Straight Lines | Class 11 Maths

Slope of a Line

📌 Definition of Slope

The inclination of a line is the angle θ (0° ≤ θ ≤ 180°) it makes with the positive direction of the x-axis, measured anticlockwise. The slope (or gradient) of a line is defined as m = tan θ, where θ ≠ 90°.

Slope of x-axis (horizontal line) = 0 (θ = 0°, tan 0° = 0)
Slope of y-axis (vertical line) = undefined (θ = 90°, tan 90° not defined)
Slope between two points P(x₁, y₁) and Q(x₂, y₂): m = (y₂ – y₁) / (x₂ – x₁)

🎯 Slope Formula

m = (y₂ – y₁) / (x₂ – x₁) = tan θ

Parallel lines: m₁ = m₂
Perpendicular lines: m₁ × m₂ = –1

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IMAGE SPACE — Inclination of a line

Add diagram: line making angle θ with x-axis, showing acute and obtuse inclination cases

🔵 Parallel Lines

Two non-vertical lines l₁ and l₂ are parallel if and only if their slopes are equal: m₁ = m₂.

🔴 Perpendicular Lines

Two non-vertical lines are perpendicular if and only if: m₁ × m₂ = –1 (negative reciprocals).

🟣 Angle Between Two Lines

tan θ = |(m₂ – m₁) / (1 + m₁m₂)|, where θ is the acute angle between lines with slopes m₁ and m₂.

⚫ Collinear Points

Points A, B, C are collinear if slope of AB = slope of BC. Area of triangle formed = 0.

📝 MCQ Practice

Slope of a Line — Test Yourself

1 Find the slope of the line passing through the points (3, –2) and (–1, 4).

A 2/3
B –2/3
C –3/2
D 3/2

Solution: m = (y₂ – y₁)/(x₂ – x₁) = (4–(–2))/(–1–3) = 6/(–4) = –3/2 ✓

2 Line through (–2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (x, 24). Find x.

A 0
B 2
C 4
D –4

Solution: m₁ = (8–6)/(4–(–2)) = 2/6 = 1/3
m₂ = (24–12)/(x–8) = 12/(x–8)
Since perpendicular: m₁ × m₂ = –1 → (1/3) × 12/(x–8) = –1
→ 4/(x–8) = –1 → x–8 = –4 → x = 4 ✓

3 If the angle between two lines is π/4 and slope of one line is 1/2, what is one possible slope of the other line?

A 1/2
B 3
C –1/2
D 2

Solution: tan(π/4) = 1 = |(m – 1/2)/(1 + m/2)|
Case 1: (m – 1/2)/(1 + m/2) = 1 → m – 1/2 = 1 + m/2 → m/2 = 3/2 → m = 3
Case 2: gives m = –1/3
Answer: 3 (or –1/3) ✓

Equations of a Line

📌 Five Standard Forms

A straight line can be expressed algebraically in multiple standard forms depending on what information about the line is given:

Form	Equation	When to use
Horizontal / Vertical	y = a or x = b	Line parallel to an axis
Point-Slope	y – y₀ = m(x – x₀)	Slope m and one point given
Two-Point	y – y₁ = [(y₂–y₁)/(x₂–x₁)](x – x₁)	Two points on the line given
Slope-Intercept	y = mx + c	Slope m and y-intercept c known
Intercept Form	x/a + y/b = 1	x-intercept a and y-intercept b known
General Form	Ax + By + C = 0	Any line (A, B not both zero)

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IMAGE SPACE — Different forms of equations of a line

Add diagram showing: intercept form with (a, 0) and (0, b) on axes, slope-intercept y = mx + c graph

💡 Key Insight: All these forms can be converted to the general form Ax + By + C = 0. The y-intercept form is most useful for graphing; intercept form is best when both intercepts are known.

📝 MCQ Practice

Equations of a Line — Test Yourself

4 Find the equation of the line through (–2, 3) with slope –4.

A 4x + y + 5 = 0
B 4x – y + 5 = 0
C x + 4y + 5 = 0
D 4x + y – 5 = 0

Solution: Point-slope form: y – y₀ = m(x – x₀)
y – 3 = –4(x – (–2)) = –4(x + 2)
y – 3 = –4x – 8 → 4x + y + 5 = 0 ✓
Check: at (–2, 3): 4(–2) + 3 + 5 = –8 + 8 = 0 ✓

5 A line makes x-intercept –3 and y-intercept 2. Its equation is:

A 3x – 2y + 6 = 0
B 2x – 3y + 6 = 0
C 2x + 3y + 6 = 0
D 3x + 2y – 6 = 0

Solution: Intercept form: x/a + y/b = 1, where a = –3, b = 2
x/(–3) + y/2 = 1
Multiply by –6: 2x – 3y = –6 → 2x – 3y + 6 = 0 ✓

6 The equation of the line through (1, –1) and (3, 5) is:

A 3x + y – 4 = 0
B 3x – y + 4 = 0
C 3x – y – 4 = 0
D x – 3y – 4 = 0

Solution: Slope = (5–(–1))/(3–1) = 6/2 = 3
Using point (1, –1): y + 1 = 3(x – 1)
y + 1 = 3x – 3 → 3x – y – 4 = 0 ✓
Check at (3, 5): 9 – 5 – 4 = 0 ✓

Distance of a Point From a Line

📌 Perpendicular Distance Formula

The perpendicular distance d from a point P(x₁, y₁) to the line Ax + By + C = 0 is the shortest distance, measured along the perpendicular from P to the line.

This is derived using the area of triangle formed by the line and the coordinate axes.
The formula works for any line in general form Ax + By + C = 0.
Distance is always positive (absolute value in numerator).

🎯 Distance Formulas

d = |Ax₁ + By₁ + C| / √(A² + B²)

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

d = |C₁ – C₂| / √(A² + B²)

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IMAGE SPACE — Perpendicular distance diagram

Add diagram: point P(x₁, y₁), perpendicular PM dropped to line L: Ax + By + C = 0, showing distance d = PM

💡 Worked Example

Distance of point (3, –5) from line 3x – 4y – 26 = 0:

A = 3, B = –4, C = –26; Point = (3, –5)
d = |3(3) + (–4)(–5) + (–26)| / √(9 + 16) = |9 + 20 – 26| / √25 = |3| / 5 = 3/5

Distance between parallel lines 3x – 4y + 7 = 0 and 3x – 4y + 5 = 0:

d = |7 – 5| / √(9 + 16) = 2/5

📝 MCQ Practice

Distance Formulas — Test Yourself

7 Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Solution: 12x + 72 = 5y – 10 → 12x – 5y + 82 = 0
A = 12, B = –5, C = 82; Point = (–1, 1)
d = |12(–1) + (–5)(1) + 82| / √(144 + 25)
= |–12 – 5 + 82| / √169 = |65| / 13 = 5 ✓

8 Find the distance between the parallel lines 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.

A 3
B 65/17
C 2
D 4

Solution: A = 15, B = 8, C₁ = –34, C₂ = 31
d = |C₁ – C₂| / √(A² + B²) = |–34 – 31| / √(225 + 64)
= 65 / √289 = 65 / 17 = 65/17 ✓

9 The perpendicular distance of the origin from the line 3x + 4y + 10 = 0 is:

A 1
B 2
C 5/2
D 10

Solution: Origin = (0, 0); Line: 3x + 4y + 10 = 0
d = |3(0) + 4(0) + 10| / √(9 + 16) = 10 / √25 = 10/5 = 2 ✓

Special Conditions & Applications

📌 Concurrent Lines, Image of a Point & Key Theorems

Several important applications arise from the theory of straight lines in coordinate geometry:

Concurrent lines: Three lines are concurrent if they all pass through a common point. Find intersection of any two, then verify it satisfies the third.
Collinear points: Three points are collinear iff slope of any two pairs is equal (or area of triangle = 0).
Image of a point in a line: If Q(h, k) is image of P(x₁, y₁) in line L, then L is the perpendicular bisector of PQ.
Foot of perpendicular: Drop a perpendicular from a point to a line; the foot can be found by solving the perpendicular line and the given line simultaneously.

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IMAGE SPACE — Image of a point in a line

Add diagram: point P(1,2), its image Q(h,k) reflected in line x–3y+4=0, midpoint on the line

📐 For image of point P(x₁, y₁) in line Ax + By + C = 0: use the formula (h – x₁)/A = (k – y₁)/B = –2(Ax₁ + By₁ + C)/(A² + B²)

💡 Key Results to Remember

Line through origin with slope m: y = mx
Condition for concurrence of 3 lines: Substitute intersection of 2 lines into the 3rd.
If A.M. and G.M. of two numbers a and b are known: solve as system a + b = 2A and ab = G².
Area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃) = ½|x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)|
Any line parallel to Ax + By + C = 0 has the form Ax + By + k = 0 for some constant k.

📝 MCQ Practice

Special Conditions — Test Yourself

10 If the three lines 2x + y – 3 = 0, 5x + ky – 3 = 0 and 3x – y – 2 = 0 are concurrent, find k.

A 2
B –2
C 3
D –3

Solution: Solve lines 1 and 3:
2x + y – 3 = 0 and 3x – y – 2 = 0
Adding: 5x – 5 = 0 → x = 1; y = 1
Point of intersection: (1, 1). Substitute into line 2:
5(1) + k(1) – 3 = 0 → 2 + k = 0 → k = –2 ✓

11 Find the equation of line parallel to 3x – 4y + 2 = 0 passing through the point (–2, 3).

A 3x – 4y – 6 = 0
B 4x + 3y – 1 = 0
C 3x – 4y + 18 = 0
D 3x + 4y + 18 = 0

Solution: Parallel line has same coefficients: 3x – 4y + k = 0
Passes through (–2, 3): 3(–2) – 4(3) + k = 0
–6 – 12 + k = 0 → k = 18
Equation: 3x – 4y + 18 = 0 ✓

12 Area of the triangle formed by the lines y = x, y = –x and x = 4 is:

A 8
B 16
C 32
D 4

Solution: Vertices: intersection of y=x and y=–x → (0, 0)
y=x and x=4 → (4, 4)
y=–x and x=4 → (4, –4)
Area = ½ |x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)|
= ½ |0(4–(–4)) + 4((–4)–0) + 4(0–4)|
= ½ |0 – 16 – 16| = ½ × 32 = 16 ✓

✦

Quick Summary

🗂️ Chapter 9 at a Glance

Slope: m = tan θ = (y₂–y₁)/(x₂–x₁). Parallel → m₁=m₂. Perpendicular → m₁m₂=–1.
Angle between lines: tan θ = |(m₂–m₁)/(1+m₁m₂)|
Point-slope form: y – y₀ = m(x – x₀)
Two-point form: y – y₁ = [(y₂–y₁)/(x₂–x₁)](x – x₁)
Slope-intercept: y = mx + c (c = y-intercept)
Intercept form: x/a + y/b = 1
Distance from point to line: d = |Ax₁+By₁+C| / √(A²+B²)
Distance between parallel lines: d = |C₁–C₂| / √(A²+B²)

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StraightLines

📌 Definition of Slope

🔵 Parallel Lines

🔴 Perpendicular Lines

🟣 Angle Between Two Lines

⚫ Collinear Points

Slope of a Line — Test Yourself

📌 Five Standard Forms

Equations of a Line — Test Yourself

📌 Perpendicular Distance Formula

💡 Worked Example

Distance Formulas — Test Yourself

📌 Concurrent Lines, Image of a Point & Key Theorems

💡 Key Results to Remember

Special Conditions — Test Yourself

🗂️ Chapter 9 at a Glance

Straight
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