MathsCBSE Class 11 · Straight Lines

Slope, Parallel and Perpendicular Lines and the Angle Between Two Lines (Class 11)

Q: What is the slope of a vertical line?

It is undefined. The inclination is 90°, and tan 90° does not exist.

Q: Can two lines have the same slope and be different?

Yes, parallel distinct lines have equal slopes and different intercepts.

Q: Why is the angle formula tan θ = |(m1 − m2)/(1 + m1 m2)|?

It comes from the tangent of the difference of the two inclinations, with the modulus taken to give the acute angle.

Q: Is collinearity in the 2026-27 syllabus?

The chapter page marks collinearity, family of lines and shifting of origin as not in the syllabus, though equal slopes remain a useful check.

Updated 17 October 2026 · Fact-checked

The slope of the line through (x1, y1) and (x2, y2) is m = (y2 − y1)/(x2 − x1) = tan θ. For ax + by + c = 0, m = −a/b. Lines are parallel if m1 = m2 and perpendicular if m1·m2 = −1. The acute angle θ between lines with slopes m1, m2 satisfies tan θ = |(m1 − m2)/(1 + m1m2)|, with θ = 90° when 1 + m1m2 = 0.

Slope is rise over run, which equals tan θ; parallel lines share the same θ and slope, and perpendicular slopes multiply to −1.

Illustration drawn for QuestionBuddy.

Understand Slope, Parallel and Perpendicular Lines and the Angle Between Lines

Slope. The inclination $θ$ of a line is the angle (between $0^{\circ}$ and $18 0^{\circ}$ ) it makes with the positive $x$ -axis, measured anticlockwise. Its slope is $m = tan θ$ (undefined for a vertical line, $θ = 9 0^{\circ}$ ). Through two points:
$ $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x_{1} \neq = x_{2}) .$ $
A horizontal line has $m = 0$ ; a line leaning right has $m > 0$ ( $θ$ acute) and leaning left has $m < 0$ ( $θ$ obtuse). From the general equation, $a x + b y + c = 0$ has $m = - \frac{a}{b}$ (for $b \neq = 0$ ). For $θ = 12 0^{\circ}$ , $m = tan 12 0^{\circ} = - 3$ .

Parallel and perpendicular lines. For non-vertical lines:

parallel $⟺ m_{1} = m_{2}$ (the same inclination);
perpendicular $⟺ m_{1} m_{2} = - 1$ , i.e. $m_{2} = - \frac{1}{m _{1}}$ .

If one line is vertical, the perpendicular line is horizontal. For $a x + b y + c = 0$ , a parallel line is $a x + b y + c^{'} = 0$ and a perpendicular line is $b x - a y + c^{''} = 0$ .

Angle between two lines. If $θ$ is the acute angle between lines with slopes $m_{1}, m_{2}$ ,
$ $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} .$ $
If $1 + m_{1} m_{2} = 0$ the angle is $9 0^{\circ}$ ; if $m_{1} = m_{2}$ it is $0^{\circ}$ (parallel). The obtuse angle is $18 0^{\circ} - θ$ .

Using slopes in geometry.

*Three points collinear:* slopes of the segments are equal (collinearity is not examined as a topic in 2026-27, but slopes still prove it).
*Parallelogram:* both pairs of opposite sides have equal slopes; check that no side is common.
*Right angle:* adjacent sides with slopes whose product is $- 1$ .
*Rhombus / rectangle / square:* additionally compare side lengths or diagonals with the distance formula.
*Midpoint theorem:* the line joining the midpoints of two sides of a triangle has the same slope as the third side.

Vertical lines. A vertical line $x = k$ has no slope; test it with angles ( $9 0^{\circ}$ ) instead of the formula.

Slope, Parallel and Perpendicular Lines and the Angle Between Lines: formulas to remember

Slope through two points: $m = (y 2 - y 1) / (x 2 - x 1)$; x1 not equal to x2.
Slope from the equation: $a x + b y + c = 0 ha s m = - a / b$; b not zero.
Parallel lines: $m 1 = m 2$; Same inclination.
Perpendicular lines: $m 1 \times m 2 = - 1$; Or one horizontal and one vertical.
Angle between lines: $t an θ = ∣ (m 1 - m 2) / (1 + m 1 m 2) ∣$; Acute angle; 90° if 1 + m1 m2 = 0.

How to solve Slope, Parallel and Perpendicular Lines and the Angle Between Lines questions

Use these steps for slope, parallel, perpendicular and angle questions.

1Find each slope from two points ( $\frac{Δ y}{Δ x}$ ) or from the equation ( $- \frac{a}{b}$ ); treat vertical lines separately.
2Compare: equal slopes mean parallel; product $- 1$ means perpendicular.
3For the angle use $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$ and take $θ = tan^{- 1}$ of the result.
4For geometry proofs, compute the slopes of all sides and, where needed, lengths with the distance formula.
5For unknowns, equate slopes (parallel) or set the product to $- 1$ (perpendicular) and solve.
6State the conclusion in words: "Hence AB is parallel to DC".

Slope, Parallel and Perpendicular Lines and the Angle Between 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.

Slope as the gradient of a road

Suppose a hill road rises 3 m for every 40 m of horizontal distance.

Find the slope of the road and the angle it makes with the horizontal.

Try it first, then show the solutionSolution

Slope $= \frac{3}{40} = 0.075$ .
$tan θ = 0.075$ , so $θ \approx 4.2 9^{\circ}$ .

Answer: Slope 0.075; angle about $4.2 9^{\circ}$ .

Memory hook: Slope is rise over run, and also tan of the angle.

Angle between two roads

Suppose two straight roads on a map have slopes $2$ and $5$ (relative to the east direction).

Find the acute angle at which they cross.

Try it first, then show the solutionSolution

$tan θ = \frac{2 - 5}{1 + 2 \cdot 5} = \frac{3}{11}$ .
So $θ = tan^{- 1} \frac{3}{11} \approx 15. 3^{\circ}$ .

Answer: About $15. 3^{\circ}$

Memory hook: Subtract the slopes on top, add 1 plus the product at the bottom.

Tricks to remember Slope, Parallel and Perpendicular Lines and the Angle Between Lines

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

Rise over run
Slope = change in y over change in x; do both subtractions in the same order.
Use it when: Slope from two points.
Flip and negate
Perpendicular slope = negative reciprocal.
Use it when: Perpendicular lines.
−a/b
For ax + by + c = 0 the slope is minus the x-coefficient over the y-coefficient.
Use it when: Equations.
Top minus, bottom plus
tan θ = |(m1 − m2)/(1 + m1 m2)|.
Use it when: Angle between lines.

Slope, Parallel and Perpendicular Lines and the Angle Between Lines: solved examples

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

Example 1: Slope of a line · 2 marks · Very short answerBoard level

Find the slope of the line through $(- 1, 4)$ and $(3, - 2)$ .

Try it first, then show the solutionSolution

$m = \frac{- 2 - 4}{3 - ( - 1 )}$ . (1 mark)
$= \frac{- 6}{4} = - \frac{3}{2}$ . (1 mark)

Answer: $- \frac{3}{2}$

Example 2: A right angle by slopes · 3 marks · Short answerBoard level

Show that the points $A (1, 2)$ , $B (4, 3)$ and $C (3, 6)$ form a right-angled triangle, and name the right angle.

Try it first, then show the solutionSolution

Slope of $A B = \frac{3 - 2}{4 - 1} = \frac{1}{3}$ . (1 mark)
Slope of $B C = \frac{6 - 3}{3 - 4} = - 3$ . (1 mark)
$\frac{1}{3} \times (- 3) = - 1$ , so $A B ⊥ B C$ and the right angle is at $B$ . (1 mark)

Answer: Right angle at $B$

Example 3: Angle between two lines · 3 marks · Short answerHarder board level

Find the acute angle between the lines $2 x - y + 3 = 0$ and $3 x + y - 7 = 0$ .

Try it first, then show the solutionSolution

$m_{1} = 2$ (from $y = 2 x + 3$ ), $m_{2} = - 3$ (from $y = - 3 x + 7$ ). (1 mark)
$tan θ = \frac{2 - ( - 3 )}{1 + 2 \times ( - 3 )} = \frac{5}{- 5} = 1$ . (1 mark)
$θ = 4 5^{\circ}$ . (1 mark)

Answer: $4 5^{\circ}$

Example 4: Finding an unknown · 3 marks · Short answerHarder board level

The line through $(k, 3)$ and $(5, - 1)$ is parallel to the line joining $(1, - 2)$ and $(4, 4)$ . Find $k$ .

Try it first, then show the solutionSolution

Slope of the second line $= \frac{4 - ( - 2 )}{4 - 1} = 2$ . (1 mark)
Slope of the first line $= \frac{- 1 - 3}{5 - k} = \frac{- 4}{5 - k}$ . (1 mark)
Parallel: $\frac{- 4}{5 - k} = 2$ , so $- 4 = 10 - 2 k$ and $k = 7$ . (1 mark)

Answer: $k = 7$

Example 5: A quadrilateral · 5 marks · Long answerHOTS / competency

Show that $A (1, 1)$ , $B (5, 2)$ , $C (6, 5)$ , $D (2, 4)$ are the vertices of a parallelogram, but that it is neither a rectangle nor a rhombus.

Try it first, then show the solutionSolution

Slopes: $A B = \frac{1}{4}$ , $D C = \frac{5 - 4}{6 - 2} = \frac{1}{4}$ ; $B C = \frac{5 - 2}{6 - 5} = 3$ , $A D = \frac{4 - 1}{2 - 1} = 3$ . (1½ marks)
Opposite sides are parallel in both pairs, so $A B C D$ is a parallelogram. (1 mark)
Rectangle test: slope $A B \times$ slope $B C = \frac{1}{4} \times 3 = \frac{3}{4} \neq = - 1$ , so no right angle. (1 mark)
Rhombus test: $A B^{2} = 16 + 1 = 17$ and $B C^{2} = 1 + 9 = 10$ , unequal, so adjacent sides differ. (1 mark)
Hence it is a parallelogram that is neither a rectangle nor a rhombus. (½ mark)

Answer: Parallelogram only

Where students lose marks in Slope, Parallel and Perpendicular Lines and the Angle Between Lines

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

Mistake: Mixing the order of subtraction
The y-difference and x-difference must be taken in the same order.
Wrong working
$m = \frac{- 2 - 4}{- 1 - 3} = \frac{- 6}{- 4} = \frac{3}{2}$ .
Right working
$\frac{- 2 - 4}{3 - ( - 1 )} = - \frac{3}{2}$ .
Do this: Write the points as $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ before substituting.
Usually costs: The answer.
Mistake: Using the reciprocal without the sign
Perpendicular slopes are negative reciprocals.
Wrong working
Slope $2$ perpendicular to slope $\frac{1}{2}$ .
Right working
Perpendicular to $2$ is $- \frac{1}{2}$ .
Do this: Check that the product is $- 1$ .
Usually costs: The answer.
Mistake: Forgetting the modulus in the angle formula
The formula gives the acute angle.
Wrong working
$tan θ = - 1$ so $θ = 13 5^{\circ}$ .
Right working
$∣ \cdot ∣ = 1$ , so $θ = 4 5^{\circ}$ (acute).
Do this: Write the modulus bars and say "acute angle".
Usually costs: 1 mark.
Mistake: Declaring a parallelogram from one pair of equal slopes
Both pairs of opposite sides are needed.
Wrong working
$A B ∥ D C$ , hence a parallelogram.
Right working
Show $A B ∥ D C$ and $B C ∥ A D$ .
Do this: Compute all four slopes.
Usually costs: The last mark.

Exam tips

Write the two points with labels before computing a slope.
Convert equations to $y = m x + c$ or use $- \frac{a}{b}$ and show it.
State the condition (equal slopes, product $- 1$ ) before substituting.
In geometry proofs compute all required slopes and lengths and finish with a sentence.
For the angle formula write the modulus and name the angle acute.

Practice questions on Slope, Parallel and Perpendicular Lines and the Angle Between 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
The slope of a line parallel to $3 x + 6 y - 5 = 0$ is:
(a) $- \frac{1}{2}$
(b) $\frac{1}{2}$
(c) $- 2$
(d) $2$
Hint
Slope is $- \frac{a}{b}$ .
Answer
(a). $- \frac{3}{6} = - \frac{1}{2}$ .
Q2 · 1 mark · MCQHarder board level
A line makes an angle of $12 0^{\circ}$ with the positive $x$ -axis. Its slope is:
(a) $3$
(b) $- 3$
(c) $\frac{1}{3}$
(d) $- \frac{1}{3}$
Hint
$m = tan θ$ .
Answer
(b). $tan 12 0^{\circ} = - tan 6 0^{\circ} = - 3$ .
Q3 · 3 marks · Short answerHarder board level
Find the acute angle between the lines $y = (2 - 3) x + 5$ and $y = (2 + 3) x - 1$ .
Hint
Note that $m_{1} m_{2} = 4 - 3 = 1$ .
Answer
$tan θ = \frac{( 2 - 3 ) - ( 2 + 3 )}{1 + 1} = \frac{2 3}{2} = 3$ , so $θ = 6 0^{\circ}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The lines $2 x + 3 y = 5$ and $3 x - 2 y = 7$ are perpendicular.
Reason (R): Their slopes are $- \frac{2}{3}$ and $\frac{3}{2}$ , whose product is $- 1$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Convert each equation to slope form.
Answer
(a). $- \frac{2}{3} \times \frac{3}{2} = - 1$ .
Q5 · 4 marks · Short answerHOTS / competency
The vertices of a triangle are $A (1, 2)$ , $B (5, 4)$ and $C (3, 8)$ . Show that the line joining the midpoints of $A B$ and $A C$ is parallel to $B C$ .
Hint
Find the midpoints first.
Answer
Midpoint of $A B$ : $M (3, 3)$ ; of $A C$ : $N (2, 5)$ . Slope $M N = \frac{5 - 3}{2 - 3} = - 2$ . Slope $B C = \frac{8 - 4}{3 - 5} = - 2$ . Equal slopes, so $M N ∥ B C$ .

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 is the slope through (x1, y1) and (x2, y2)?Show answer
(y2 − y1)/(x2 − x1).
2.What is the slope of ax + by + c = 0?Show answer
−a/b.
3.What is the condition for perpendicular lines?Show answer
m1 m2 = −1.
4.What is the slope of a line with inclination 135°?Show answer
−1.
5.What is the acute angle between lines when 1 + m1 m2 = 0?Show answer
90°.

Slope, Parallel and Perpendicular Lines and the Angle Between Lines: frequently asked questions

What is the slope of a vertical line?

It is undefined. The inclination is 90°, and tan 90° does not exist.

Can two lines have the same slope and be different?

Yes, parallel distinct lines have equal slopes and different intercepts.

Why is the angle formula tan θ = |(m1 − m2)/(1 + m1 m2)|?

It comes from the tangent of the difference of the two inclinations, with the modulus taken to give the acute angle.

Is collinearity in the 2026-27 syllabus?

The chapter page marks collinearity, family of lines and shifting of origin as not in the syllabus, though equal slopes remain a useful check.