Straight Lines – Class 11 Maths Chapter 9 Notes & NCERT PDF

Class 11MathematicsChapter 9NCERT book: Mathematics - Class 11

Notes and a simple summary of Chapter 9 of the NCERT Class 11 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 9: Straight Lines

Introduction

Chapter 9 of the NCERT Class 11 Mathematics textbook, Straight Lines, turns the plane coordinate geometry of Class 10 (distance formula, section formula, area of a triangle) into a toolkit for describing lines by equations. The chapter opens with Hans Freudenthal's reminder that geometry is "a means … to make children feel the strength of the human spirit." Every idea here — slope, parallelism, perpendicularity, the many forms of a line's equation and the perpendicular distance of a point from a line — is used again in conic sections, calculus (tangents and normals) and physics (velocity–time graphs).

Key Concepts & Theorems

9.2 Slope of a Line

The inclination θ of a line is the angle it makes with the positive direction of the x-axis, measured anticlockwise (0° ≤ θ < 180°).

Slope m = tan θ, for θ ≠ 90°. The slope of the x-axis (and any horizontal line) is 0; the slope of a vertical line is not defined.

9.2.1 Slope from two points. For a non-vertical line through (x₁, y₁) and (x₂, y₂):

m = (y₂ − y₁)/(x₂ − x₁) = (y₁ − y₂)/(x₁ − x₂), x₁ ≠ x₂.

9.2.2 Parallel and perpendicular lines.

Two non-vertical lines are parallel ⇔ their slopes are equal (m₁ = m₂).

Two non-vertical lines are perpendicular ⇔ the product of their slopes is −1 (m₁m₂ = −1), i.e. each slope is the negative reciprocal of the other.

9.2.3 Angle between two lines. If θ is the acute angle between lines with slopes m₁ and m₂ (1 + m₁m₂ ≠ 0):

tan θ = |(m₂ − m₁)/(1 + m₁m₂)|

Removing the modulus sign gives the obtuse angle φ = 180° − θ as well.

Collinearity. Points A, B, C are collinear ⇔ slope of AB = slope of BC.

9.3 Various Forms of the Equation of a Line

Horizontal and vertical lines (9.3.1): a horizontal line at distance a from the x-axis is y = a or y = −a; a vertical line at distance b from the y-axis is x = b or x = −b.

Point-slope form (9.3.2): through (x₀, y₀) with slope m: y − y₀ = m(x − x₀).

Two-point form (9.3.3): through (x₁, y₁) and (x₂, y₂): y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁).

Slope-intercept form (9.3.4): slope m, y-intercept c: y = mx + c. With x-intercept d instead: y = m(x − d).

Intercept form (9.3.5): x-intercept a and y-intercept b: x/a + y/b = 1.

Any equation Ax + By + C = 0 (A and B not both zero) is the general equation of a line. Its slope is −A/B (B ≠ 0), x-intercept −C/A and y-intercept −C/B.

9.4 Distance of a Point from a Line

The perpendicular distance of (x₁, y₁) from Ax + By + C = 0 is

d = |Ax₁ + By₁ + C| / √(A² + B²)

9.4.1 Distance between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0:

d = |C₁ − C₂| / √(A² + B²)

(The x- and y-coefficients must be identical before you apply this formula.)

Worked Examples

Example 1 (slopes). Through (3, −2) and (−1, 4): m = (4 + 2)/(−1 − 3) = −3/2. Through (3, −2) and (7, −2): m = 0 (horizontal line). Through (3, −2) and (3, 4): x₁ = x₂, so the line is vertical and its slope is undefined. Inclination 60°: m = tan 60° = √3.

Example 2 (angle between lines). The angle between two lines is π/4 and one slope is 1/2. Using tan θ = |(m₂ − ½)/(1 + m₂/2)| = 1, the two cases give m₂ = 3 or m₂ = −1/3.

Example 3 (perpendicularity). The line through (−2, 6) and (4, 8) has slope 2/6 = 1/3. The line through (8, 12) and (x, 24) has slope 12/(x − 8). Perpendicular ⇒ (1/3)·12/(x − 8) = −1 ⇒ x = 4.

Examples 5, 6 and 8 (equations).

Through (−2, 3) with slope −4: y − 3 = −4(x + 2), i.e. 4x + y + 5 = 0.

Through (1, −1) and (3, 5): slope = 6/2 = 3, so y + 1 = 3(x − 1), i.e. 3x − y − 4 = 0.

Intercepts −3 and 2: x/(−3) + y/2 = 1, i.e. 2x − 3y + 6 = 0.

Example 9 (point–line distance). Distance of (3, −5) from 3x − 4y − 26 = 0: d = |9 + 20 − 26|/√(9 + 16) = 3/5.

Example 10 (parallel lines). Distance between 3x − 4y + 7 = 0 and 3x − 4y + 5 = 0 is |7 − 5|/5 = 2/5.

Example 11 (concurrent lines). The lines 2x + y − 3 = 0 and 3x − y − 2 = 0 meet at (1, 1). For 5x + ky − 3 = 0 to pass through (1, 1): 5 + k − 3 = 0 ⇒ k = −2.

Formula Sheet

m = tan θ = (y₂ − y₁)/(x₂ − x₁)

Parallel: m₁ = m₂; perpendicular: m₁m₂ = −1

tan θ = |(m₂ − m₁)/(1 + m₁m₂)|

Point-slope: y − y₀ = m(x − x₀)

Two-point: y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁)

Slope-intercept: y = mx + c; with x-intercept d: y = m(x − d)

Intercept: x/a + y/b = 1

General: Ax + By + C = 0, slope −A/B

d = |Ax₁ + By₁ + C|/√(A² + B²); parallel lines d = |C₁ − C₂|/√(A² + B²)

Exam Focus

Exercises 9.1–9.3 and the Miscellaneous Exercise repeatedly test: finding slopes and checking collinearity; writing equations from given conditions; converting a general equation into slope-intercept and intercept forms; and distances.

Outline — "Reduce x + 7y = 0 / 6x + 3y − 5 = 0 into slope-intercept form" type: isolate y → y = −(A/B)x − C/B; state slope and y-intercept clearly.

Outline — "Find the distance of a point from a line": (1) write the line as Ax + By + C = 0; (2) substitute the point into the left side; (3) take the modulus; (4) divide by √(A² + B²); simplify.

Outline — "Prove three points are collinear": compute slope AB and slope BC and show they are equal (or show area of triangle is zero).

Key Terms

Slope / gradient — प्रवणता (ढाल)

Inclination — झुकाव (आनति)

Intercept — अंतःखंड

Parallel lines — समांतर रेखाएँ

Perpendicular lines — लंब रेखाएँ

Collinear points — संरेख बिंदु

Concurrent lines — संगामी रेखाएँ

Point-slope form — बिंदु-ढाल रूप

Two-point form — दो-बिंदु रूप

Slope-intercept form — ढाल-अंतःखंड रूप

Intercept form — अंतःखंड रूप

General equation — व्यापक समीकरण

Perpendicular distance — लंबवत दूरी

Abscissa / ordinate — भुज / कोटि

Common Mistakes

Writing the slope of a vertical line as 0 (it is undefined; a horizontal line has slope 0).

Mixing the order of points: (y₂ − y₁)/(x₁ − x₂) gives the wrong sign — keep both differences in the same order.

Forgetting the modulus in the angle and distance formulas, and ending up with a "negative distance".

Applying the parallel-lines distance formula when the coefficients of x and y are not identical (e.g. 3x − 4y + 7 = 0 and 6x − 8y + 5 = 0 must first be scaled).

Using slope −B/A instead of −A/B for Ax + By + C = 0.

In the intercept form, forgetting the sign of a negative intercept (x-intercept −3 gives x/(−3)).

💡 Key Learning Points

✓Slope m = tan θ = (y₂ − y₁)/(x₂ − x₁); horizontal lines have slope 0, vertical lines have undefined slope.
✓Parallel lines have equal slopes; perpendicular lines satisfy m₁m₂ = −1; the acute angle between lines satisfies tan θ = |(m₂ − m₁)/(1 + m₁m₂)|.
✓A line can be written in point-slope, two-point, slope-intercept, intercept and general (Ax + By + C = 0) forms, and converted between them.
✓The perpendicular distance of (x₁, y₁) from Ax + By + C = 0 is |Ax₁ + By₁ + C|/√(A² + B²).
✓The distance between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is |C₁ − C₂|/√(A² + B²).

👨‍🏫 Teaching Tips

→Start with a real ramp or staircase: measure rise and run, then connect rise/run to tan θ and the idea of slope.
→Use graph paper or GeoGebra to let students drag a line and watch its slope change sign as the inclination passes 90°.
→Build a one-page 'forms of a line' chart where every form is derived from the point-slope form, so students see one idea rather than five formulas.
→Practise converting one line, e.g. 3x − 4y + 10 = 0, into slope-intercept and intercept forms in the same lesson.
→Verify the distance formula on a simple case (distance of the origin from x = 3 should be 3) before using it generally.
→Pair Example 11 (concurrency) with solving simultaneous equations to link algebra and geometry.
→Give 'error-spotting' tasks where a worked solution uses the wrong sign in the slope formula.

📋 Assessment Questions

Find the slope of the line through (3, −2) and (−1, 4) and state whether the line rises or falls.
Find the equation of the line through (1, −1) and (3, 5) and write it in general form.
Find the value of x if the line through (−2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (x, 24).
Find the perpendicular distance of (3, −5) from 3x − 4y − 26 = 0.
Show that three given points are collinear using slopes.

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