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Class 11 · Chapter 9 · Coordinate Geometry unit (12 of 80 marks)

Straight Lines Class 11: notes and important questions

Revision notes, 17 board-style questions with the step marks shown, and a four-step route from the basics to 95+, each step ending in a short checkpoint.

  • 17 questions
  • 5 multiple choice, 1 assertion–reason, 3 very short answer, 4 short answer, 2 long answer, 2 case study
  • About 11 hours to master

Coordinate Geometry unit: 12 of 80 theory marks (Straight Lines, Conic Sections, Introduction to Three Dimensional Geometry).

Revision notes

Straight Lines: revision notes

1. Slope

The slope of a non-vertical line making angle \(\theta\) with the positive \(x\)-axis is \(m = \tan\theta\). Through \((x_1, y_1)\) and \((x_2, y_2)\): \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\). Horizontal lines have slope \(0\); vertical lines have no slope.

  • Parallel lines: \(m_1 = m_2\). Perpendicular lines: \(m_1m_2 = -1\).
  • Three points are collinear if the slope between the first two equals the slope between the last two.

2. Angle between two lines

If \(\theta\) is the acute angle between lines of slopes \(m_1, m_2\) (\(1 + m_1m_2 \neq 0\)): \(\tan\theta = \left|\dfrac{m_2 - m_1}{1 + m_1m_2}\right|\).

3. Forms of the equation of a line

  • Parallel to the axes: \(x = a\), \(y = b\).
  • Point-slope: \(y - y_1 = m(x - x_1)\). Slope-intercept: \(y = mx + c\).
  • Two-point: \(y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)\).
  • Intercept form: \(\dfrac xa + \dfrac yb = 1\) (intercepts \(a\) and \(b\)).
  • General form \(Ax + By + C = 0\): slope \(-\dfrac AB\), \(x\)-intercept \(-\dfrac CA\), \(y\)-intercept \(-\dfrac CB\).

4. Distances

Distance of \((x_1, y_1)\) from \(Ax + By + C = 0\): \(d = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\). Distance between the parallel lines \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\): \(\dfrac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}\) (make the \(x\) and \(y\) coefficients identical first).

Worked example 1

Find the equation of the line through \((2, -1)\) parallel to \(3x - y + 4 = 0\).

Slope \(3\): \(y + 1 = 3(x - 2)\), i.e. \(3x - y - 7 = 0\).

Worked example 2

Find the distance of \((1, 2)\) from \(4x + 3y - 20 = 0\).

\(\dfrac{|4 + 6 - 20|}{5} = 2\).

Worked example 3

Find the acute angle between \(y = \sqrt3x\) and the \(x\)-axis.

\(\tan\theta = \sqrt3\), so \(\theta = 60^\circ\).

Common errors

  • Slope written as \(\dfrac{x_2 - x_1}{y_2 - y_1}\), or with the points in a different order on top and bottom.
  • Forgetting the modulus in the angle and distance formulae.
  • Distance between parallel lines without first making the coefficients of \(x\) and \(y\) the same.
  • Slope of \(Ax + By + C = 0\) taken as \(\dfrac AB\) instead of \(-\dfrac AB\).

Exam tips

  • Give final equations in the general form \(Ax + By + C = 0\) with integer coefficients.
  • For perpendicular lines, write the new slope as \(-\dfrac1m\) before substituting.

Topics in this chapter: Slope · Forms of the equation of a line · Distances · Angle between two lines.

Route to 95: four steps

Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.

Step 1

Secure the basics

You can find slopes, test parallel and perpendicular lines and write simple line equations.

Read first: 1. Slope; 3. Forms of the equation of a line 5 practice questions · checkpoint: 3 questions, 3 marks, pass 80%
Practise step 1
Step 2

Exam standard

You can use every form of the line equation, test collinearity and find distances from a point to a line.

Read first: 2. Angle between two lines; 3. Forms; 4. Distances; Worked examples 1-3 6 practice questions · checkpoint: 3 questions, 8 marks, pass 80%
Practise step 2
Step 3

Full marks on long answers

You can combine line equations, distances and areas in triangle problems.

Read first: 3. Two-point form; 4. Distance of a point from a line 3 practice questions · checkpoint: 2 questions, 9 marks, pass 80%
Practise step 3
Step 4

95+ stretch (HOTS)

You can find lines at a given angle, reflections of points and distances between parallel lines.

Read first: 2. Angle between lines; 4. Parallel lines; Common errors 3 practice questions · checkpoint: 3 questions, 9 marks, pass 80%
Practise step 4

Practice questions

Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 2 of the 17 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.

Q1·1 mark·Multiple choiceSlope

The slope of the line through \((2, 3)\) and \((6, 11)\) is

  1. (a)\(\dfrac12\)
  2. (b)\(2\)
  3. (c)\(8\)
  4. (d)\(-2\)
Q2·1 mark·Multiple choiceSlope

The slope of a line inclined at \(135^\circ\) to the positive \(x\)-axis is

  1. (a)\(-1\)
  2. (b)\(1\)
  3. (c)\(\sqrt3\)
  4. (d)\(-\dfrac{1}{\sqrt3}\)
Q3·1 mark·Multiple choiceForms of the equation of a line

The line through \((1, -2)\) with slope \(3\) is

  1. (a)\(x - 3y - 7 = 0\)
  2. (b)\(3x - y + 5 = 0\)
  3. (c)\(3x - y - 5 = 0\)
  4. (d)\(3x + y - 1 = 0\)

Where marks are lost in Straight Lines

  • Slope of Ax + By + C = 0 taken as A/B. Fix: rearrange to y = −(A/B)x − C/B.
  • Modulus dropped in the distance formula, giving a negative distance. Fix: |Ax₁ + By₁ + C|/√(A² + B²).
  • Parallel-line distance with unequal coefficients (3x − 4y + 7 and 6x − 8y − 6). Fix: divide the second by 2 first.
  • Only one line found when the angle condition allows two. Fix: tan θ = ± (m₂ − m₁)/(1 + m₁m₂) gives two slopes.
  • Final equation left with fractions. Fix: multiply through to the general form with integers.

Test Straight Lines against the clock

Want it against the clock? Take a timed 30-mark chapter test on Straight Lines, new questions each time (CBSE Essentials or the free trial).