Straight Lines

Straight Lines - Comprehensive CBSE Class 11 Notes

Introduction and Historical Context

**Analytical Geometry** is the systematic study of geometry using algebraic methods. René Descartes pioneered this approach in his 1637 publication 'La Géométry'. This chapter extends the coordinate geometry concepts from earlier classes to study the simplest geometric figure—the **straight line**.

**Essential Recall Formulas from Earlier Classes:**

**Distance Formula:** Distance PQ between P(x₁, y₁) and Q(x₂, y₂) is: PQ = √[(x₂ - x₁)² + (y₂ - y₁)²]

**Section Formula (Internal Division):** A point dividing line segment joining (x₁, y₁) and (x₂, y₂) in ratio m:n has coordinates: ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))

**Midpoint Formula:** When m = n, midpoint is ((x₁ + x₂)/2, (y₁ + y₂)/2)

**Area of Triangle:** With vertices (x₁, y₁), (x₂, y₂), (x₃, y₃), area = ½|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

**Collinearity Condition:** Three points are collinear if and only if the area of the triangle formed by them is zero.

Slope of a Line

Definition of Inclination and Slope

**Inclination (θ)** is the angle made by a line with the positive direction of the x-axis, measured counterclockwise. The range of inclination is **0° ≤ θ ≤ 180°**.

**Slope (m)** is defined as: **m = tan θ**, where θ ≠ 90°

Key observations:

Slope of x-axis = 0 (since θ = 0°, tan 0° = 0)

Slope of y-axis is **undefined** (since θ = 90°, tan 90° is undefined)

Slope is denoted by **m**

Slope When Two Points Are Given

When a non-vertical line passes through points P(x₁, y₁) and Q(x₂, y₂), the slope is calculated as:

**m = (y₂ - y₁)/(x₂ - x₁)**, where x₁ ≠ x₂

**Derivation:** Consider line l with inclination θ passing through P(x₁, y₁) and Q(x₂, y₂). Draw perpendicular QR to x-axis and PM perpendicular to RQ.

When θ is acute: In triangle MPQ, tan θ = MQ/MP = (y₂ - y₁)/(x₂ - x₁)

When θ is obtuse: θ = 180° - ∠MPQ, so m = tan θ = tan(180° - ∠MPQ) = (y₂ - y₁)/(x₂ - x₁)

In both cases, the formula is the same.

**Example:** Find slope of line through (3, -2) and (-1, 4):

m = [4 - (-2)]/[-1 - 3] = 6/(-4) = -3/2

Conditions for Parallelism and Perpendicularity

**Parallel Lines Condition:** Two non-vertical lines l₁ and l₂ with slopes m₁ and m₂ are parallel if and only if:

**m₁ = m₂**

Proof: If l₁ ∥ l₂, then their inclinations are equal: α = β, therefore tan α = tan β, so m₁ = m₂.

**Perpendicular Lines Condition:** Two non-vertical lines l₁ and l₂ with slopes m₁ and m₂ are perpendicular if and only if:

**m₁ · m₂ = -1** or equivalently **m₂ = -1/m₁**

Proof: If l₁ ⊥ l₂, then β = α + 90°, so tan β = tan(α + 90°) = -cot α = -1/tan α, therefore m₁ · m₂ = -1.

**Example:** Line through (-2, 6) and (4, 8) has slope m₁ = (8 - 6)/(4 + 2) = 2/6 = 1/3. If a perpendicular line passes through (8, 12) and (x, 24), then m₂ = 12/(x - 8). For perpendicularity: (1/3) · 12/(x - 8) = -1, giving x - 8 = -36, so x = -28.

Angle Between Two Lines

Let L₁ and L₂ be non-vertical lines with slopes m₁ and m₂, and inclinations α₁ and α₂ respectively.

**Formula for Acute Angle θ:**

**tan θ = |(m₂ - m₁)/(1 + m₁m₂)|**, where 1 + m₁m₂ ≠ 0

Or without absolute value:

**tan θ = (m₂ - m₁)/(1 + m₁m₂)**

The obtuse angle φ = 180° - θ.

**Derivation:** If θ = α₂ - α₁, then tan θ = tan(α₂ - α₁) = (tan α₂ - tan α₁)/(1 + tan α₁ tan α₂) = (m₂ - m₁)/(1 + m₁m₂)

**Example:** If angle between two lines is π/4 and slope of one line is 1/2, find the other slope.

Using tan(π/4) = 1: |m - 1/2|/(1 + m/2) = 1

This gives: m - 1/2 = 1 + m/2 or m - 1/2 = -(1 + m/2)

Solving: m = 3 or m = -1/3

Various Forms of the Equation of a Line

Horizontal and Vertical Lines

**Horizontal Line (parallel to x-axis):** All points have the same y-coordinate. Equation is **y = a** or **y = -a**, where a is the distance from x-axis.

**Vertical Line (parallel to y-axis):** All points have the same x-coordinate. Equation is **x = b** or **x = -b**, where b is the distance from y-axis.

**Example:** Line through (-2, 3) parallel to x-axis: **y = 3**; parallel to y-axis: **x = -2**

Point-Slope Form

When a non-vertical line has slope m and passes through fixed point P₀(x₀, y₀), any point P(x, y) on the line satisfies:

**y - y₀ = m(x - x₀)**

This is obtained directly from the slope definition: (y - y₀)/(x - x₀) = m

**Example:** Equation of line through (-2, 3) with slope -4:

y - 3 = -4(x + 2)

y - 3 = -4x - 8

4x + y + 5 = 0

Two-Point Form

When a line passes through P₁(x₁, y₁) and P₂(x₂, y₂), the equation is:

**(y - y₁)/(y₂ - y₁) = (x - x₁)/(x₂ - x₁)** or **y - y₁ = [(y₂ - y₁)/(x₂ - x₁)](x - x₁)**

This uses the fact that slopes P₁P = P₁P₂ for any point P on the line.

**Example:** Line through (1, -1) and (3, 5):

(y + 1)/(5 + 1) = (x - 1)/(3 - 1)

(y + 1)/6 = (x - 1)/2

2(y + 1) = 6(x - 1)

2y + 2 = 6x - 6

3x - y - 4 = 0

Slope-Intercept Form

**Case 1 - y-intercept known (c):** When line with slope m cuts the y-axis at point (0, c):

**y = mx + c**

where c is the y-intercept.

**Case 2 - x-intercept known (d):** When line with slope m cuts the x-axis at point (d, 0):

**y = m(x - d)**

**Example:** Line with tan θ = 1/2 and y-intercept -3/2:

y = (1/2)x - 3/2 or x - 2y - 3 = 0

Intercept Form

When a line makes x-intercept a and y-intercept b, it passes through (a, 0) and (0, b). The equation is:

**x/a + y/b = 1**

where a ≠ 0 and b ≠ 0.

**Derivation:** Using two-point form with (a, 0) and (0, b):

(y - 0)/(b - 0) = (x - a)/(0 - a)

y/b = (x - a)/(-a)

-ay = b(x - a)

-ay = bx - ab

bx + ay = ab

x/a + y/b = 1

**Example:** Line with x-intercept -3 and y-intercept 2:

x/(-3) + y/2 = 1

-2x + 3y = 6

2x - 3y + 6 = 0

General Form of Line Equation

Any equation of the form **Ax + By + C = 0** where A and B are not both zero is the general form of a line equation.

**Key Point:** Any line can be expressed in general form, and conversely, any linear equation represents a straight line.

Distance of a Point From a Line

Perpendicular Distance Formula

The perpendicular distance from point P(x₀, y₀) to line Ax + By + C = 0 is:

**d = |Ax₀ + By₀ + C|/√(A² + B²)**

This represents the shortest distance from the point to the line, measured along the perpendicular.

**Derivation Outline:**

Foot of perpendicular from P(x₀, y₀) to line Ax + By + C = 0 lies on both the given line and the perpendicular through P

Perpendicular has slope B/A (negative reciprocal of line's slope -A/B)

Find intersection of two lines to get foot of perpendicular

Calculate distance using distance formula

**Example:** Distance from point (1, 3) to line 3x + 4y - 15 = 0:

d = |3(1) + 4(3) - 15|/√(9 + 16) = |3 + 12 - 15|/5 = |0|/5 = 0

This means point (1, 3) lies on the line.

**Example:** Distance from (2, 5) to 2x - 3y + 4 = 0:

d = |2(2) - 3(5) + 4|/√(4 + 9) = |4 - 15 + 4|/√13 = |-7|/√13 = 7/√13 = 7√13/13

Distance Between Two Parallel Lines

For parallel lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 (where a₁/a₂ = b₁/b₂):

If written as ax + by + c₁ = 0 and ax + by + c₂ = 0:

**d = |c₁ - c₂|/√(a² + b²)**

**Alternative Method:** Take any point on first line, find its distance to second line.

**Example:** Distance between 2x + 3y - 4 = 0 and 2x + 3y + 5 = 0:

d = |-4 - 5|/√(4 + 9) = 9/√13 = 9√13/13

Important Applications and Results

**Area of Triangle using Line Distances:** If three vertices of a triangle are A, B, C and one side BC lies on line ax + by + c = 0, then area = (1/2) × base BC × perpendicular distance of A from BC.

**Concurrency of Lines:** Three or more lines meet at a point if the coordinates of intersection of any two satisfy the equation of the third.

**Reflection of a Point Across a Line:** Can be found using perpendicular distance principles.

Key Problem-Solving Strategies

**For Finding Line Equations:**

1. Identify what information is given: points, slope, intercepts, inclination, or relationships

2. Choose appropriate form: point-slope, two-point, intercept, or general

3. Simplify to standard form (usually general form Ax + By + C = 0)

**For Geometric Problems:**

1. Use slope conditions for parallelism (m₁ = m₂) and perpendicularity (m₁m₂ = -1)

2. Use collinearity condition (equal areas = 0) to prove points lie on a line

3. Use distance formula to verify right angles or congruent sides

**Common Mistakes to Avoid:**

Forgetting that slope is undefined for vertical lines (vertical line through x₀ is x = x₀, not a function)

Incorrect sign when using intercept form (negative intercepts)

Not checking if 1 + m₁m₂ = 0 when using angle formula (formula doesn't apply when lines are perpendicular)

Confusion between "distance on axis" and "intercept" (intercept is where line cuts axis, distance is position of that point)

Forgetting absolute value in distance formula, which always gives non-negative distance

Exam-Important Points Summary

**Slope** is the tangent of inclination; undefined for vertical lines

**Parallel lines** have equal slopes; **perpendicular lines** have slopes whose product is -1

**Angle between lines** uses tan θ = |(m₂ - m₁)/(1 + m₁m₂)|

**Five main forms** of line equation: horizontal (y = a), vertical (x = b), point-slope, two-point, intercept

**General form** Ax + By + C = 0 is universal for all non-vertical lines

**Distance from point to line** = |Ax₀ + By₀ + C|/√(A² + B²) is a critical formula

**Collinearity** can be verified using slope equality or area method