Slope of a Line | Class 11 NCERT Maths Explained with Formulas, Derivations and JEE/Board Applications

Introduction

The slope of a line is one of the most essential tools in coordinate geometry. Whenever we want to describe the inclination or steepness of a straight line on the Cartesian plane, we use its slope. It provides a numerical measure of how steeply a line rises or falls as we move along the x-axis. In higher mathematics, slope becomes the foundation for calculus concepts like derivatives, as the derivative of a curve at a point is simply the slope of the tangent line at that point.

For Class 11 Mathematics, the slope of a line is not just a simple definition—it unlocks the understanding of parallelism, perpendicularity, equations of lines, angle calculation, and collinearity of points. These are all important topics for JEE, KCET, COMEDK, and CBSE/ISC board exams.

Definition of Slope

Suppose a straight line makes an angle θ with the positive direction of the x-axis. Then the slope of the line, usually denoted by m, is defined as:

m = tan θ

This definition works whenever θ ≠ 90° + n·180°, because tangent is undefined at these angles. From this simple definition, we get:

If θ = 0°, line is parallel to the x-axis, slope = tan 0° = 0.
If θ = 90°, line is parallel to the y-axis, slope undefined.
If 0° < θ < 90°, slope is positive.
If 90° < θ < 180°, slope is negative.

Thus, slope encodes both the direction and the nature of inclination of a line.

Slope Between Two Points

If a line passes through two points P(x₁, y₁) and Q(x₂, y₂), its slope can be calculated using the coordinate formula:

m = (y₂ – y₁) / (x₂ – x₁), provided x₂ ≠ x₁

This formula simply measures the vertical change divided by the horizontal change. In physics, this is often described as “rise over run.”

Derivation: Draw the line PQ. The horizontal difference is (x₂ – x₁), and vertical difference is (y₂ – y₁). The angle θ made with the x-axis satisfies tan θ = opposite/adjacent = (y₂ – y₁)/(x₂ – x₁). Thus slope = tan θ.

Types of Slopes

Positive slope – line rises from left to right. Example: y = 3x + 2, slope = 3.
Negative slope – line falls from left to right. Example: y = –2x + 5, slope = –2.
Zero slope – horizontal line, parallel to x-axis. Example: y = 4, slope = 0.
Undefined slope – vertical line, parallel to y-axis. Example: x = –2, slope undefined.

Special Results about Slopes

Parallel lines: Two lines are parallel if and only if they have the same slope. If m₁ and m₂ are slopes, then m₁ = m₂.

Perpendicular lines: Two lines are perpendicular if and only if the product of their slopes equals –1. That is, m₁ × m₂ = –1.

Collinearity condition: If three points A, B, C are collinear, then slope AB = slope BC. This gives a very fast way to test collinearity.

Angle Between Two Lines in Terms of Slopes

Let two lines have slopes m₁ and m₂. Then the angle θ between them satisfies:

tan θ = |(m₁ – m₂)/(1 + m₁m₂)|

If m₁ = m₂, then lines are parallel and θ = 0°.
If m₁m₂ = –1, then lines are perpendicular and θ = 90°.

Slope from General Form of Line

For a line given in the general form ax + by + c = 0, rearranging:

y = (–a/b)x – c/b

This is of the form y = mx + constant, where slope m = –a/b.

Thus slope can be extracted directly without plotting or a two-point formula.

Detailed Solved Examples

Example 1: Find slope of line through (2, 3) and (5, 11).
Solution: m = (11 – 3)/(5 – 2) = 8/3.

Example 2: Find slope of line joining (–4, 7) and (2, –5).
Solution: m = (–5 – 7)/(2 – (–4)) = (–12)/6 = –2.

Example 3: Find slope of line parallel to 2x + 3y – 5 = 0.
Solution: Slope = –a/b = –2/3. So slope of any parallel line is –2/3.

Example 4: Find slope of line perpendicular to 7x – 4y + 2 = 0.
Solution: Slope of given line = –a/b = –7/(–4) = 7/4. Perpendicular slope = –1/m = –4/7.

Example 5: Are points (1, 2), (3, 6), (5, 10) collinear?
Slope of AB = (6 – 2)/(3 – 1) = 4/2 = 2.
Slope of BC = (10 – 6)/(5 – 3) = 4/2 = 2. Equal slopes → points collinear.

Example 6: Find angle between y = 2x + 3 and y = –½x + 4.
Slopes: m₁ = 2, m₂ = –½. Since m₁m₂ = –1, lines perpendicular, angle = 90°.

Example 7: Find slope of line 5x – 3y + 7 = 0.
Solution: Rearranged as y = (5/3)x + 7/3. Slope = 5/3.

Example 8: Find slope of line through (0, –2) and (4, –2).
Solution: m = (–2 – (–2))/(4 – 0) = 0/4 = 0. Horizontal line slope = 0.

Example 9: Find slope of line through (3, –2) and (3, 7).
Solution: m = (7 – (–2))/(3 – 3) = 9/0 → undefined slope.

Example 10: Find slope of line perpendicular to line through (1, 2) and (3, 6).
Slope of given line = (6 – 2)/(3 – 1) = 4/2 = 2. Perpendicular slope = –1/2.

Advanced Applications

Equation of a line: If slope m and point (x₁, y₁) are known, equation is y – y₁ = m(x – x₁).

Angle between lines: With slopes, tan θ = |(m₁ – m₂)/(1 + m₁m₂)|.

Parallel/perpendicular conditions: Slopes equal or product –1 helps quickly verify relationships.

Collinearity test: Equal slopes confirm if three points lie on the same straight line.

Physics connection: In a distance-time graph, slope gives velocity; in velocity-time graph, slope gives acceleration.

Mistakes Students Make

Forgetting slope undefined for vertical lines.
Mixing conditions: parallel (m₁ = m₂) vs perpendicular (m₁m₂ = –1).
Wrongly rearranging ax + by + c = 0 to slope form.
Ignoring denominator zero in slope formula.

Exam Weightage

CBSE/Boards: 3–4 marks, direct slope calculation or collinearity questions.
JEE Main: 1 MCQ, 2–4 marks.
KCET/COMEDK: 1–2 MCQs, each 2–3 marks.
JEE Advanced: 1 or more questions as part of 4–6 mark problems in coordinate geometry.