A B$ meets the y-axis.
(c) Find the ratio $A P : P B$ .

Try it first, then show the solutionSolution

(a) $m = \frac{7 - 1}{4 + 2} = 1$ ; $y - 1 = x + 2$ , so $x - y + 3 = 0$ .
(b) $x = 0$ : $y = 3$ . $P (0, 3)$ .
(c) Using $k : 1$ on the x-coordinate: $\frac{4 k - 2}{k + 1} = 0 \Rightarrow k = \frac{1}{2}$ . Ratio $1 : 2$ .

Answer: (a) $x - y + 3 = 0$ (b) $(0, 3)$ (c) $1 : 2$

Example 6: Case study: a burning candle · 4 marks · Case-basedHOTS / competency

A candle 30 cm tall burns down steadily by 2.5 cm each hour. Let $h$ cm be its height after $t$ hours.
(a) Write $h$ in the form $h = m t + c$ and explain what $m$ and $c$ mean.
(b) Find the height after 4 hours.
(c) After how long does the candle burn out?

Try it first, then show the solutionSolution

(a) $h = - 2.5 t + 30$ . $c = 30$ is the starting height (the h-intercept); $m = - 2.5$ is the change in height per hour, negative because it shrinks.
(b) $h = - 10 + 30 = 20$ cm.
(c) $h = 0$ : $2.5 t = 30$ , $t = 12$ hours.

Answer: (a) $h = - 2.5 t + 30$ (b) 20 cm (c) 12 hours

Where students lose marks in Equation of a Line in Different Forms

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

Mistake: Leaving fractional coefficients
Examiners expect $a x + b y + c = 0$ with integers.
Wrong working
$y = - \frac{5}{3} x$
Right working
$5 x + 3 y = 0$
Do this: Multiply through by the denominator as a final step.
Usually costs: Reported in 2024 Q10(ii) and 2025 Q7(ii).
Mistake: Sign error with a negative point
$x - x_{1}$ with $x_{1} = - 2$ is $x + 2$ .
Wrong working
Through $(- 2, 5)$ : $y - 5 = m (x - 2)$ .
Right working
$y - 5 = m (x + 2)$ .
Do this: Write $x - (- 2)$ first, then simplify.
Usually costs: The equation, and any later part using it.
Mistake: Adding a constant to a line through the origin
A line through $(0, 0)$ has no constant term.
Wrong working
Line through the origin: $3 x + 4 y + 5 = 0$ .
Right working
$3 x + 4 y = 0$ .
Do this: Check $(0, 0)$ in your answer.
Usually costs: Reported in 2025 Q1(iv).

Exam tips

Final answers like $3 x - y + 4 = 0$ (2024 Q10(ii)) and $5 x + 3 y = 0$ (2025 Q7(ii)) were expected with integer coefficients.
In graph questions, read intercepts carefully: a point on the x-axis is $(p, 0)$ ; the 2025 report lists swapped coordinates.
Use whichever given point has simpler numbers in $y - y_{1} = m (x - x_{1})$ .
Substitute the given point(s) into your final equation to check it.

Practice questions on Equation of a Line in Different Forms

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 y-intercept of $4 x - 3 y = 12$ is:
(a) 4
(b) −4
(c) 3
(d) −3
Hint
Put $x = 0$ .
Answer
(b) −4
Q2 · 2 marks · Short answerBoard level
Find the equation of the line with slope $- \frac{1}{2}$ through $(4, - 3)$ .
Hint
Multiply by 2 to clear the fraction.
Answer
$x + 2 y + 2 = 0$
Q3 · 3 marks · Short answerHarder board level
Find the equation of the line through $(1, - 2)$ and $(- 3, 6)$ . Does it pass through the origin?
Hint
$m = - 2$ .
Answer
$2 x + y = 0$ ; yes, there is no constant term.
Q4 · 2 marks · Short answerHarder board level
The point $(2, - 1)$ lies on $y = 3 x + c$ . Find $c$ and the x-intercept of the line.
Hint
Substitute, then put $y = 0$ .
Answer
$c = - 7$ ; x-intercept $\frac{7}{3}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The line $2 x + 3 y = 0$ passes through the origin.
Reason (R): A line whose equation has no constant term passes through the origin.
(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
Substitute $(0, 0)$ .
Answer
(c)
Q6 · 3 marks · Short answerHarder board level
A line meets the x-axis at $(- 4, 0)$ and the y-axis at $(0, 6)$ . Find its equation.
Hint
$m = \frac{6 - 0}{0 + 4}$ ; $c = 6$ .
Answer
$y = \frac{3}{2} x + 6$ , that is $3 x - 2 y + 12 = 0$ .
Q7 · 4 marks · Short answerHOTS / competency
A line passes through $(2, 3)$ and its y-intercept is twice its slope. Find its equation.
Hint
$y = m x + 2 m$ ; substitute $(2, 3)$ .
Answer
$3 = 4 m$ , $m = \frac{3}{4}$ , $c = \frac{3}{2}$ : $3 x - 4 y + 6 = 0$ .

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.What does $c$ mean in $y = m x + c$ ?Show answer
The y-intercept: the line meets the y-axis at $(0, c)$ .
2.Write the point-slope form.Show answer
$y - y_{1} = m (x - x_{1})$
3.Equation of the horizontal line through $(3, - 2)$ ?Show answer
$y = - 2$
4.Equation of the line through the origin with slope 4?Show answer
$y = 4 x$
5.Where does $x + 2 y = 6$ meet the y-axis?Show answer
$(0, 3)$

Equation of a Line in Different Forms: frequently asked questions

What are the forms of the equation of a line in ICSE Class 10?

The CISCE syllabus names the slope-intercept form y = mx + c and the two-point form, which uses (y − y₁) = m(x − x₁) after finding m from the two points.

What is the y-intercept of a line?

It is c in y = mx + c: the y-coordinate of the point where the line crosses the y-axis, that is, the point on the line where x = 0.

Why should I write the final equation with integer coefficients?

It is the standard form examiners expect, such as 3x + 2y − 4 = 0 rather than y = −3x/2 + 2. CISCE reports list fractional coefficients as a reason for lost marks.