MathsICSE Class 10 · Equation of a Line

Parallel and Perpendicular Lines: Slope Conditions and Equations (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Two non-vertical lines are parallel when their slopes are equal (m₁ = m₂) and perpendicular when the product of their slopes is −1 (m₁m₂ = −1), so the perpendicular slope is the negative reciprocal. To write such a line through a point, find the new slope and use y − y₁ = m(x − x₁).

Parallel lines share one slope (m₁ = m₂); for perpendicular lines flip the fraction and change the sign, so the product is −1.

Illustration drawn for QuestionBuddy.

Understand Parallel and Perpendicular Lines

The CISCE syllabus asks for the conditions for two lines to be parallel or perpendicular, with simple applications.

Parallel lines. Parallel lines have the same inclination, so the same slope:

$m_{1} = m_{2} .$

For $a x + b y + c = 0$ , any parallel line can be written $a x + b y + k = 0$ (same $a$ and $b$ , a different constant). Then substitute the given point to find $k$ .

Perpendicular lines. If neither line is vertical, they are perpendicular exactly when

$m_{1} m_{2} = - 1$ , that is $m_{2} = - \frac{1}{m _{1}}$ .

Flip the fraction and change the sign: slope $\frac{3}{5}$ becomes $- \frac{5}{3}$ ; slope $- 2$ becomes $\frac{1}{2}$ . A line perpendicular to $a x + b y + c = 0$ can be written $b x - a y + k = 0$ .

Special cases. Horizontal lines ( $m = 0$ ) are perpendicular to vertical lines (slope not defined); the rule $m_{1} m_{2} = - 1$ cannot be used for them, but $y = k$ and $x = h$ are always perpendicular.

Typical questions.

1. **Find $k$ ** so that two lines are parallel or perpendicular. Find each slope in terms of $k$ and apply the condition. In 2024 Q10(ii)(a) the slopes were $k$ and 3, so $3 k = - 1$ and $k = - \frac{1}{3}$ ; the examiners reported candidates who equated the slopes or set the product to $+ 1$ .
2. Equation through a point, parallel or perpendicular to a given line (2024 Q10(ii)(b), 2025 Q7(ii)).
3. Proving shapes with slopes: parallel opposite sides give a parallelogram; a product of $- 1$ gives a right angle (rectangle, right-angled triangle).
4. Foot of a perpendicular: write the perpendicular line, then solve the two equations simultaneously.

Through the origin. A line through $(0, 0)$ has no constant term. 2025 Q1(iv) asked for the line through the origin parallel to $3 x + 4 y + 7 = 0$ : the answer is $3 x + 4 y = 0$ .

Parallel and Perpendicular Lines: formulas to remember

Parallel lines: $m_{1} = m_{2}$; Same slope.
Perpendicular lines: $m_{1} m_{2} = - 1 ⟺ m_{2} = - \frac{1}{m _{1}}$; Neither line vertical.
Parallel to ax + by + c = 0: $a x + b y + k = 0$; Find $k$ from the given point.
Perpendicular to ax + by + c = 0: $b x - a y + k = 0$; Swap $a$ and $b$ , change one sign.
Axes-parallel lines: $y = k ⊥ x = h$; Horizontal and vertical lines.

How to solve Parallel and Perpendicular Lines questions

Get the slopes first, apply the condition, then write the equation.

1Write each line in the form $y = m x + c$ (or use $m = - \frac{a}{b}$ ) to read its slope.
2Parallel: new slope = old slope. Perpendicular: new slope = $- \frac{1}{old slope}$ .
3For an unknown $k$ : set $m_{1} = m_{2}$ or $m_{1} m_{2} = - 1$ and solve.
4For an equation: use $y - y_{1} = m (x - x_{1})$ with the given point.
5Give the answer as $a x + b y + c = 0$ with integer coefficients.
6Check: substitute the point, and confirm the slope condition.

Parallel and Perpendicular Lines 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.

Parallel lines share a slope

Suppose two straight rail tracks on a site plan lie along $y = 2 x + 1$ and $y = 2 x - 4$ , and a third track must run parallel to them through the depot at $(1, 1)$ .

Find the equation of the third track.

Try it first, then show the solutionSolution

The existing tracks have slope 2, so the new one does too.
$y - 1 = 2 (x - 1) \Rightarrow y = 2 x - 1$ , that is $2 x - y - 1 = 0$ .

Answer: $2 x - y - 1 = 0$

Memory hook: Parallel = same steepness, different starting height.

Shortest path is perpendicular

Imagine a straight wall along $3 x + 4 y = 25$ on a plan (units in metres) and a pipe to be laid from the origin to the wall by the shortest route.

Find the pipe's line, where it meets the wall and its length.

Try it first, then show the solutionSolution

Shortest route is perpendicular: wall slope $- \frac{3}{4}$ , pipe slope $\frac{4}{3}$ , through the origin: $4 x - 3 y = 0$ .
Solve with $3 x + 4 y = 25$ : $y = \frac{4 x}{3}$ , so $3 x + \frac{16 x}{3} = 25$ , $x = 3$ , $y = 4$ .
Length $= 3^{2} + 4^{2} = 5$ m.

Answer: Pipe along $4 x - 3 y = 0$ , meeting the wall at $(3, 4)$ ; 5 m long.

Memory hook: The shortest path to a line meets it at a right angle.

Tricks to remember Parallel and Perpendicular Lines

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

Flip and switch
Perpendicular slope: flip the fraction, switch the sign. $\frac{2}{3} \to - \frac{3}{2}$ .
Use it when: Every perpendicular question.
Product −1, not +1
Perpendicular slopes have opposite signs, so their product is negative.
Use it when: Avoiding the error reported in 2024 Q10(ii).
Keep a and b for parallel
Parallel to $3 x + 4 y + 7 = 0$ means $3 x + 4 y + k = 0$ ; only the constant changes.
Use it when: Quick parallel equations.

Parallel and Perpendicular Lines: solved examples

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

Example 1: Perpendicular slope · 1 mark · MCQBoard level

The slope of a line perpendicular to $y = 4 x - 1$ is:
(a) 4
(b) −4
(c) $\frac{1}{4}$
(d) $- \frac{1}{4}$

Try it first, then show the solutionSolution

Given slope 4; perpendicular slope $= - \frac{1}{4}$ (check: $4 \times - \frac{1}{4} = - 1$ ).

Answer: (d) $- \frac{1}{4}$

Example 2: Parallel line through a point · 3 marks · Short answerBoard level

Find the equation of the line through $(3, - 2)$ parallel to $2 x - 5 y + 7 = 0$ .

Try it first, then show the solutionSolution

Parallel line: $2 x - 5 y + k = 0$ .
Through $(3, - 2)$ : $6 + 10 + k = 0$ , so $k = - 16$ .
Equation: $2 x - 5 y - 16 = 0$ .

Answer: $2 x - 5 y - 16 = 0$

Example 3: Finding k for parallel and perpendicular · 3 marks · Short answerHarder board level

Find $k$ if the lines $k x + 2 y - 5 = 0$ and $3 x - y + 4 = 0$ are (a) parallel (b) perpendicular.

Try it first, then show the solutionSolution

Slopes: $m_{1} = - \frac{k}{2}$ and $m_{2} = 3$ .
(a) $- \frac{k}{2} = 3 \Rightarrow k = - 6$ .
(b) $- \frac{k}{2} \times 3 = - 1 \Rightarrow k = \frac{2}{3}$ .

Answer: (a) $k = - 6$ (b) $k = \frac{2}{3}$

Example 4: Perpendicular to a segment · 4 marks · Short answerHarder board level

Find the equation of the line through $(- 1, 4)$ perpendicular to the line joining $A (2, 3)$ and $B (5, - 3)$ .

Try it first, then show the solutionSolution

$m_{A B} = \frac{- 3 - 3}{5 - 2} = - 2$ ; perpendicular slope $\frac{1}{2}$ .
$y - 4 = \frac{1}{2} (x + 1) \Rightarrow 2 y - 8 = x + 1$ .
$x - 2 y + 9 = 0$ . Check $(- 1, 4)$ : $- 1 - 8 + 9 = 0$ ✓.

Answer: $x - 2 y + 9 = 0$

Example 5: A right-angled isosceles triangle · 4 marks · Short answerHOTS / competency

Show that the triangle with vertices $A (1, 2)$ , $B (4, 6)$ and $C (8, 3)$ is right-angled and isosceles, and find its area.

Try it first, then show the solutionSolution

$m_{A B} = \frac{4}{3}$ , $m_{B C} = \frac{3 - 6}{8 - 4} = - \frac{3}{4}$ . Product $= - 1$ , so $∠ B = 90°$ .
$A B = 9 + 16 = 5$ and $B C = 16 + 9 = 5$ : isosceles.
Area $= \frac{1}{2} \times 5 \times 5 = 12.5$ square units.

Answer: Right angle at $B$ ; $A B = B C = 5$ ; area 12.5 square units.

Example 6: Case study: a road, a footpath and a cycle lane · 4 marks · Case-basedHOTS / competency

On a town plan, a straight road meets the x-axis at $P (- 6, 0)$ and the y-axis at $Q (0, 4)$ .
(a) Find the slope of the road.
(b) A footpath from the origin meets the road at right angles. Find its equation.
(c) A cycle lane through $(0, - 2)$ runs parallel to the road. Find its equation.

Try it first, then show the solutionSolution

(a) $m = \frac{4 - 0}{0 + 6} = \frac{2}{3}$ .
(b) Slope $- \frac{3}{2}$ , through the origin: $y = - \frac{3}{2} x$ , that is $3 x + 2 y = 0$ .
(c) Slope $\frac{2}{3}$ , y-intercept $- 2$ : $y = \frac{2}{3} x - 2$ , that is $2 x - 3 y - 6 = 0$ .

Answer: (a) $\frac{2}{3}$ (b) $3 x + 2 y = 0$ (c) $2 x - 3 y - 6 = 0$

Where students lose marks in Parallel and Perpendicular Lines

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

Mistake: Product of slopes set to +1, or slopes equated
Perpendicular means $m_{1} m_{2} = - 1$ .
Wrong working
Slopes $k$ and 3 perpendicular: $3 k = 1$ , $k = \frac{1}{3}$ (or $k = 3$ ).
Right working
$3 k = - 1$ , $k = - \frac{1}{3}$ .
Do this: Write the condition $m_{1} m_{2} = - 1$ before substituting.
Usually costs: Reported in 2024 Q10(ii)(a).
Mistake: Using the perpendicular condition for a parallel line
Parallel lines keep the same slope.
Wrong working
Through the origin parallel to $3 x + 4 y + 7 = 0$ : $4 x - 3 y = 0$ .
Right working
$3 x + 4 y = 0$ .
Do this: Underline 'parallel' or 'perpendicular' in the question.
Usually costs: Reported in 2025 Q1(iv).
Mistake: Using a point that is not on the required line
The equation must pass through the stated point.
Wrong working
Line $R S$ through the origin written using point $P$ .
Right working
Use $(0, 0)$ : $5 x + 3 y = 0$ .
Do this: Circle the point the new line passes through.
Usually costs: Reported in 2025 Q7(ii).

Exam tips

2024 Q10(ii) (3 marks): $k = - \frac{1}{3}$ , then the parallel line through $(- 1, 1)$ : $3 x - y + 4 = 0$ .
2025 Q7(ii) (3 marks): slope of $P Q$ was $\frac{3}{5}$ , so the perpendicular through the origin was $5 x + 3 y = 0$ .
Always convert to $y = m x + c$ (or use $- \frac{a}{b}$ ) before comparing slopes.
State the condition used in words: 'since the lines are perpendicular, $m_{1} m_{2} = - 1$ '.

Practice questions on Parallel and Perpendicular Lines

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
Which line is parallel to $y = - 2 x + 5$ ?
(a) $y = 2 x + 1$
(b) $4 x + 2 y - 3 = 0$
(c) $x - 2 y = 0$
(d) $2 x - y = 4$
Hint
Find each slope.
Answer
(b): its slope is $- \frac{4}{2} = - 2$ .
Q2 · 2 marks · Short answerBoard level
Find the slope of a line perpendicular to $3 x - 6 y + 1 = 0$ .
Hint
The given slope is $\frac{1}{2}$ .
Answer
$- 2$
Q3 · 3 marks · Short answerHarder board level
Find $a$ if the lines $a x + 3 y = 4$ and $6 x - 2 y = 5$ are perpendicular.
Hint
Slopes $- \frac{a}{3}$ and 3.
Answer
$- \frac{a}{3} \times 3 = - 1$ , so $a = 1$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The line through $(1, 1)$ perpendicular to $y = x$ is $x + y = 2$ .
Reason (R): If a line has slope $m$ , a line perpendicular to it has slope $- \frac{1}{m}$ .
(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
Slope of $x + y = 2$ is $- 1$ .
Answer
(c)
Q5 · 2 marks · Short answerHarder board level
Write the equations of the lines through $(2, 5)$ that are (a) parallel to the x-axis (b) perpendicular to the x-axis.
Hint
Horizontal and vertical lines.
Answer
(a) $y = 5$ (b) $x = 2$
Q6 · 4 marks · Short answerHOTS / competency
Find the foot of the perpendicular from $(5, 5)$ to the line $x + y = 4$ , and the length of the perpendicular.
Hint
The perpendicular has slope 1 and passes through $(5, 5)$ : $y = x$ .
Answer
Solving $y = x$ and $x + y = 4$ : foot $(2, 2)$ ; length $9 + 9 = 32$ .
Q7 · 4 marks · Short answerHarder board level
The line through $A (1, k)$ and $B (3, 7)$ is parallel to the line through $C (0, 2)$ and $D (2, 6)$ . Find $k$ , and the equation of the line through $B$ perpendicular to $A B$ .
Hint
$m_{C D} = 2$ .
Answer
$\frac{7 - k}{2} = 2$ , so $k = 3$ . Perpendicular slope $- \frac{1}{2}$ : $x + 2 y - 17 = 0$ .

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.Condition for parallel lines?Show answer
$m_{1} = m_{2}$
2.Condition for perpendicular lines?Show answer
$m_{1} m_{2} = - 1$
3.Perpendicular slope to $- \frac{2}{5}$ ?Show answer
$\frac{5}{2}$
4.Line through the origin parallel to $x - 2 y + 3 = 0$ ?Show answer
$x - 2 y = 0$
5.Are $x = 3$ and $y = - 1$ perpendicular?Show answer
Yes: one is vertical, the other horizontal.

Parallel and Perpendicular Lines: frequently asked questions

What is the condition for two lines to be perpendicular?

The product of their slopes is −1: m₁m₂ = −1. So the slope of a perpendicular line is the negative reciprocal, −1/m. A horizontal and a vertical line are also perpendicular.

How do I write a line parallel to ax + by + c = 0 through a point?

Write ax + by + k = 0 with the same a and b, substitute the point to find k, and give the final equation.

Why can the rule m₁m₂ = −1 not be used for vertical lines?

A vertical line has no defined slope, so there is nothing to multiply. Lines x = h and y = k are perpendicular by inspection.