Equation of a Line ICSE Class 10: Slope, Forms and Parallel and Perpendicular Lines

What Equation of a Line is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Equation of a Line is in the 2. Algebra (vii) Co-ordinate Geometry (b)(ii) Equation of a line part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

Slope (gradient) of a line

If a line makes an angle $\theta$ with the positive direction of the x-axis, its slope is $m=\tan \theta$. Through two points $(x_1,y_1)$ and $(x_2,y_2)$:
$m=\frac{y_2-y_1}{x_2-x_1}.$

• Rising from left to right: $m>0$. Falling: $m<0$.
• A horizontal line (parallel to the x-axis): $m=0$.
• A vertical line (parallel to the y-axis): slope not defined.
• Standard angles: $\theta =30^{\circ },45^{\circ },60^{\circ }$ give $m=\frac{1}{\sqrt{3}},1,\sqrt{3}$.

Keep the same order in the top and bottom: $(y_2-y_1)$ over $(x_2-x_1)$.

Worked example

Find the slope of the line through $(2,-1)$ and $(5,8)$. Find the inclination of a line whose slope is $\sqrt{3}$.

Try it first, then show the solutionSolution

1. $m=\frac{8-(-1)}{5-2}=\frac{9}{3}=3$.
2. $\tan \theta =\sqrt{3}$, so $\theta =60^{\circ }$.

Answer: Slope 3; inclination 60°

Slope-intercept form y = mx + c

$y=mx+c$ is a line with slope $m$ and y-intercept $c$: it crosses the y-axis at $(0,c)$, the point on the line where $x=0$.

To read $m$ and $c$ from an equation like $ax+by+k=0$, make $y$ the subject: $y=-\frac{a}{b}x-\frac{k}{b}$. So $m=-\frac{a}{b}$ and $c=-\frac{k}{b}$.

Worked example

Find the slope and y-intercept of $3x-2y+6=0$.

Try it first, then show the solutionSolution

1. $2y=3x+6$, so $y=\frac{3}{2}x+3$.
2. $m=\frac{3}{2}$, $c=3$: the line meets the y-axis at $(0,3)$.

Answer: m = 3/2, c = 3

Equation through a point with a given slope, or through two points

Through $(x_1,y_1)$ with slope $m$:
$y-y_1=m(x-x_1).$
Through two points: first find $m$ from the two points, then use either point in the same form.

Always finish by clearing fractions and collecting terms, e.g. $x+2y-4=0$. The examiners expect whole-number coefficients.

Worked example

Find the equation of the line through $(1,-2)$ and $(4,7)$.

Try it first, then show the solutionSolution

1. $m=\frac{7-(-2)}{4-1}=3$.
2. $y+2=3(x-1)$.
3. $y+2=3x-3$, so $3x-y-5=0$.
4. Check with $(4,7)$: $12-7-5=0$ ✓

Answer: 3x − y − 5 = 0

Parallel and perpendicular lines

For two lines with slopes $m_1$ and $m_2$ (neither vertical):

• Parallel: $m_1=m_2$.
• Perpendicular: $m_1{m}_2=-1$, i.e. $m_2=-\frac{1}{m_1}$ (turn the fraction upside down and change its sign).

A line parallel to $ax+by+k=0$ has the form $ax+by+k^{\prime }=0$; if it passes through the origin, $k^{\prime }=0$.
A line parallel to the x-axis is $y=b$; parallel to the y-axis is $x=a$. These two kinds are perpendicular to each other even though one slope is not defined.

Worked example

Find the equation of the line through $(2,-3)$ perpendicular to $2x-3y+5=0$.

Try it first, then show the solutionSolution

1. Given line: $y=\frac{2}{3}x+\frac{5}{3}$, slope $\frac{2}{3}$.
2. Perpendicular slope $=-\frac{3}{2}$.
3. $y+3=-\frac{3}{2}(x-2)$; $2y+6=-3x+6$.
4. $3x+2y=0$: no constant term, so the line passes through the origin.

Answer: 3x + 2y = 0

Collinear points

Three points $A$, $B$, $C$ are collinear (on one line) if the slope of $AB$ equals the slope of $BC$. Equal slopes and a common point $B$ mean one line.

Use this to test points, or to find an unknown coordinate that makes three points collinear.

Worked example

Show that $(1,-1)$, $(3,3)$ and $(-2,-7)$ are collinear.

Try it first, then show the solutionSolution

1. Slope from $(1,-1)$ to $(3,3)$: $\frac{4}{2}=2$.
2. Slope from $(3,3)$ to $(-2,-7)$: $\frac{-10}{-5}=2$.
3. Equal slopes with a common point, so the points are collinear.

Answer: Collinear (both slopes are 2)

Points on the axes and intercepts

Where a line meets the x-axis, $y=0$; where it meets the y-axis, $x=0$. Write these points as $(a,0)$ and $(0,b)$; examiners report candidates swapping them.

A line through the origin has no constant term: $y=mx$, or $ax+by=0$.

Worked example

Find where $2x+3y=12$ meets the axes.

Try it first, then show the solutionSolution

1. x-axis: $y=0$, $2x=12$, $x=6$: $(6,0)$.
2. y-axis: $x=0$, $3y=12$, $y=4$: $(0,4)$.

Answer: (6, 0) and (0, 4)

Applications: perpendicular bisector, median, altitude

• Perpendicular bisector of $AB$: through the mid-point of $AB$, with slope $-\frac{1}{m_{AB}}$.
• Median from $A$ in triangle $ABC$: through $A$ and the mid-point of $BC$.
• Altitude from $A$: through $A$, perpendicular to $BC$.

Each needs one point and one slope, then $y-y_1=m(x-x_1)$.

Worked example

Find the perpendicular bisector of the join of $A(1,2)$ and $B(5,-2)$.

Try it first, then show the solutionSolution

1. Mid-point: $(3,0)$.
2. $m_{AB}=\frac{-4}{4}=-1$, so the perpendicular slope is $1$.
3. $y-0=1(x-3)$: $x-y-3=0$.

Answer: x − y − 3 = 0

Maths in real life: Equation of a Line

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.

Topics in this chapter

• Slope of a Line
• Equation of a Line in Different Forms
• Parallel and Perpendicular Lines

Equation of a Line formula sheet

Slope through two points

$m=\frac{y_2-y_1}{x_2-x_1}$

Same order on top and bottom.

Slope and inclination

$m=\tan \theta$

θ measured from the positive x-axis.

Slope-intercept form

$y=mx+c$

c is the y-intercept: the line passes through (0, c).

Point-slope form

$y-y_1=m(x-x_1)$

For two points, find m first.

Slope of ax + by + k = 0

$m=-\frac{a}{b}$

y-intercept is −k/b.

Parallel lines

$m_1=m_2$

Collinear points: slope AB = slope BC.

Perpendicular lines

$m_1{m}_2=-1$

Not for a vertical line; x = a and y = b are always perpendicular.

How the board asks Equation of a Line

The question patterns to expect, and the method that works for each.

Tricks to remember Equation of a Line

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

• Rise over run

  Slope = rise ÷ run = change in y ÷ change in x. y goes upstairs.

  Use it when: Never putting x on top.

• Flip and switch

  Perpendicular slope: flip the fraction, switch the sign. 2/3 becomes −3/2.

  Use it when: Perpendicular-line questions.

• Minus a over b

  For ax + by + k = 0, the slope is −a/b ('minus x-coefficient over y-coefficient').

  Use it when: Reading slopes quickly.

• No constant, through the origin

  A line through the origin has no number on its own: ax + by = 0.

  Use it when: MCQs about lines through O.

Equation of a Line: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Equation of a Line

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

• ✗Mistake: Wrong condition for perpendicular lines

  CISCE examiners' comments, 2024: for perpendicular lines, candidates equated the slopes or set their product to 1, getting k = 3 or k = 1/3.

  Wrong working

  Perpendicular: $m_1=m_2$, or $m_1{m}_2=1$

  Right working

  Perpendicular: $m_1{m}_2=-1$

  ✓Do this: Parallel: equal. Perpendicular: product −1 ('flip and switch').

  Usually costs: Most of a 3-mark part.

• ✗Mistake: Leaving fractions in the final equation

  CISCE examiners' comments, 2024 and 2025: equations were left with fractional coefficients, for example y = −5x/3 instead of 5x + 3y = 0.

  ✓Do this: Multiply through by the denominator and write ax + by + c = 0 with whole numbers.

  Usually costs: 1 mark.

• ✗Mistake: Swapping the coordinates of axis points

  CISCE examiners' comments, 2025: candidates wrote B as (−2, 0) instead of (0, −2), and wrote the intercept points of a line with x and y swapped.

  ✓Do this: On the x-axis the point is (a, 0); on the y-axis it is (0, b).

  Usually costs: The coordinates mark and the slope that depends on it.

• ✗Mistake: Sign errors in the slope

  CISCE examiners' comments, 2025: the slope of a diagonal was found as −5/7 instead of 5/7, and some candidates left a slope as 0/4 or wrote 0 ÷ 4 = 4, then did not state that the lines were parallel.

  Wrong working

  $m=\frac{0}{4}=4$

  Right working

  $m=\frac{0}{4}=0$: both sides are horizontal, so they are parallel

  ✓Do this: Keep the order the same top and bottom, and check against the graph: a line rising to the right has a positive slope. Always write the conclusion.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Forgetting that a line through the origin has no constant

  CISCE examiners' comments, 2025: in the MCQ on a line through the origin parallel to a given line, candidates kept a constant term, or used the perpendicular condition instead of the parallel one.

  ✓Do this: Parallel to ax + by + k = 0 and through O: ax + by = 0.

  Usually costs: 1 mark.

• ✗Mistake: Not using equal slopes for collinear points

  CISCE examiners' comments, 2024: candidates were confused about the collinearity condition, or made errors simplifying the slope equation.

  ✓Do this: Collinear means slope AB = slope BC. Cross-multiply carefully.

  Usually costs: 1 mark in an MCQ.

Practice questions on Equation 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.

1. Q1 · 1 mark · MCQBoard level

   A line has slope $\sqrt{3}$. Its inclination to the positive x-axis is:
   (a) $30^{\circ }$
   (b) $45^{\circ }$
   (c) $60^{\circ }$
   (d) $90^{\circ }$

   Hint

   tan θ = √3.

   Answer

   (c)

2. Q2 · 1 mark · MCQBoard level

   The y-intercept of the line $2x+5y=10$ is:
   (a) $5$
   (b) $2$
   (c) $-2$
   (d) $10$

   Hint

   Put x = 0.

   Answer

   (b)

3. Q3 · 1 mark · MCQHarder board level

   The line through the origin perpendicular to $2x-5y+7=0$ is:
   (a) $2x-5y=0$
   (b) $5x+2y=0$
   (c) $5x-2y=0$
   (d) $5x+2y+7=0$

   Hint

   Slope of the given line is 2/5.

   Answer

   (b): slope $-\frac{5}{2}$ and no constant term.

4. Q4 · 1 mark · Assertion-reasonHarder board level

   Assertion (A): The lines $x=3$ and $y=-2$ are perpendicular.
   Reason (R): Two lines are perpendicular only when the product of their slopes is $-1$.
   (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

   What is the slope of x = 3?

   Answer

   (a): a vertical and a horizontal line are perpendicular, but the vertical one has no slope, so 'only when' is false.

5. Q5 · 4 marks · Short answerHarder board level

   $A(2,-3)$ and $B(a,5)$ are such that $AB$ is parallel to $4x-y+2=0$. Find $a$ and the equation of $AB$.

   Hint

   Slope of AB = 4.

   Answer

   $\frac{5+3}{a-2}=4$, so $a=4$. $y+3=4(x-2)$: $4x-y-11=0$.

6. Q6 · 3 marks · Short answerBoard level

   Find the equation of the line with slope $-\frac{2}{3}$ and y-intercept 4.

   Hint

   y = mx + c, then clear fractions.

   Answer

   $y=-\frac{2}{3}x+4$, i.e. $2x+3y-12=0$.

7. Q7 · 3 marks · Long answerBoard level

   Find the slope and the y-intercept of $5x-2y-8=0$, and the point where it meets the y-axis.

   Hint

   Make y the subject.

   Answer

   $y=\frac{5}{2}x-4$: slope $\frac{5}{2}$, y-intercept $-4$, point $(0,-4)$.

8. Q8 · 4 marks · Long answerHarder board level

   Show that $(2,1)$, $(-1,-5)$ and $(4,5)$ are collinear, and find the equation of the line through them.

   Hint

   Two slopes, then point-slope form.

   Answer

   Slopes $\frac{-6}{-3}=2$ and $\frac{4}{2}=2$: equal, so collinear. $y-1=2(x-2)$: $2x-y-3=0$.

9. Q9 · 4 marks · Long answerHOTS / competency

   Triangle $ABC$ has $A(3,5)$, $B(-1,1)$ and $C(5,-1)$. Find the equations of the altitude and the median from $B$. Explain why they are different lines, using the lengths $BA$ and $BC$.

   Hint

   The altitude and median from B coincide only if BA = BC.

   Answer

   $m_{AC}=-3$, so the altitude has slope $\frac{1}{3}$: $x-3y+4=0$. Mid-point of $AC$ is $(4,2)$; median slope $\frac{1}{5}$: $x-5y+6=0$. They differ because $BA=\sqrt{32}$ and $BC=\sqrt{40}$ are unequal; the median from $B$ is perpendicular to $AC$ only when $BA=BC$.

10. Q10 · 4 marks · Long answerHarder board level

    The line $2x-3y=6$ meets the x-axis at $A$ and the y-axis at $B$. Find the equation of the perpendicular bisector of $AB$ and the point where it meets the y-axis.

    Hint

    A(3, 0), B(0, −2).

    Answer

    Mid-point $(1.5,-1)$; $m_{AB}=\frac{2}{3}$; perpendicular slope $-\frac{3}{2}$. $y+1=-\frac{3}{2}(x-1.5)$ gives $6x+4y-5=0$. It meets the y-axis at $\left(0,\frac{5}{4}\right)$.

11. Q11 · 1 mark · MCQHOTS / competency

    If $(p,0)$, $(0,q)$ and $(1,1)$ are collinear, where $p$ and $q$ are non-zero, then:
    (a) $p+q=1$
    (b) $\frac{1}{p}+\frac{1}{q}=1$
    (c) $pq=1$
    (d) $p=q$

    Hint

    Equate two slopes, then divide by pq.

    Answer

    (b): $-\frac{q}{p}=1-q$ gives $pq=p+q$; dividing by $pq$ gives $\frac{1}{q}+\frac{1}{p}=1$.

12. Q12 · 5 marks · Long answerHOTS / competency

    $A(1,1)$, $B(4,2)$, $C(3,5)$ and $D(0,4)$ are the vertices of a quadrilateral. (a) Using slopes, show that $ABCD$ is a rectangle. (b) Show that it is in fact a square. (c) Show that its diagonals are perpendicular and find the equation of $BD$.

    Hint

    Four slopes, two lengths, two diagonal slopes.

    Answer

    (a) $m_{AB}=m_{DC}=\frac{1}{3}$, $m_{BC}=m_{AD}=-3$; opposite sides parallel and $\frac{1}{3}\times (-3)=-1$, so all angles are right angles: a rectangle. (b) $AB=BC=\sqrt{10}$: a square. (c) $m_{AC}=2$, $m_{BD}=-\frac{1}{2}$, product $-1$. $BD$: $y-2=-\frac{1}{2}(x-4)$, i.e. $x+2y-8=0$.

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. 1.Slope through (x₁, y₁) and (x₂, y₂)?Show answer

   (y₂ − y₁) ÷ (x₂ − x₁)

2. 2.What is c in y = mx + c?Show answer

   The y-intercept: the line passes through (0, c).

3. 3.Slope of ax + by + k = 0?Show answer

   −a/b

4. 4.Condition for parallel lines?Show answer

   Equal slopes.

5. 5.Condition for perpendicular lines?Show answer

   Product of slopes = −1.

6. 6.Slope of a line inclined at 60°?Show answer

   √3

7. 7.Slope of a horizontal line?Show answer

   0

8. 8.Test for three collinear points?Show answer

   Slope AB = slope BC.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you find a slope without mixing the order?
• ☐Can you read m and c from 3x − 2y + 6 = 0?
• ☐Can you write the perpendicular slope at once?
• ☐Do you clear fractions in every final equation?
• ☐Do you write axis points as (a, 0) and (0, b)?
• ☐Can you find the median, altitude and perpendicular bisector of a triangle?

Equation of a Line: frequently asked questions