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Straight Lines — Mathematics Class 11 Notes (CBSE & HBSE)

Free NCERT Mathematics notes for Straight Lines (Class 11) on Siksha Sarovar, aligned to CBSE and Haryana Board (HBSE). This chapter is broken into 4 topics with clear explanations, formulas, solved examples and board-pattern practice — free to read, no sign-up required.

Board exam focus — Straight Lines (CBSE & HBSE)

Coordinate geometry workhorse — slopes, line forms, distances. CBSE expects 6–8 marks; HBSE 5–7 marks. Heavy formula chapter — memorise the five forms of line equation.

Slope, Angle, Parallel & Perpendicular Lines

Slope of a Line

For a non-vertical line, slope m = tan θ, where θ is the inclination (angle made with positive x-axis, 0 ≤ θ < π).

For a line through (x₁, y₁) and (x₂, y₂): m = (y₂ − y₁) / (x₂ − x₁) (x₁ ≠ x₂).

Important Slopes

LineSlope
Horizontal (parallel to x-axis)0
Vertical (parallel to y-axis)undefined
Making 45° with x-axis1
Making 135° with x-axis−1

Angle Between Two Lines

If m₁ and m₂ are slopes: tan φ = |(m₁ − m₂) / (1 + m₁ m₂)| (acute angle).

For obtuse angle, take 180° − φ.

Parallel & Perpendicular Conditions

  • Parallel: m₁ = m₂.
  • Perpendicular: m₁ · m₂ = −1.
  • Vertical + horizontal: always perpendicular (one undefined, other 0).

Collinearity

Three points (x₁, y₁), (x₂, y₂), (x₃, y₃) are collinear iff: (y₂ − y₁)(x₃ − x₂) = (y₃ − y₂)(x₂ − x₁).

Or equivalently, the area of the triangle they form = 0: (1/2)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| = 0.

Useful Formulas

  • Distance between (x₁, y₁) and (x₂, y₂): d = √[(x₂−x₁)² + (y₂−y₁)²].
  • Section formula (internal division in ratio m : n):
  • ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n)).

  • Mid-point: ((x₁+x₂)/2, (y₁+y₂)/2).
CBSE Tip: Always compute slope and angle in radians or degrees consistently — don't mix tan 45° with tan(π/4) results.

Various Forms of a Line

1. Point-Slope Form

Line through (x₁, y₁) with slope m: y − y₁ = m(x − x₁).

2. Two-Point Form

Line through (x₁, y₁) and (x₂, y₂): (y − y₁) = [(y₂ − y₁)/(x₂ − x₁)] (x − x₁).

3. Slope-Intercept Form

Line with slope m and y-intercept c: y = mx + c.

4. Intercept Form

Line with x-intercept a and y-intercept b (a, b ≠ 0): x/a + y/b = 1.

5. Normal Form (Perpendicular Form)

Line such that the perpendicular from origin makes angle ω with positive x-axis and has length p: x cos ω + y sin ω = p.

Here p > 0 and 0 ≤ ω < 2π.

General Form

Any straight line can be written as: Ax + By + C = 0, with A and B not both zero.

From this:

  • Slope = −A/B (B ≠ 0).
  • x-intercept = −C/A, y-intercept = −C/B.
  • Normal form length p = |C| / √(A² + B²), with cos ω = ±A/√(A²+B²), sin ω = ±B/√(A²+B²).

Converting Between Forms

  • Ax + By + C = 0 → y = (−A/B)x − C/B → identify slope and intercept.
  • Two-point form → simplify to slope-intercept by computing slope first.

Special Lines

  • y = k → horizontal at height k.
  • x = h → vertical at x = h.
  • y = x → 45° line through origin.
  • y = −x → 135° line through origin.
HBSE Tip: Choose the form that matches given data: - Point + slope → point-slope. - Two points → two-point. - Slope + y-intercept → slope-intercept. - Two intercepts → intercept form.

Distance from Point to Line, Between Parallel Lines

Distance of a Point from a Line

Distance from point P(x₁, y₁) to line Ax + By + C = 0: d = |Ax₁ + By₁ + C| / √(A² + B²).

Distance Between Two Parallel Lines

Lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0: d = |C₁ − C₂| / √(A² + B²).

IMPORTANT: Coefficients of x and y must match (or be made to match by scaling) before applying this formula.

Foot of Perpendicular

Given point P(x₁, y₁) and line ℓ : Ax + By + C = 0:

The foot of perpendicular Q has coordinates such that PQ ⊥ ℓ and Q lies on ℓ. Use parametric formulas:

x = x₁ − A · (Ax₁ + By₁ + C)/(A² + B²) y = y₁ − B · (Ax₁ + By₁ + C)/(A² + B²)

Reflection of a Point in a Line

Reflection P' = 2Q − P, where Q is the foot of perpendicular.

Distance Between Parallel Lines (Verification)

For 3x + 4y + 5 = 0 and 3x + 4y − 10 = 0: d = |5 − (−10)| / √(9 + 16) = 15/5 = 3.

Use of Distance Formulas

  • To find altitude of a triangle (distance from vertex to opposite side).
  • To find the equation of the line equidistant from two parallel lines.
  • In optimisation problems ("closest point on line to given point").

Quick Check

If P lies on the line, Ax₁ + By₁ + C = 0 ⇒ distance = 0 (sanity check before computation).

CBSE Tip: When parallel lines have different coefficient sets, divide one by a constant to match — otherwise the formula gives a wrong result.

Family of Lines & Concurrency

Family of Lines Through Intersection

Given two lines ℓ₁ : a₁x + b₁y + c₁ = 0 and ℓ₂ : a₂x + b₂y + c₂ = 0, any line through their intersection can be written as: (a₁x + b₁y + c₁) + k(a₂x + b₂y + c₂) = 0, where k is a real parameter.

Choosing k uniquely determines a line in the family. Useful when intersection point is hard to compute first.

Concurrent Lines

Three lines are concurrent iff they pass through a common point.

For three lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, a₃x + b₃y + c₃ = 0, they are concurrent iff the determinant:

| a₁ b₁ c₁ |
| a₂ b₂ c₂ | = 0
| a₃ b₃ c₃ |

Area of Triangle Given Three Lines

If three lines form a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃): Area = (1/2)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

Bisectors of Angles Between Two Lines

For lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0: Angle bisectors satisfy (a₁x + b₁y + c₁)/√(a₁² + b₁²) = ± (a₂x + b₂y + c₂)/√(a₂² + b₂²). The + sign gives one bisector; the − sign gives the other (perpendicular to it).

Useful Constructions

  • Line parallel to ax + by + c = 0 through (x₁, y₁): a(x − x₁) + b(y − y₁) = 0.
  • Line perpendicular to ax + by + c = 0 through (x₁, y₁): b(x − x₁) − a(y − y₁) = 0.

Locus Problems

A point (x, y) moves such that its distance from (a, b) equals its distance from line ℓ: parabola. Its distances from two fixed points are equal: perpendicular bisector.

HBSE Tip: When asked to show three lines are concurrent, use the determinant condition — quicker than solving two equations for the intersection point.

Frequently asked questions

Are these Straight Lines notes free?

Yes — the Straight Lines notes for Mathematics (Class 11) on Siksha Sarovar are completely free to read, with no account required.

Do these notes follow CBSE and HBSE?

Yes. The Straight Lines notes are NCERT-aligned and include guidance for both CBSE and Haryana Board (HBSE), with important questions and MCQs for revision.

What does the Straight Lines chapter cover?

Concept explanations, key formulas and definitions, fully solved examples and board-pattern practice questions for Straight Lines.