1. Chapter at a Glance
A road climbing a hill, a ramp for wheelchairs, the path of light in a straight beam — straight lines are everywhere. In coordinate geometry a line is described by an equation, and its steepness by a single number, the slope. This chapter builds the different forms of the equation of a line, conditions for parallel and perpendicular lines, the angle between two lines, and the distance of a point from a line — basic tools for all of analytical geometry and for JEE.
- Recap: distance, section and midpoint formulas.
- Inclination and slope of a line; slope through two points.
- Conditions for parallel and perpendicular lines; collinearity.
- Angle between two lines.
- Equations: horizontal/vertical, point-slope, two-point, slope-intercept, intercept form.
- General equation Ax + By + C = 0 and its slope and intercepts.
- Distance of a point from a line; distance between parallel lines.
2. Slope of a Line
The inclination θ of a line is the angle it makes with the positive x-axis, measured anticlockwise (0° ≤ θ < 180°). Its slope is m = tan θ (θ ≠ 90°). A vertical line has undefined slope.
| Slope | Inclination | Line |
|---|---|---|
| m > 0 | 0° < θ < 90° | Rises left to right |
| m < 0 | 90° < θ < 180° | Falls |
| m = 0 | θ = 0° | Horizontal |
| Undefined | θ = 90° | Vertical |
- Non-vertical lines are parallel ⇔ m₁ = m₂.
- Non-vertical lines are perpendicular ⇔ m₁m₂ = −1.
- Points A, B, C are collinear ⇔ slope AB = slope BC.
This gives the acute angle θ; the obtuse angle is 180° − θ.
Find the acute angle between two lines with slopes m₁ = −2 and m₂ = 1/3.
Solution: tan θ = |(1/3 + 2)/(1 − 2/3)| = |(7/3)/(1/3)| = 7 → θ = tan⁻¹ 7 ≈ 81.9°.
3. Forms of the Equation of a Line
| Form | Given | Equation |
|---|---|---|
| Horizontal / vertical | Through (a, b) | y = b / x = a |
| Point-slope | Point (x₀, y₀), slope m | y − y₀ = m(x − x₀) |
| Two-point | (x₁, y₁), (x₂, y₂) | y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁) |
| Slope-intercept | Slope m, y-intercept c | y = mx + c |
| Slope, x-intercept d | m, d | y = m(x − d) |
| Intercept form | x-intercept a, y-intercept b | x/a + y/b = 1 |
Find the equation of the line through (−2, 3) with slope −4.
Solution: y − 3 = −4(x + 2) → 4x + y + 5 = 0.
Find the line through (1, −1) and (3, 5).
Solution: m = 6/2 = 3. y + 1 = 3(x − 1) → 3x − y − 4 = 0.
Find the line that cuts off intercepts 3 and −2 on the axes.
Solution: x/3 + y/(−2) = 1 → 2x − 3y = 6.
4. General Equation
Every line can be written as Ax + By + C = 0 (A, B not both zero), and every such equation is a line.
| Quantity | Value (B ≠ 0 where needed) |
|---|---|
| Slope | −A/B |
| y-intercept | −C/B |
| x-intercept | −C/A |
| Parallel line | Ax + By + k = 0 |
| Perpendicular line | Bx − Ay + k = 0 |
Find the line through (2, 3) perpendicular to 3x − 4y + 5 = 0.
Solution: Perpendicular form 4x + 3y + k = 0. Through (2, 3): 8 + 9 + k = 0 → k = −17. 4x + 3y − 17 = 0.
5. Distances
Distance of (x₁, y₁) from Ax + By + C = 0:
Ax + By + C₁ = 0 and Ax + By + C₂ = 0: d = |C₁ − C₂| / √(A² + B²).
Find the distance of (3, −5) from 3x − 4y − 26 = 0, and between 3x − 4y + 7 = 0 and 3x − 4y + 5 = 0.
Solution: |9 + 20 − 26|/5 = 3/5. Parallel lines: |7 − 5|/5 = 2/5.
6. Solved Problems (Board Pattern)
Show that (−2, −1), (4, 0), (3, 3) and (−3, 2) are vertices of a parallelogram.
Solution: Slopes: AB = 1/6, CD = (2 − 3)/(−6) = 1/6 → AB ∥ CD. BC = 3/(−1) = −3, DA = (−1 − 2)/(−2 + 3) = −3 → BC ∥ DA. Both pairs parallel → parallelogram. ∎
Find the perpendicular bisector of the segment joining (1, 0) and (3, 5).
Solution: Midpoint (2, 5/2); slope of segment 5/2 → perpendicular slope −2/5. y − 5/2 = −(2/5)(x − 2) → 4x + 10y − 33 = 0.
Find the foot of the perpendicular from (−1, 3) to 3x − 4y − 16 = 0.
Solution: Perpendicular through (−1, 3): 4x + 3y − 5 = 0. Solve with 3x − 4y = 16: x = 68/25, y = −49/25. Foot (68/25, −49/25).
7. Master Formula Table
| Result | Formula |
|---|---|
| Slope | m = tan θ = (y₂ − y₁)/(x₂ − x₁) |
| Parallel / perpendicular | m₁ = m₂ / m₁m₂ = −1 |
| Angle between lines | tan θ = |(m₂ − m₁)/(1 + m₁m₂)| |
| Point-slope | y − y₀ = m(x − x₀) |
| Slope-intercept | y = mx + c |
| Intercept form | x/a + y/b = 1 |
| General form | Ax + By + C = 0, slope −A/B |
| Point to line | |Ax₁ + By₁ + C|/√(A² + B²) |
| Parallel lines | |C₁ − C₂|/√(A² + B²) |
8. Master Summary Table
| Given | Best form | Example use |
|---|---|---|
| Point + slope | Point-slope | Tangent-like problems |
| Two points | Two-point | Side of a triangle |
| Intercepts | Intercept form | Area of triangle with axes |
| Parallel/perpendicular to a line | Ax + By + k / Bx − Ay + k | Altitudes, bisectors |
| Distance | Point-line formula | Heights of triangles |
9. Mind Map
- tan θ
- ∥ and ⊥ tests
- Point-slope, two-point
- y = mx + c, x/a + y/b = 1
- Ax + By + C = 0
- Slope −A/B
- Point to line
- Parallel lines
10. Tips and Tricks
- To write a line parallel to Ax + By + C = 0, keep A and B and change only C.
- To write a perpendicular line, swap A and B and change one sign.
- Area of the triangle a line x/a + y/b = 1 forms with the axes = ½|ab|.
- Slope of Ax + By + C = 0 is −A/B, not A/B.
- Forgetting the modulus in the distance formula.
- Using the parallel-lines formula when x and y coefficients are not identical (make them equal first).
11. Important One-liners
- Slope = tan(inclination).
- Horizontal lines have slope 0; vertical lines have undefined slope.
- Perpendicular slopes multiply to −1.
- Every linear equation in x and y represents a straight line.
- Distance from the origin to Ax + By + C = 0 is |C|/√(A² + B²).
12. Exam-pattern Questions with Detailed Answers
The slope of the line through (3, −2) and (−1, 4) is:
- −3/2
- 3/2
- −2/3
- 2/3
Show answer
(4 + 2)/(−1 − 3) = −3/2.
Answer: (A)
Find the slope and intercepts of 3x + 2y − 12 = 0.
Show answer
Slope −3/2; x-intercept 4; y-intercept 6.
Answer: −3/2; 4; 6
Write the equation of the line with inclination 45° and y-intercept 2.
Show answer
m = 1: y = x + 2.
Answer: y = x + 2
Find x if the line through (x, −1) and (2, 1) is parallel to the line through (5, 3) and (4, 1).
Show answer
Second slope (1 − 3)/(4 − 5) = 2. First: 2/(2 − x) = 2 → x = 1.
Answer: x = 1
Find the angle between x − √3y = 5 and √3x + y = 1.
Show answer
m₁ = 1/√3, m₂ = −√3; m₁m₂ = −1 → 90°.
Answer: 90°
Find the equation of the line through (−3, 5) perpendicular to the line through (2, 5) and (−3, 6).
Show answer
Given slope (6 − 5)/(−5) = −1/5; perpendicular slope 5. y − 5 = 5(x + 3) → 5x − y + 20 = 0.
Answer: 5x − y + 20 = 0
Find the distance between 15x + 8y − 34 = 0 and 15x + 8y + 31 = 0.
Show answer
|−34 − 31|/17 = 65/17.
Answer: 65/17
Find the line whose intercepts are equal and which passes through (2, 3).
Show answer
x/a + y/a = 1 → x + y = a; 2 + 3 = 5 → x + y = 5.
Answer: x + y = 5
Find the equation of the line passing through (2, 2) and cutting off intercepts on the axes whose sum is 9.
Show answer
x/a + y/(9 − a) = 1 with (2, 2): 2/a + 2/(9 − a) = 1 → 18 = a(9 − a) → a² − 9a + 18 = 0 → a = 3 or 6. Lines: x/3 + y/6 = 1 (2x + y = 6) and x/6 + y/3 = 1 (x + 2y = 6).
Answer: 2x + y = 6 or x + 2y = 6
Find the equation of the altitude from A of △ABC with A(2, 3), B(4, −1), C(1, 2), and its length.
Show answer
BC slope (2 + 1)/(1 − 4) = −1; BC: x + y − 3 = 0. Altitude slope 1 through A: y − 3 = x − 2 → x − y + 1 = 0. Length = |2 + 3 − 3|/√2 = √2.
Answer: x − y + 1 = 0; length √2
Find the points on the x-axis at distance 4 from the line x/3 + y/4 = 1.
Show answer
Line 4x + 3y − 12 = 0. |4a − 12|/5 = 4 → 4a − 12 = ±20 → a = 8 or −2. Points (8, 0), (−2, 0).
Answer: (8, 0) and (−2, 0)
If p is the length of the perpendicular from the origin to x/a + y/b = 1, prove that 1/p² = 1/a² + 1/b².
Show answer
Line bx + ay − ab = 0. p = |−ab|/√(a² + b²) → p² = a²b²/(a² + b²) → 1/p² = (a² + b²)/(a²b²) = 1/b² + 1/a². ∎
Answer: Proved
13. Quick Revision Summary
- m = tan θ = rise/run; parallel ⇔ equal slopes; perpendicular ⇔ m₁m₂ = −1.
- Angle between lines: tan θ = |(m₂ − m₁)/(1 + m₁m₂)|.
- Forms: point-slope, two-point, y = mx + c, x/a + y/b = 1.
- General Ax + By + C = 0: slope −A/B.
- Distance from a point: |Ax₁ + By₁ + C|/√(A² + B²); parallel lines: |C₁ − C₂|/√(A² + B²).