Class 11Mathematics · Coordinate GeometryFull chapter

Straight Lines

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter
Slope of a Line and Angle Between Two LinesPoint-Slope Form and Two-Point FormSlope-Intercept Form and Intercept FormNormal Form of a LineGeneral Equation of a LineDistance of a Point from a Line

Slope of a Line and Angle Between Two Lines

Quick answer The slope of a line measures its steepness as the tangent of its inclination; the angle between two lines is found from a tan formula built out of their slopes.

The inclination of a line is the angle θ it makes with the positive direction of the x-axis, measured anticlockwise, where 0° ≤ θ < 180°. The slope (or gradient) of the line is defined as m = tanθ. A line parallel to the x-axis has θ = 0°, so m = 0; a line parallel to the y-axis has θ = 90°, where tanθ is undefined, so such a line has no slope.

If a line passes through two points A(x1, y1) and B(x2, y2) with x1 ≠ x2, its slope is m = (y2 − y1)/(x2 − x1). This value does not depend on which point is taken first, since reversing both differences leaves the ratio unchanged.

Two lines with slopes m1 and m2 are parallel if and only if m1 = m2, and are perpendicular if and only if m1m2 = −1. Three points are collinear exactly when the slope of the segment joining any two of them equals the slope of the segment joining another pair that shares a common point with the first.

If two lines have slopes m1 and m2, the angle θ between them satisfies tanθ = (m2 − m1)/(1 + m1m2). Taking the absolute value gives the acute angle between the lines; the obtuse angle is 180° minus this acute angle.

Worked example. Find the angle which the line joining the points (3, −1) and (4, −2) makes with the positive x-axis.

Slope m = (−2 − (−1))/(4 − 3) = −1/1 = −1. Since tanθ = −1 with 0° ≤ θ < 180°, θ = 135°.

Worked example (collinearity). Show that the points A(1, 4), B(3, −2) and C(−1, 10) are collinear.

Slope of AB = (−2 − 4)/(3 − 1) = −6/2 = −3. Slope of BC = (10 − (−2))/(−1 − 3) = 12/(−4) = −3. Since slope of AB = slope of BC and B is common to both, the points A, B, C lie on the same straight line, so they are collinear.

Slope from inclination m = tanθ 0° ≤ θ < 180°, θ ≠ 90°
Slope from two points m = (y₂ − y₁)/(x₂ − x₁) x1 ≠ x2
Parallel lines m₁ = m₂
Perpendicular lines m₁m₂ = −1
Angle between two lines tanθ = |(m₂ − m₁)/(1 + m₁m₂)|
Remember
  • Slope m = tanθ, where θ is the inclination of the line with the positive x-axis (0°≤θ<180°, θ≠90°).
  • A horizontal line has slope 0; a vertical line has no defined slope.
  • m1=m2 is the condition for parallel lines; m1·m2=−1 is the condition for perpendicular lines.
  • Three points are collinear if and only if the slope of the segment joining any two of them equals the slope of the segment joining another pair sharing that same point.

Point-Slope Form and Two-Point Form

Quick answer The point-slope form writes a line's equation from one point and its slope; the two-point form extends this to a line determined by two given points.

If a line passes through a fixed point (x0, y0) and has slope m, then for any other point (x, y) on the line, m = (y − y0)/(x − x0). Rearranging gives the point-slope form: y − y0 = m(x − x0).

When two points A(x1, y1) and B(x2, y2) on the line are known instead of the slope, the slope is first computed as m = (y2 − y1)/(x2 − x1) and then substituted into the point-slope form using either point. This gives the two-point form: y − y1 = [(y2 − y1)/(x2 − x1)](x − x1).

Worked example (point-slope form). Find the equation of the line passing through (2, 3) with slope −4.

Using y − y0 = m(x − x0): y − 3 = −4(x − 2) ⟹ y − 3 = −4x + 8 ⟹ 4x + y − 11 = 0.

Worked example (two-point form). Find the equation of the line passing through (1, 1) and (−2, 4).

Slope m = (4 − 1)/(−2 − 1) = 3/(−3) = −1. Using the point (1, 1): y − 1 = −1(x − 1) ⟹ y − 1 = −x + 1 ⟹ x + y − 2 = 0.

Point-slope form y − y₀ = m(x − x₀)
Two-point form y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁) x1 ≠ x2
Remember
  • Point-slope form needs one point and the slope; two-point form needs two points (slope is computed first).
  • Either given point may be substituted in the two-point form — both lead to the same final equation.
  • These forms are the starting point for deriving every other form of a line's equation.
  • Always simplify the final equation to the general form Ax+By+C=0 for a unique, comparable answer.

Slope-Intercept Form and Intercept Form

Quick answer The slope-intercept form uses the y-intercept, while the intercept form uses both axis intercepts to write a line's equation directly.

If a line has slope m and cuts the y-axis at the point (0, c), substituting (0, c) into the point-slope form y − c = m(x − 0) gives the slope-intercept form: y = mx + c, where c is the y-intercept. A line through the origin (c = 0) simply reduces to y = mx.

If a line makes non-zero intercepts a and b on the x-axis and y-axis respectively, it passes through (a, 0) and (0, b). Its slope is (b − 0)/(0 − a) = −b/a, and the point-slope form through (a, 0) gives, after simplification, the intercept form: x/a + y/b = 1.

Worked example (slope-intercept form). Find the slope and y-intercept of the line 3x + 2y − 12 = 0, and state its slope-intercept form.

2y = −3x + 12 ⟹ y = (−3/2)x + 6. So the slope is −3/2 and the y-intercept is 6; the slope-intercept form is y = (−3/2)x + 6.

Worked example (intercept form). Find the equation of the line which makes intercepts 3 and −2 on the x-axis and y-axis respectively.

Using x/a + y/b = 1 with a = 3, b = −2: x/3 + y/(−2) = 1. Multiplying throughout by 6: 2x − 3y = 6, i.e. 2x − 3y − 6 = 0.

Slope-intercept form y = mx + c c is the y-intercept
Intercept form x/a + y/b = 1 a, b ≠ 0 are the x- and y-intercepts
Remember
  • Slope-intercept form y=mx+c is the most convenient for reading off slope and y-intercept directly.
  • Intercept form x/a + y/b = 1 needs both intercepts to be non-zero; it cannot be used for lines through the origin or lines parallel to an axis.
  • To convert a general equation to intercept form, isolate the constant on one side and divide.
  • a and b in the intercept form can be negative, indicating the axis is cut on the negative side.

Normal Form of a Line

Quick answer The normal form describes a line using the length p of the perpendicular from the origin and the angle ω that perpendicular makes with the x-axis.

Let a line be at perpendicular distance p (p > 0) from the origin, and let the perpendicular from the origin to the line make an angle ω with the positive x-axis (0 ≤ ω < 360°). The foot of this perpendicular is the point (p cosω, p sinω), and the line through this point perpendicular to the segment from the origin has the normal form: x cosω + y sinω = p.

To convert a general equation Ax + By + C = 0 into normal form, write it as Ax + By = −C, then divide throughout by √(A2 + B2), choosing the sign of the divisor so that the right-hand side becomes positive (since p must be positive).

Worked example. Reduce the equation √3x + y − 8 = 0 to normal form, and find p and ω.

Write it as √3x + y = 8. Here A = √3, B = 1, so √(A2 + B2) = √(3 + 1) = 2. Dividing by 2: (√3/2)x + (1/2)y = 4. Since the right-hand side is already positive, this is the normal form with cosω = √3/2 and sinω = 1/2, giving ω = 30°, and p = 4. So the normal form is x cos30° + y sin30° = 4.

Normal form x cosω + y sinω = p p>0 is the distance of the line from the origin, 0≤ω<360°
Reduction from general form (A/√(A²+B²))x + (B/√(A²+B²))y = −C/√(A²+B²) sign of √(A²+B²) chosen to make RHS positive
Remember
  • p is always taken positive; ω is adjusted (via the sign chosen while dividing) so this holds.
  • cosω and sinω are read off directly as the coefficients of x and y once the equation is normalised.
  • The normal form gives the perpendicular distance from the origin to a line directly as p.
  • Dividing by +√(A²+B²) or −√(A²+B²) is decided solely by which choice makes the constant term positive.

General Equation of a Line

Quick answer Every straight line can be written as Ax+By+C=0 with A, B not both zero; this single form contains every special form as a particular case.

Every equation of the first degree in x and y, namely Ax + By + C = 0 (where A and B are not both zero), represents a straight line, and conversely every straight line can be written in this general form.

From the general form, the slope and the x- and y-intercepts can be read off directly: slope m = −A/B (when B ≠ 0), x-intercept = −C/A (when A ≠ 0), y-intercept = −C/B (when B ≠ 0).

Special cases are worth noting: if A = 0, the equation becomes By + C = 0, i.e. y = constant, a line parallel to the x-axis; if B = 0, it becomes Ax + C = 0, i.e. x = constant, a line parallel to the y-axis; if C = 0, the line Ax + By = 0 passes through the origin.

Worked example. Find the slope, x-intercept and y-intercept of the line 3x + 2y − 12 = 0, and hence write its equation in intercept form.

Here A = 3, B = 2, C = −12. Slope m = −A/B = −3/2. x-intercept = −C/A = 12/3 = 4. y-intercept = −C/B = 12/2 = 6. So in intercept form, the equation is x/4 + y/6 = 1.

General form Ax + By + C = 0 A and B not both zero
Slope from general form m = −A/B B≠0
x-intercept x-intercept = −C/A A≠0
y-intercept y-intercept = −C/B B≠0
Remember
  • Ax+By+C=0 is the master form; point-slope, two-point, slope-intercept, intercept and normal forms are all special cases of it.
  • Slope from the general form: m = −A/B (coefficient of x over coefficient of y, negated).
  • A=0 gives a line parallel to the x-axis; B=0 gives a line parallel to the y-axis; C=0 gives a line through the origin.
  • Converting between forms is mostly algebraic rearrangement — the underlying line does not change.

Distance of a Point from a Line

Quick answer The perpendicular distance from a point to a line can be found directly from the line's general equation, and a similar formula gives the distance between two parallel lines.

The perpendicular distance of a point P(x1, y1) from the line Ax + By + C = 0 is d = |Ax1 + By1 + C| / √(A2 + B2). The modulus is used because distance is always non-negative, regardless of which side of the line the point lies on.

For two parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 (same A, B, so same slope), the perpendicular distance between them simplifies to d = |C1 − C2| / √(A2 + B2). Both equations must first be written with identical coefficients of x and y before this formula is applied.

Worked example (point to line). Find the distance of the point (3, −5) from the line 3x − 4y − 26 = 0.

Here A = 3, B = −4, C = −26, x1 = 3, y1 = −5. d = |3(3) − 4(−5) − 26| / √(32 + (−4)2) = |9 + 20 − 26| / √25 = |3| / 5 = 3/5.

Worked example (between parallel lines). Find the distance between the parallel lines 3x − 4y + 7 = 0 and 3x − 4y + 5 = 0.

Here A = 3, B = −4, C1 = 7, C2 = 5. d = |7 − 5| / √(32 + (−4)2) = 2/5.

Distance of a point from a line d = |Ax₁ + By₁ + C| / √(A² + B²)
Distance between parallel lines d = |C₁ − C₂| / √(A² + B²) lines must have identical A, B
Distance of origin from a line d = |C| / √(A² + B²)
Remember
  • Always write the line in general form Ax+By+C=0 before applying the distance formula.
  • The modulus sign ensures the distance is never negative, whichever side of the line the point lies on.
  • The distance-between-parallel-lines formula applies only once both lines share identical A and B coefficients.
  • Distance of the origin from Ax+By+C=0 is the special case |C|/√(A²+B²).

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

m = tanθ
Slope from inclination
m = (y₂ − y₁)/(x₂ − x₁)
Slope from two points
m₁ = m₂
Parallel lines
m₁m₂ = −1
Perpendicular lines
tanθ = |(m₂ − m₁)/(1 + m₁m₂)|
Angle between two lines
y − y₀ = m(x − x₀)
Point-slope form
y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁)
Two-point form
y = mx + c
Slope-intercept form
x/a + y/b = 1
Intercept form
x cosω + y sinω = p
Normal form
(A/√(A²+B²))x + (B/√(A²+B²))y = −C/√(A²+B²)
Reduction from general form
Ax + By + C = 0
General form
m = −A/B
Slope from general form
x-intercept = −C/A
x-intercept
y-intercept = −C/B
y-intercept
d = |Ax₁ + By₁ + C| / √(A² + B²)
Distance of a point from a line
d = |C₁ − C₂| / √(A² + B²)
Distance between parallel lines
d = |C| / √(A² + B²)
Distance of origin from a line

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Slope of a line easy

What is the slope of the line joining the points (2, −3) and (−4, 1)?

Q2 Angle between two lines medium

Two lines have slopes 1 and −1. What is the angle between them?

Q3 Point-slope form easy

Find the equation of the line passing through (1, 2) with slope 3.

Q4 Two-point form easy

What is the equation of the line joining the origin and the point (2, 2)?

Q5 Slope-intercept form medium

What are the slope and y-intercept of the line 2x − 3y + 6 = 0?

Q6 Intercept form medium

Find the equation of the line that makes intercepts 4 and −3 on the x-axis and y-axis respectively.

Q7 Normal form hard

When x + y − 4 = 0 is reduced to normal form x cosω + y sinω = p, what is the value of p?

Q8 General equation of a line easy

Which of these lines is parallel to the y-axis?

Q9 Condition for perpendicular lines medium

How are the lines 2x + 3y − 5 = 0 and 3x − 2y + 7 = 0 related?

Q10 Distance of a point from a line medium

Find the distance of the point (2, 3) from the line 3x − 4y − 1 = 0.

Q11 Distance between parallel lines medium

Find the distance between the parallel lines 3x + 4y − 7 = 0 and 3x + 4y + 8 = 0.

Q12 Collinearity of points easy

Are the points (1, 4), (3, −2) and (−1, 10) collinear?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Find the slope of the line which makes an angle of 30° with the positive direction of the y-axis, measured anticlockwise.Slope of a line

If a line makes an angle of 30° with the positive y-axis (measured anticlockwise), its inclination with the positive x-axis is θ = 90° + 30° = 120°.

Slope m = tanθ = tan120° = tan(180° − 60°) = −tan60° = −√3.

2 Without using the distance formula, show that the points A(−2, −1), B(4, 0), C(3, 3) and D(−3, 2) are the vertices of a parallelogram.Slope and parallelism

Two pairs of opposite sides of a quadrilateral are parallel exactly when they have equal slopes.

Slope of AB = (0 − (−1))/(4 − (−2)) = 1/6. Slope of DC = (3 − 2)/(3 − (−3)) = 1/6. Since slope of AB = slope of DC, AB is parallel to DC.

Slope of BC = (3 − 0)/(3 − 4) = 3/(−1) = −3. Slope of AD = (2 − (−1))/(−3 − (−2)) = 3/(−1) = −3. Since slope of BC = slope of AD, BC is parallel to AD.

As both pairs of opposite sides AB, DC and BC, AD are parallel, ABCD is a parallelogram.

3 Find the equation of the line passing through the point (−4, 3) with slope 1/2.Point-slope form

Using the point-slope form y − y0 = m(x − x0) with (x0, y0) = (−4, 3) and m = 1/2:

y − 3 = (1/2)(x − (−4)) = (1/2)(x + 4).

Multiplying both sides by 2: 2y − 6 = x + 4, so x − 2y + 10 = 0.

4 Find the equation of the line which cuts off an intercept of −3 units on the x-axis and has slope −2.Point-slope and intercept

An x-intercept of −3 means the line passes through the point (−3, 0).

Using the point-slope form: y − 0 = −2(x − (−3)) = −2(x + 3) = −2x − 6.

So the equation is y = −2x − 6, i.e. 2x + y + 6 = 0.

5 Find the distance between the parallel lines 3x − 4y + 7 = 0 and 3x − 4y + 5 = 0.Distance between parallel lines

Both lines already have identical coefficients of x and y: A = 3, B = −4, with C1 = 7 and C2 = 5.

d = |C1 − C2| / √(A2 + B2) = |7 − 5| / √(9 + 16) = 2/5.

6 The perpendicular from the origin to a line meets it at the point (−2, 9). Find the equation of the line.General equation of a line

Let P(−2, 9) be the foot of the perpendicular from the origin O. The segment OP is perpendicular to the required line.

Slope of OP = (9 − 0)/(−2 − 0) = −9/2. Since the line is perpendicular to OP, its slope m satisfies m × (−9/2) = −1, giving m = 2/9.

Using the point-slope form through P(−2, 9): y − 9 = (2/9)(x − (−2)) = (2/9)(x + 2).

Multiplying both sides by 9: 9y − 81 = 2x + 4, so 2x − 9y + 85 = 0.

Previous-year board questions 4

Q1 Find the equation of the line passing through the point (2, 2√3) and inclined with the x-axis at an angle of 75°. CBSE 2020 3 marks

The slope of the required line is m = tan75°. Using tan75° = tan(45° + 30°) = (tan45° + tan30°)/(1 − tan45°tan30°) = (1 + 1/√3)/(1 − 1/√3) = (√3 + 1)/(√3 − 1).

Rationalising: m = (√3 + 1)2 / ((√3)2 − 12) = (3 + 2√3 + 1)/2 = (4 + 2√3)/2 = 2 + √3.

Using the point-slope form through (2, 2√3): y − 2√3 = (2 + √3)(x − 2).

Expanding: y − 2√3 = (2 + √3)x − 2(2 + √3) = (2 + √3)x − 4 − 2√3.

So y = (2 + √3)x − 4, i.e. (2 + √3)x − y − 4 = 0.

Q2 The line through the points (h, 3) and (4, 1) intersects the line 7x − 9y − 19 = 0 at right angle. Find the value of h. CBSE 2019 3 marks

The slope of 7x − 9y − 19 = 0 is m1 = −A/B = −7/(−9) = 7/9.

The slope of the line through (h, 3) and (4, 1) is m2 = (1 − 3)/(4 − h) = −2/(4 − h).

Since the two lines are perpendicular, m1m2 = −1: (7/9) × (−2/(4 − h)) = −1.

This gives −14/(9(4 − h)) = −1, so −14 = −9(4 − h) = −36 + 9h, giving 9h = 22, so h = 22/9.

Q3 Find the equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (−2, 3). CBSE 2022 2 marks

Any line parallel to 3x − 4y + 2 = 0 has the same coefficients of x and y, so it can be written as 3x − 4y + k = 0 for some constant k.

Since this line passes through (−2, 3): 3(−2) − 4(3) + k = 0 ⟹ −6 − 12 + k = 0 ⟹ k = 18.

So the required equation is 3x − 4y + 18 = 0.

Q4 Find the distance of the point (−1, 1) from the line 12(x + 6) = 5(y − 2). CBSE 2023 2 marks

Expanding the given equation: 12x + 72 = 5y − 10, so 12x − 5y + 82 = 0. Here A = 12, B = −5, C = 82.

Distance from (−1, 1): d = |12(−1) − 5(1) + 82| / √(122 + (−5)2) = |−12 − 5 + 82| / √(144 + 25) = |65| / √169 = 65/13 = 5.

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