MathsCBSE Class 11 · Chapter 9

Straight Lines Class 11: Slope, Equations of a Line and Solved Questions

Updated 14 October 2026 · Fact-checked

Straight Lines is Chapter 9 of CBSE Class 11 Maths. For 2026-27 it covers a recall of coordinate geometry, the slope of a line, conditions for parallel and perpendicular lines, the angle between two lines, the point-slope, two-point, slope-intercept and intercept forms with horizontal and vertical lines, and the distance of a point from a line.

What Straight Lines is about

Coordinate geometry turns pictures into algebra. In Class 10 you found distances, mid-points and areas using coordinates. Here the object of study is the straight line.

The key number for a line is its slope $m = tan θ$ , where θ is the angle the line makes with the positive x-axis. From two points, $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ . Slopes tell you at once whether two lines are parallel ( $m_{1} = m_{2}$ ) or perpendicular ( $m_{1} m_{2} = - 1$ ), and give the angle between two lines through $tan θ = \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}}$ .

A line is fixed by a point and a slope, or by two points, or by its slope and y-intercept, or by its two intercepts, giving the point-slope, two-point, slope-intercept and intercept forms. Finally, the perpendicular distance of a point $(x_{1}, y_{1})$ from the line $A x + B y + C = 0$ is $\frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$ . For 2026-27, the normal form and the general equation of a line as topics are assessed only formatively, and collinearity, family of lines and shifting of origin are not in the syllabus.

Weightage: This chapter is part of Unit III, Coordinate Geometry, which the syllabus gives 12 of the 80 theory marks in the school annual exam (CBSE adds that there is no fixed chapter-wise weightage).Source: CBSE Mathematics (041) syllabus 2026-27, Class XI course structure, Unit III Coordinate Geometry

Key concepts

Recall: distance, section formula and area

NCERT section 9.1

From earlier classes, for points $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ :

Distance: $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Section formula: the point dividing PQ internally in the ratio m : n is $(\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n})$ ; the mid-point is $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ .
Area of a triangle with vertices $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , $(x_{3}, y_{3})$ :

$\frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$

If this area is 0, the three points lie on one line. A quadrilateral's area can be found by splitting it into two triangles.

Worked example

Find the point on the x-axis equidistant from (5, 4) and (1, 2), and the area of the triangle with vertices (1, 2), (4, 6), (−2, 3). [3 marks]

Try it first, then show the solutionSolution

Let the point be (x, 0): $(x - 5)^{2} + 16 = (x - 1)^{2} + 4 \Rightarrow - 10 x + 41 = - 2 x + 5 \Rightarrow x = 4.5$
Area $= \frac{1}{2} ∣1 (6 - 3) + 4 (3 - 2) + (- 2) (2 - 6) ∣$
$= \frac{1}{2} ∣3 + 4 + 8∣ = 7.5$ square units

Answer: (4.5, 0); 7.5 square units

Slope of a line

NCERT section 9.2 and 9.2.1

The inclination θ of a line is the angle it makes with the positive direction of the x-axis, measured anticlockwise, with $0° \leq θ < 180°$ . The slope (gradient) is
$m = tan θ, θ \neq = 90°$

The x-axis and lines parallel to it have slope 0. Vertical lines ( $θ = 90°$ ) have undefined slope.
Acute θ gives positive slope (line rises to the right); obtuse θ gives negative slope.

From two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ with $x_{1} \neq = x_{2}$ :
$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
(rise over run). It does not matter which point you call the first.

Worked example

Find the slope of the line through (2, −3) and (6, 5), and the slope of a line making 60° with the positive y-axis, measured anticlockwise. [2 marks]

Try it first, then show the solutionSolution

$m = \frac{5 - ( - 3 )}{6 - 2} = \frac{8}{4} = 2$
60° anticlockwise from the y-axis means 90° + 60° = 150° from the x-axis: $m = tan 150° = - \frac{1}{3}$

Answer: 2; $- \frac{1}{3}$

Parallel and perpendicular lines

NCERT section 9.2.2

For two non-vertical lines with slopes $m_{1}$ and $m_{2}$ :

Parallel exactly when $m_{1} = m_{2}$ (equal inclinations).
Perpendicular exactly when $m_{1} m_{2} = - 1$ , i.e. $m_{2} = - \frac{1}{m _{1}}$ (negative reciprocals).

A horizontal line (slope 0) is perpendicular to every vertical line (undefined slope).

Uses:

A triangle is right-angled if two sides have slopes with product −1.
A quadrilateral is a parallelogram if both pairs of opposite sides have equal slopes.
Three points A, B, C lie on a line if slope AB = slope BC.

Worked example

Without using the Pythagoras theorem, show that A(2, 1), B(4, 5) and C(6, 4) form a right-angled triangle. [2 marks]

Try it first, then show the solutionSolution

Slope AB $= \frac{5 - 1}{4 - 2} = 2$ ; slope BC $= \frac{4 - 5}{6 - 4} = - \frac{1}{2}$
Product $= 2 \times (- \frac{1}{2}) = - 1$ , so AB ⊥ BC: right angle at B

Answer: Right-angled at B.

Angle between two lines

NCERT section 9.2.3

If two lines have slopes $m_{1}$ and $m_{2}$ (with $1 + m_{1} m_{2} \neq = 0$ ), the acute angle θ between them satisfies
$tan θ = \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}}$

Why: if the inclinations are $α_{1}$ and $α_{2}$ , the angle between the lines is $α_{2} - α_{1}$ (or its supplement), and $tan (α_{2} - α_{1}) = \frac{t a n α _{2} - t a n α _{1}}{1 + t a n α _{1} t a n α _{2}}$ . The modulus picks the acute angle; the obtuse angle is $180° - θ$ .

If $1 + m_{1} m_{2} = 0$ , the lines are perpendicular (θ = 90°).
When the angle is given and one slope is unknown, removing the modulus gives two cases, so there are usually two answers.

Worked example

Find the acute angle between the lines with slopes $\frac{1}{2}$ and 3. [2 marks]

Try it first, then show the solutionSolution

$tan θ = \frac{3 - \frac{1}{2}}{1 + \frac{3}{2}} = \frac{5/2}{5/2} = 1$
θ = 45°

Answer: 45°

Horizontal, vertical, point-slope and two-point forms

NCERT section 9.3.1 to 9.3.3

A point (x, y) lies on a line exactly when its coordinates satisfy the line's equation.

Horizontal line at distance a from the x-axis: $y = a$ (or $y = - a$ below). Vertical line: $x = b$ . The x-axis is $y = 0$ and the y-axis is $x = 0$ .
Point-slope form: the line through $(x_{0}, y_{0})$ with slope m is

$y - y_{0} = m (x - x_{0})$
Reason: for any other point (x, y) on it, $\frac{y - y _{0}}{x - x _{0}} = m$ .

Two-point form: through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :

$y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$
(find the slope, then use point-slope).

Worked example

Find the equation of the line (a) through (−2, 3) with slope $\frac{3}{4}$ (b) through (1, −2) and (3, 4). [2 marks]

Try it first, then show the solutionSolution

(a) $y - 3 = \frac{3}{4} (x + 2) \Rightarrow 4 y - 12 = 3 x + 6 \Rightarrow 3 x - 4 y + 18 = 0$
(b) Slope $= \frac{4 + 2}{3 - 1} = 3$ : $y + 2 = 3 (x - 1) \Rightarrow y = 3 x - 5$

Answer: (a) $3 x - 4 y + 18 = 0$ (b) $y = 3 x - 5$

Slope-intercept and intercept forms

NCERT section 9.3.4 and 9.3.5

Slope-intercept form: a line with slope m meeting the y-axis at (0, c) is

$y = m x + c$
c is the y-intercept. If instead it meets the x-axis at (d, 0), it is $y = m (x - d)$ .

Intercept form: a line with x-intercept a and y-intercept b (both non-zero) is

$\frac{x}{a} + \frac{y}{b} = 1$
It passes through (a, 0) and (0, b). Intercepts can be negative.

Converting an equation like $A x + B y + C = 0$ : to get slope and y-intercept, make y the subject: $y = - \frac{A}{B} x - \frac{C}{B}$ . To get intercepts, divide so the right side is 1. (Studying $A x + B y + C = 0$ as a topic, the general equation, is formative-only in 2026-27, but you still use it in calculations.)

Worked example

Find the equation of the line that cuts off equal intercepts on the axes and passes through (3, −1). [2 marks]

Try it first, then show the solutionSolution

Equal intercepts a: $\frac{x}{a} + \frac{y}{a} = 1$ , i.e. $x + y = a$
Through (3, −1): $a = 3 - 1 = 2$ , so $x + y = 2$

Answer: $x + y = 2$

Distance of a point from a line

NCERT section 9.4 and 9.4.1

The perpendicular distance of $P (x_{1}, y_{1})$ from the line $A x + B y + C = 0$ is
$d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$

Proof idea (NCERT): the line meets the axes at $Q (- \frac{C}{A}, 0)$ and $R (0, - \frac{C}{B})$ . Compute the area of triangle PQR in two ways: by the coordinate formula, and as $\frac{1}{2} \times QR \times d$ . Equating gives the formula.

Distance between parallel lines $A x + B y + C_{1} = 0$ and $A x + B y + C_{2} = 0$ :
$d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}}$
Make the x and y coefficients identical first (multiply one equation if needed).

Foot of the perpendicular from $(x_{1}, y_{1})$ : move from P along the normal direction (A, B) by $k = - \frac{A x _{1} + B y _{1} + C}{A ^{2} + B ^{2}}$ : foot $= (x_{1} + k A, y_{1} + k B)$ ; the mirror image is $(x_{1} + 2 k A, y_{1} + 2 k B)$ .

Worked example

Find the distance of (2, −1) from $3 x + 4 y - 7 = 0$ , and the distance between the parallel lines $6 x - 8 y + 5 = 0$ and $3 x - 4 y - 5 = 0$ . [3 marks]

Try it first, then show the solutionSolution

$d = \frac{∣3 ( 2 ) + 4 ( - 1 ) - 7∣}{9 + 16} = \frac{∣ - 5∣}{5} = 1$
Write the second line as $6 x - 8 y - 10 = 0$ so the coefficients match
$d = \frac{∣5 - ( - 10 ) ∣}{36 + 64} = \frac{15}{10} = 1.5$

Answer: 1 unit; 1.5 units

Maths in real life: Straight Lines

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.

Linear model from two data points

Suppose a taxi company charges ₹100 for a 4 km ride and ₹160 for an 8 km ride, with the fare a linear function of distance.

Find the fare F in terms of distance d, and the fare for 12 km.

Try it first, then show the solutionSolution

Two points (4, 100) and (8, 160): slope $= \frac{60}{4} = 15$ (₹ per km)
$F - 100 = 15 (d - 4) \Rightarrow F = 15 d + 40$
$F (12) = 180 + 40 = 220$

Answer: $F = 15 d + 40$ ; ₹220

Memory hook: Slope = rate; intercept = fixed charge.

Distance from a point to a line

Imagine a straight road along the line $4 x - 3 y = 0$ on a town map (units in km), and a house at (5, 0).

How far is the house from the road, measured along the shortest path?

Try it first, then show the solutionSolution

Shortest path is perpendicular: $d = \frac{∣4 ( 5 ) - 3 ( 0 ) ∣}{16 + 9}$
$= \frac{20}{5} = 4$ km

Answer: 4 km

Memory hook: Shortest distance to a line is always perpendicular.

Demand line

Suppose a shop sells 600 kg of rice a week at ₹50 per kg and 500 kg at ₹60 per kg, and demand is linear in price.

How much would it sell at ₹55 per kg?

Try it first, then show the solutionSolution

Slope $= \frac{500 - 600}{60 - 50} = - 10$ kg per rupee
Demand $D = 600 - 10 (p - 50)$ ; at p = 55, $D = 600 - 50 = 550$

Answer: 550 kg

Memory hook: Negative slope: higher price, lower demand.

Topics in this chapter

Straight Lines formula sheet

Slope: $m = tan θ = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Parallel and perpendicular: $parallel: m_{1} = m_{2}, perpendicular: m_{1} m_{2} = - 1$
Angle between two lines: $tan θ = \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}}$
Point-slope and two-point forms: $y - y_{0} = m (x - x_{0}), y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$
Slope-intercept and intercept forms: $y = m x + c, \frac{x}{a} + \frac{y}{b} = 1$
Distance of a point from a line: $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$
Distance between parallel lines: $d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}}$

How the school exam asks Straight Lines

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

Question type	How it is asked	Method
Coordinate recall	Areas of triangles and quadrilaterals, equidistant points, vertices of special triangles (Exercise 9.1).	Use the distance and area formulae; split quadrilaterals into triangles.
Slopes, parallel and perpendicular	Find slopes; show right angles or parallelograms without distance formula; slopes from inclinations.	Compare slopes; product −1 for perpendicular.
Angle between lines	Find the angle; find a slope given the angle and another slope; lines through a point at a given angle.	Use the tan formula; remove the modulus with ±.
Equations of lines	Lines from a point and slope, two points, intercepts, perpendicular bisectors, medians; linear models such as length versus temperature (Exercise 9.2).	Choose the form that matches the given data; simplify to a neat equation.
Distances	Distance of a point from a line, between parallel lines, length of an altitude, foot of perpendicular, points at a given distance (Exercise 9.3).	Write the line as $A x + B y + C = 0$ first; keep the modulus and consider both signs where asked for points.

Tricks to remember Straight 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 ÷ change in x, in the same order for both.
Use it when: Slope from two points.
Flip and change sign
Perpendicular slope: turn the fraction upside down and change its sign: 2/3 → −3/2.
Use it when: Perpendicular lines.
Difference over one-plus-product
tan of the angle = (m₂ − m₁) ÷ (1 + m₁m₂), then take the modulus for the acute angle.
Use it when: Angle between lines.
Plug in, take modulus, divide by root
Distance: put the point into Ax + By + C, take the modulus, divide by $A^{2} + B^{2}$ .
Use it when: Distance formula.
Over a, over b, equals one
Intercepts a and b: $\frac{x}{a} + \frac{y}{b} = 1$ .
Use it when: Intercept form.

Straight Lines: exam-level solved questions

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

Example 1: Perpendicular slope · 1 mark · MCQExam level

A line perpendicular to $2 x - 3 y + 4 = 0$ has slope:
(a) $\frac{2}{3}$
(b) $- \frac{2}{3}$
(c) $\frac{3}{2}$
(d) $- \frac{3}{2}$

Try it first, then show the solutionSolution

Given slope $\frac{2}{3}$ ; the perpendicular slope is $- \frac{3}{2}$ .

Answer: (d)

Example 2: Intercept form · 1 mark · Assertion-reasonExam level

Assertion (A): The line $4 x - 3 y = 12$ has x-intercept 3 and y-intercept −4.
Reason (R): Dividing by 12 gives $\frac{x}{3} + \frac{y}{- 4} = 1$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

R is correct algebra and gives the intercepts directly.

Answer: (a)

Example 3: Concurrent lines · 2 marks · Very short answerHarder exam level

Find p so that the lines $2 x + y - 5 = 0$ , $x - y - 1 = 0$ and $p x + 3 y - 10 = 0$ meet at one point. [2 marks]

Try it first, then show the solutionSolution

Adding the first two: $3 x - 6 = 0$ , x = 2, y = 1
The third must pass through (2, 1): $2 p + 3 - 10 = 0 \Rightarrow p = 3.5$

Answer: $p = \frac{7}{2}$

Example 4: Intercepts with a given sum · 3 marks · Short answerHarder exam level

Find the equations of the lines through (3, 4) whose intercepts on the axes add up to 14. [3 marks]

Try it first, then show the solutionSolution

Let the intercepts be a and 14 − a: $\frac{3}{a} + \frac{4}{14 - a} = 1$
$3 (14 - a) + 4 a = a (14 - a) \Rightarrow a^{2} - 13 a + 42 = 0 \Rightarrow a = 6$ or 7
Lines: $\frac{x}{6} + \frac{y}{8} = 1$ (i.e. $4 x + 3 y = 24$ ) and $\frac{x}{7} + \frac{y}{7} = 1$ (i.e. $x + y = 7$ )

Answer: $4 x + 3 y = 24$ and $x + y = 7$

Example 5: Ratio of division by a line · 3 marks · Short answerHarder exam level

In what ratio does the line $x + 2 y = 3$ divide the segment joining (−1, 0) and (5, 6)? [3 marks]

Try it first, then show the solutionSolution

Let the ratio be k : 1; the point is $(\frac{5 k - 1}{k + 1}, \frac{6 k}{k + 1})$
It lies on the line: $\frac{5 k - 1 + 12 k}{k + 1} = 3 \Rightarrow 17 k - 1 = 3 k + 3 \Rightarrow k = \frac{2}{7}$
Ratio 2 : 7

Answer: 2 : 7 (internally)

Example 6: Unknown coordinate from perpendicularity · 2 marks · Very short answerHarder exam level

The line through (h, 2) and (3, 5) is perpendicular to $x - 3 y + 4 = 0$ . Find h. [2 marks]

Try it first, then show the solutionSolution

Given slope $\frac{1}{3}$ , so the required slope is −3
$\frac{5 - 2}{3 - h} = - 3 \Rightarrow 3 = - 9 + 3 h \Rightarrow h = 4$

Answer: h = 4

Example 7: Distance between parallel lines in slope form · 3 marks · Short answerHOTS / competency

Show that the distance between the parallel lines $y = m x + c_{1}$ and $y = m x + c_{2}$ is $\frac{∣ c _{1} - c _{2} ∣}{1 + m ^{2}}$ , and use it for $y = 2 x + 3$ and $y = 2 x - 7$ . [3 marks]

Try it first, then show the solutionSolution

Write the lines as $m x - y + c_{1} = 0$ and $m x - y + c_{2} = 0$ : the same A = m and B = −1
Distance $= \frac{∣ c _{1} - c _{2} ∣}{m ^{2} + 1}$
Here m = 2, $c_{1} = 3$ , $c_{2} = - 7$ : $\frac{10}{5} = 25$

Answer: Proved; $25$ units

Example 8: Points at a given distance · 3 marks · Short answerHOTS / competency

Find the points on the x-axis whose distance from the line $3 x + 4 y = 12$ is 3 units. [3 marks]

Try it first, then show the solutionSolution

Let the point be (x, 0): $\frac{∣3 x - 12∣}{5} = 3 \Rightarrow ∣3 x - 12∣ = 15$
$3 x - 12 = 15 \Rightarrow x = 9$ ; $3 x - 12 = - 15 \Rightarrow x = - 1$
Two points, one on each side of the line

Answer: (9, 0) and (−1, 0)

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Laying a pipeline

Suppose on a site plan (units in metres) a straight water main runs along $3 x + 4 y - 24 = 0$ . A new house is at H(2, 1) and a tank at T(8, 6).

(i) · 1 mark
Find the shortest distance from the house to the main.
Answer
Answer: 2.8 m
$\frac{∣6 + 4 - 24∣}{5} = \frac{14}{5}$
(ii) · 1 mark
Find the slope of a pipe laid perpendicular to the main.
Answer
Answer: $\frac{4}{3}$
Main slope $- \frac{3}{4}$ ; negative reciprocal.
(iii) · 1 mark
Find the equation of the line HT.
Answer
Answer: $5 x - 6 y - 4 = 0$
Slope $\frac{5}{6}$ : $y - 1 = \frac{5}{6} (x - 2)$
(iv) · 1 mark
Are H and T on the same side of the main?
Answer
Answer: No
$3 x + 4 y - 24$ is −14 at H and +24 at T: opposite signs.

Thermometer calibration

Suppose a student finds that a metal rod is 120.00 cm long at 20 °C and 120.18 cm at 110 °C, and length L is a linear function of temperature C.

(i) · 1 mark
Find the slope of the L-C line.
Answer
Answer: 0.002 cm per °C
$\frac{0.18}{90}$
(ii) · 1 mark
Write L in terms of C.
Answer
Answer: $L = 0.002 C + 119.96$
$L - 120 = 0.002 (C - 20)$
(iii) · 1 mark
Find the length at 0 °C.
Answer
Answer: 119.96 cm
The L-intercept.
(iv) · 1 mark
At what temperature is the rod 120.10 cm?
Answer
Answer: 70 °C
$0.002 C = 0.14$

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.

Find the equation of the perpendicular bisector of the segment joining (1, 3) and (5, 7). [2 marks] (2 marks)

Step	Marks
Mid-point (3, 5); slope of the segment $= \frac{7 - 3}{5 - 1} = 1$ , so the bisector's slope is −1	1
$y - 5 = - 1 (x - 3) \Rightarrow x + y = 8$	1

Find the equations of the lines through (2, 1) that make an angle of 45° with the line $x - 3 y = 4$ . [3 marks] (3 marks)

Step	Marks
Slope of the given line $= \frac{1}{3}$ . For the required slope m: $\frac{m - \frac{1}{3}}{1 + \frac{m}{3}} = 1$ , i.e. $\frac{3 m - 1}{3 + m} = 1$	1
$3 m - 1 = 3 + m \Rightarrow m = 2$ ; $3 m - 1 = - (3 + m) \Rightarrow 4 m = - 2 \Rightarrow m = - \frac{1}{2}$	1
Lines: $y - 1 = 2 (x - 2)$ , i.e. $2 x - y - 3 = 0$ ; and $y - 1 = - \frac{1}{2} (x - 2)$ , i.e. $x + 2 y - 4 = 0$	1

Triangle ABC has vertices A(5, 0), B(0, 0) and C(6, 8). (a) Find the equation of BC. (b) Find the length of the altitude from A. (c) Find the area of the triangle in two ways. (d) Find the equation of the altitude from A. [5 marks] (5 marks)

Step	Marks
(a) Slope BC $= \frac{8}{6} = \frac{4}{3}$ , through the origin: $y = \frac{4}{3} x$ , i.e. $4 x - 3 y = 0$	1
(b) Altitude $= \frac{∣4 ( 5 ) - 3 ( 0 ) ∣}{16 + 9} = \frac{20}{5} = 4$	1
(c) BC $= 36 + 64 = 10$ ; area $= \frac{1}{2} \times 10 \times 4 = 20$	1
Coordinate formula: $\frac{1}{2} ∣5 (0 - 8) + 0 (8 - 0) + 6 (0 - 0) ∣ = \frac{1}{2} \times 40 = 20$ ✓	1
(d) Perpendicular to BC: slope $- \frac{3}{4}$ ; through A: $y = - \frac{3}{4} (x - 5)$ , i.e. $3 x + 4 y - 15 = 0$	1

Where students lose marks in Straight Lines

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

Mistake: Mixing the order in the slope formula
Taking $y_{2} - y_{1}$ but $x_{1} - x_{2}$ .
Wrong working
Slope through (2, −3), (6, 5): $\frac{5 + 3}{2 - 6} = - 2$
Right working
$\frac{5 + 3}{6 - 2} = 2$
Do this: Subtract in the same order top and bottom.
Usually costs: 1 mark and every later step.
Mistake: Using slope for a vertical line
Dividing by zero without noticing.
Do this: If $x_{1} = x_{2}$ , the line is $x = x_{1}$ ; its slope is undefined.
Usually costs: 1 mark.
Mistake: Forgetting the second answer for an angle condition
Dropping the ± after removing the modulus.
Do this: Solve both $\frac{m _{2} - m _{1}}{1 + m _{1} m _{2}} = tan θ$ and $= - tan θ$ .
Usually costs: 1-2 marks.
Mistake: Distance between parallel lines without matching coefficients
Using $6 x - 8 y + 5 = 0$ and $3 x - 4 y - 5 = 0$ directly.
Do this: Multiply so that both have the same A and B, then use $∣ C_{1} - C_{2} ∣$ .
Usually costs: 1-2 marks.
Mistake: Dropping the modulus in the distance formula
Getting a negative distance.
Do this: Distance is never negative; keep |…|.
Usually costs: 1 mark.
Mistake: Wrong angle for a line described from the y-axis
Using the angle with the y-axis as θ.
Do this: θ is always measured from the positive x-axis; convert first.
Usually costs: 1 mark.

NCERT exercise guide: Straight Lines

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 9.1 (recall, slopes and angles)

What it tests: Areas, equidistant points, slopes from points and angles, right triangles and parallelograms by slopes, angle between lines, unknown slopes from a given angle.

How to approach it: Distance and area formulae; slope comparisons; the tan formula with ±.

Same kind as Exercise 9.1

The slope of one line is three times the slope of another, and the tangent of the angle between them is $\frac{1}{2}$ . Find the slopes.

Show the solutionSolution

$\frac{3 m - m}{1 + 3 m ^{2}} = \frac{1}{2} \Rightarrow 4∣ m ∣ = 1 + 3 m^{2} \Rightarrow 3 m^{2} - 4∣ m ∣ + 1 = 0$
$∣ m ∣ = 1$ or $\frac{1}{3}$

Answer: Slopes 1 and 3, or $\frac{1}{3}$ and 1 (or the negatives of these pairs).

Exercise 9.2 (equations of lines)

What it tests: Lines from given conditions, medians, perpendiculars, equal or related intercepts, linear models, lines whose segment between the axes is bisected.

How to approach it: Pick the matching form; for intercept conditions, start from $\frac{x}{a} + \frac{y}{b} = 1$ .

Same kind as Exercise 9.2

The perpendicular from the origin to a line meets it at (2, −4). Find the line.

Show the solutionSolution

Slope of the perpendicular $= \frac{- 4}{2} = - 2$ , so the line's slope is $\frac{1}{2}$
$y + 4 = \frac{1}{2} (x - 2) \Rightarrow x - 2 y = 10$

Answer: $x - 2 y = 10$

Exercise 9.3 and Miscellaneous Exercise (forms and distances)

What it tests: Converting to slope-intercept and intercept forms, distances from points and between parallel lines, perpendicular lines with given intercepts, feet of perpendiculars, altitudes, concurrency, lines at given angles, images in a line.

How to approach it: Rearrange carefully; use the distance formula; solve simultaneous equations for intersections.

Same kind as Exercise 9.3

Find the foot of the perpendicular and the image of (1, 2) in the line $x + y - 7 = 0$ .

Show the solutionSolution

$k = - \frac{1 + 2 - 7}{1 + 1} = 2$ : foot $= (1 + 2, 2 + 2) = (3, 4)$ (check 3 + 4 = 7)
Image $= (1 + 4, 2 + 4) = (5, 6)$

Answer: Foot (3, 4); image (5, 6)

Practice questions on Straight 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 · MCQExam level
The slope of the line through (3, −2) and (−1, 4) is:
(a) $\frac{3}{2}$
(b) $- \frac{3}{2}$
(c) $\frac{2}{3}$
(d) $- \frac{2}{3}$
Hint
Rise over run.
Answer
(b): $\frac{4 + 2}{- 1 - 3} = - \frac{3}{2}$
Q2 · 1 mark · MCQExam level
The distance between the lines $3 x + 4 y = 5$ and $3 x + 4 y = 15$ is:
(a) 10
(b) 2
(c) 4
(d) 1
Hint
$\frac{∣ C _{1} - C _{2} ∣}{5}$ .
Answer
(b) 2
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): The lines $y = 2 x + 3$ and $4 x - 2 y + 7 = 0$ are parallel.
Reason (R): Both have slope 2.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Rearrange the second.
Answer
(a)
Q4 · 2 marks · Very short answerExam level
Find the equation of the line through (0, −3) making an angle of 150° with the positive x-axis.
Hint
m = tan 150°.
Answer
$y = - \frac{1}{3} x - 3$ , i.e. $x + 3 y + 33 = 0$ .
Q5 · 2 marks · Very short answerHarder exam level
Find the equation of the line through (1, 1) perpendicular to $2 x - 3 y + 4 = 0$ .
Hint
Slope $- \frac{3}{2}$ .
Answer
$y - 1 = - \frac{3}{2} (x - 1) \Rightarrow 3 x + 2 y = 5$ .
Q6 · 2 marks · Very short answerHarder exam level
Using slopes only, show that A(1, 1), B(5, 2), C(6, 5) and D(2, 4) form a parallelogram.
Hint
Opposite sides.
Answer
Slopes: AB $= \frac{1}{4}$ , DC $= \frac{5 - 4}{6 - 2} = \frac{1}{4}$ ; BC $= \frac{3}{1} = 3$ , AD $= \frac{3}{1} = 3$ . Both pairs of opposite sides are parallel.
Q7 · 2 marks · Very short answerHarder exam level
Find the acute angle between $y = x$ and $y = 3 x$ .
Hint
Inclinations 45° and 60°.
Answer
15° (from the inclinations, or $tan θ = \frac{3 - 1}{1 + 3} = 2 - 3$ ).
Q8 · 3 marks · Short answerHarder exam level
Find the equation of the median through R of triangle PQR with P(3, −1), Q(−1, 5), R(6, 4).
Hint
Mid-point of PQ.
Answer
Mid-point of PQ is (1, 2). Slope from (1, 2) to (6, 4) is $\frac{2}{5}$ : $y - 2 = \frac{2}{5} (x - 1)$ , i.e. $2 x - 5 y + 8 = 0$ (check R: 12 − 20 + 8 = 0).
Q9 · 3 marks · Short answerHarder exam level
Find the distance of the point (2, 3) from the line $12 (x - 1) = 5 (y + 2)$ .
Hint
Rewrite as Ax + By + C = 0.
Answer
$12 x - 12 = 5 y + 10 \Rightarrow 12 x - 5 y - 22 = 0$ ; $d = \frac{∣24 - 15 - 22∣}{144 + 25} = \frac{13}{13} = 1$ .
Q10 · 2 marks · Very short answerHOTS / competency
A variable point P moves so that the sum of its distances from the lines $x = 0$ and $y = 0$ , in the first quadrant, is always 6. Show that P moves on a straight line and find it.
Hint
Distances are x and y.
Answer
In the first quadrant the distances are x and y, so $x + y = 6$ : a straight line with both intercepts 6.
Q11 · 3 marks · Short answerHOTS / competency
Find the area of the triangle formed by the lines $y = 2 x$ , $y = - 2 x$ and $y = k$ (k > 0).
Hint
Find the vertices.
Answer
Vertices (0, 0), $(\frac{k}{2}, k)$ , $(- \frac{k}{2}, k)$ . Base k (on y = k), height k: area $= \frac{1}{2} \times k \times k = \frac{k ^{2}}{2}$ .

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 of a line inclined at 135°?Show answer
−1.
2.Condition for two lines to be perpendicular?Show answer
$m_{1} m_{2} = - 1$ .
3.Equation of the y-axis?Show answer
x = 0.
4.Slope and y-intercept of $2 x + 3 y = 6$ ?Show answer
$- \frac{2}{3}$ and 2.
5.Intercepts of $\frac{x}{4} - \frac{y}{5} = 1$ ?Show answer
x-intercept 4, y-intercept −5.
6.Distance of the origin from $3 x + 4 y = 10$ ?Show answer
2.
7.Angle between $y = x$ and $y = 3 x$ ?Show answer
15°.
8.Which forms are formative-only in 2026-27?Show answer
The normal form and the general equation as a topic.

Best order to study it

Revise distance, section and area formulae.
Learn inclination and slope, including from two points.
Learn the conditions for parallel and perpendicular lines.
Learn and practise the angle-between-lines formula.
Learn horizontal, vertical, point-slope and two-point forms.
Learn slope-intercept and intercept forms and converting between forms.
Learn the distance of a point from a line and between parallel lines.
Finish with mixed problems: feet of perpendiculars, images, linear models.

Quick revision checklist

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

Can you find the area of a triangle from its vertices?
Can you find a slope from an angle or from two points?
Can you test for parallel and perpendicular lines?
Can you find the angle between two lines?
Can you find a slope when the angle and another slope are given?
Can you write a line through a point with a given slope?
Can you write a line through two points?
Can you write and use the intercept form?
Can you find the distance of a point from a line?
Can you find the distance between two parallel lines?

Straight Lines: frequently asked questions

Which parts of Straight Lines are removed for 2026-27?

Collinearity as a separate section, the normal form, the general equation as a topic, family of lines through an intersection, and shifting of origin were removed from the NCERT book. CBSE lists the normal form and general equation as assessed only formatively. The slope, angle, standard forms and distance formulae remain.

Why is the slope of a vertical line undefined?

Its inclination is 90° and tan 90° is undefined; equivalently, x_2 - x_1 = 0 in the slope formula, and division by zero is not allowed.

How do I choose which form of the equation to use?

Match the given data: a point and slope → point-slope; two points → two-point; slope and y-intercept → slope-intercept; two intercepts → intercept form.

Can I still use equations like 3x + 4y − 7 = 0?

Yes. Lines are routinely written this way, and the distance formula uses it. What is formative-only is the separate study of the general equation and its conversions to the normal form.

How is the Class 11 Maths exam set?

It is a school annual exam of 80 marks (3 hours) plus 20 marks of internal assessment; Unit III, Coordinate Geometry, which includes this chapter, carries 12 marks. CBSE gives no Class 11 sample paper, so schools usually mix MCQs, assertion-reason, short and long answers and case-based questions.

Based on: CBSE Mathematics (041) syllabus 2026-27 (Classes XI-XII); NCERT Mathematics, Textbook for Class XI, Chapter 9 (Reprint 2026-27)