Chapter Introduction
Straight Lines is Chapter 9 of CBSE Class 11 Mathematics, falling under the Coordinate Geometry unit. This chapter builds on the coordinate geometry basics from Class 10 and introduces the algebraic study of straight lines. You will learn to find the slope, write equations in various forms, compute angles and distances, and apply conditions for parallel and perpendicular lines. This chapter typically carries 6-9 marks in board exams and is crucial for coordinate geometry in JEE.
The key idea: every geometric property of a straight line can be expressed as an algebraic equation, and every algebraic equation of the form Ax + By + C = 0 represents a straight line.
Key Definitions
| Term | Definition |
|---|---|
| Slope (m) | The tangent of the angle a line makes with the positive x-axis; m = tan(theta) |
| Inclination | The angle theta (0 <= theta < 180) that a line makes with the positive direction of the x-axis |
| Gradient | Another name for slope |
| Intercept | The distance from the origin to where the line cuts the x-axis (x-intercept) or y-axis (y-intercept) |
| Locus | The set of all points satisfying a given geometric condition |
| Normal form | Equation of a line expressed using the perpendicular distance p from the origin and the angle alpha the perpendicular makes with the x-axis |
Detailed Concepts
1. Slope of a Line
When inclination theta is given: m = tan(theta)
- If theta = 0 degrees: m = 0 (horizontal line)
- If theta = 90 degrees: slope is undefined (vertical line)
When two points are given: m = (y2 - y1)/(x2 - x1)
When the equation is given: For Ax + By + C = 0, slope m = -A/B
2. Angle Between Two Lines
If two lines have slopes m1 and m2:
| **tan(theta) = | (m2 - m1)/(1 + m1*m2) | ** |
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Special cases:
- If m1*m2 = -1: lines are perpendicular, angle = 90 degrees
- If m1 = m2: lines are parallel (or coincident), angle = 0 degrees
- If 1 + m1*m2 = 0: lines are perpendicular
| For lines in general form: If a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then tan(theta) = | (a1b2 - a2b1)/(a1a2 + b1b2) |
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3. Conditions for Parallel and Perpendicular Lines
| Condition | For slopes | For general form |
|---|---|---|
| Parallel | m1 = m2 | A1/A2 = B1/B2 != C1/C2 |
| Coincident | m1 = m2 and same y-intercept | A1/A2 = B1/B2 = C1/C2 |
| Perpendicular | m1 * m2 = -1 | A1A2 + B1B2 = 0 |
| Intersecting | m1 != m2 | A1/A2 != B1/B2 |
4. Forms of Line Equation
| Form | Equation | When to use |
|---|---|---|
| Point-Slope | y - y1 = m(x - x1) | When one point and slope are known |
| Two-Point | (y - y1)/(y2 - y1) = (x - x1)/(x2 - x1) | When two points are known |
| Slope-Intercept | y = mx + c | When slope and y-intercept are known |
| Intercept | x/a + y/b = 1 | When x-intercept a and y-intercept b are known |
| Normal | x cos(alpha) + y sin(alpha) = p | When perpendicular distance p from origin and angle alpha are known |
| General | Ax + By + C = 0 | Standard form; all other forms reduce to this |
Converting to general form: Every form can be rewritten as Ax + By + C = 0. For example, y = mx + c becomes mx - y + c = 0, and x/a + y/b = 1 becomes bx + ay - ab = 0.
5. Distance Formulae
Distance of a point from a line:
| d = | Ax1 + By1 + C | / sqrt(A^2 + B^2) |
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Distance between parallel lines:
For Ax + By + C1 = 0 and Ax + By + C2 = 0:
| d = | C1 - C2 | / sqrt(A^2 + B^2) |
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Important: the coefficients of x and y must be identical before applying this formula. If they are not, first normalize one equation.
6. Locus
The locus of a point is the path traced by a point moving under a given geometric condition. To find the equation of the locus:
1. Let the moving point be (x, y) or (h, k). 2. Express the given geometric condition as an equation in x and y. 3. Simplify to get the locus equation.
Common loci:
- Point equidistant from two given points: perpendicular bisector of the segment joining them
- Point equidistant from two intersecting lines: angle bisectors of those lines
- Point at a fixed distance from a given point: circle
Forms of Line Equation — Detailed
| Form | Given Data | Equation | Slope | y-intercept |
|---|---|---|---|---|
| Point-Slope | (x1, y1), m | y - y1 = m(x - x1) | m | y1 - m*x1 |
| Two-Point | (x1, y1), (x2, y2) | (y-y1)/(y2-y1) = (x-x1)/(x2-x1) | (y2-y1)/(x2-x1) | (x2y1 - x1y2)/(x2-x1) |
| Slope-Intercept | m, c | y = mx + c | m | c |
| Intercept | a, b | x/a + y/b = 1 | -b/a | b |
| Normal | p, alpha | x cos alpha + y sin alpha = p | -cos(alpha)/sin(alpha) | p/sin(alpha) |
Important Numericals
Example 1: Finding the equation of a line
Find the equation of the line passing through (2, 3) with slope 4.
Using point-slope form: y - 3 = 4(x - 2)
y - 3 = 4x - 8
4x - y - 5 = 0
Example 2: Angle between lines
Find the angle between lines y = 2x + 3 and y = 3x - 1.
m1 = 2, m2 = 3
| tan(theta) = | (3 - 2)/(1 + 2*3) | = | 1/7 | = 1/7 |
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theta = arctan(1/7)
Example 3: Distance of a point from a line
Find the distance of point (1, 2) from line 3x + 4y - 12 = 0.
| d = | 3(1) + 4(2) - 12 | / sqrt(9 + 16) = | 3 + 8 - 12 | / 5 = | -1 | /5 = 1/5 |
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Example 4: Parallel line through a point
Find the equation of the line parallel to 2x + 3y - 7 = 0 and passing through (1, 2).
Slope of given line: m = -2/3. Parallel line has same slope.
y - 2 = (-2/3)(x - 1)
3y - 6 = -2x + 2
2x + 3y - 8 = 0
Example 5: Distance between parallel lines
Find the distance between 3x + 4y + 5 = 0 and 3x + 4y + 15 = 0.
Coefficients of x and y are already same.
| d = | 5 - 15 | / sqrt(9 + 16) = 10/5 = 2 |
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Common Mistakes to Avoid
1. Undefined slope: A vertical line has undefined slope, not "infinite slope." Write "slope is undefined" or "the line is parallel to y-axis."
| 2. Sign errors in angle formula: tan(theta) = | (m2 - m1)/(1 + m1*m2) | . The absolute value ensures a positive angle. If 1 + m1*m2 = 0, the lines are perpendicular; do not divide by zero. |
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3. Confusing parallel and coincident: If A1/A2 = B1/B2 = C1/C2, lines are coincident (same line). If A1/A2 = B1/B2 != C1/C2, lines are parallel and distinct. Always check all three ratios.
| 4. Not normalizing before distance between parallel lines: The formula d = | C1 - C2 | /sqrt(A^2 + B^2) requires that the coefficients of x and y are the same in both equations. If 2x + 4y + 3 = 0 and x + 2y + 5 = 0, first normalize the second to 2x + 4y + 10 = 0. |
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5. Wrong intercept form: x/a + y/b = 1 means x-intercept is a and y-intercept is b. Do not write x/a + y/b = 0 or confuse the signs.
6. Slope from general form: For Ax + By + C = 0, slope is m = -A/B (not A/B or -B/A).
7. Locus problems without simplification: After writing the geometric condition, always simplify to standard form. Do not leave the locus equation unsimplified.
Exam Tips
- For 1-mark questions, know: slope formula, conditions for parallel/perpendicular, distance formula.
- For 3-mark questions, practice: finding line equations in all forms, angle between lines, distance calculations.
- For 5-mark questions, practice: locus problems, converting between forms, and problems combining multiple concepts (e.g., find intersection, then distance from a point).
- When finding the equation of a line, state which form you are using.
- For distance problems, always write the formula first, then substitute.
- In problems with multiple conditions (e.g., "parallel to one line and passing through a point"), handle one condition at a time.
- Draw a rough sketch for geometry problems — it helps verify your answer.
NCERT Exercise Problem Patterns
| Exercise | Topic | Typical Question Type |
|---|---|---|
| 9.1 | Slope, angle between lines | Find slope, check parallel/perpendicular |
| 9.2 | Forms of line equation | Point-slope, two-point, slope-intercept, intercept form |
| 9.3 | General equation, distance | Convert forms, distance of point from line |
| 9.4 | Distance between parallel lines, locus | Distance between lines, locus problems |
Derivations to Remember
Derivation 1: Angle between two lines
Let lines L1 and L2 make angles alpha1 and alpha2 with x-axis, so m1 = tan(alpha1), m2 = tan(alpha2).
The angle between them: theta = alpha2 - alpha1
tan(theta) = tan(alpha2 - alpha1) = (tan(alpha2) - tan(alpha1))/(1 + tan(alpha1)tan(alpha2)) = (m2 - m1)/(1 + m1m2)
| Since we want the acute angle: tan(theta) = | (m2 - m1)/(1 + m1*m2) |
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Derivation 2: Distance of a point from a line
For line Ax + By + C = 0, find the distance of P(x1, y1).
The perpendicular from P meets the line at Q. The equation of PQ (perpendicular to line) has slope B/A.
| Solving the line and the perpendicular through P simultaneously gives Q, then distance PQ = | Ax1 + By1 + C | /sqrt(A^2 + B^2). |
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Derivation 3: Normal form derivation
The line at distance p from origin, with perpendicular making angle alpha with x-axis.
The foot of perpendicular from origin on the line is (p cos(alpha), p sin(alpha)).
Since the line is perpendicular to the line from origin to (p cos(alpha), p sin(alpha)), its equation is x cos(alpha) + y sin(alpha) = p.
Additional Solved Examples
Example 6: Perpendicular line through a point
Find the equation of the line perpendicular to 3x - 4y + 7 = 0 and passing through (2, 1).
Slope of given line: m1 = 3/4. Perpendicular slope: m2 = -4/3
y - 1 = (-4/3)(x - 2)
3y - 3 = -4x + 8
4x + 3y - 11 = 0
Example 7: Intersection of lines and distance
Find the intersection of 2x + 3y = 6 and 3x - y = 5, then its distance from origin.
From second equation: y = 3x - 5. Substituting: 2x + 3(3x - 5) = 6 => 11x = 21 => x = 21/11
y = 3(21/11) - 5 = 63/11 - 55/11 = 8/11. Intersection: (21/11, 8/11)
Distance from origin: sqrt((21/11)^2 + (8/11)^2) = sqrt(441 + 64)/11 = sqrt(505)/11
Example 8: Converting to normal form
Convert 3x + 4y - 12 = 0 to normal form.
Divide by sqrt(9 + 16) = 5: (3/5)x + (4/5)y = 12/5
So cos(alpha) = 3/5, sin(alpha) = 4/5, p = 12/5
alpha = arctan(4/3), distance from origin = 12/5 = 2.4
Practice Checklist
- Find equations of lines using all 5 forms (one problem each)
- Solve 3 angle-between-lines problems
- Solve 3 distance problems (point from line, between parallel lines)
- Attempt 2 locus problems
- Practice converting between forms (at least 4 conversions)
- Draw a rough sketch for every problem
Quick Revision Notes
1. Slope: m = (y2-y1)/(x2-x1) = tan(theta); for Ax+By+C=0, m = -A/B
3. Parallel: m1 = m2 (or A1/A2 = B1/B2); Perpendicular: m1*m2 = -1 (or A1A2+B1B2=0) 4. Point-Slope: y - y1 = m(x - x1) 5. Two-Point: (y-y1)/(y2-y1) = (x-x1)/(x2-x1) 6. Slope-Intercept: y = mx + c 7. Intercept: x/a + y/b = 1 8. Normal: x cos(alpha) + y sin(alpha) = p 9. General: Ax + By + C = 0
12. Locus: express geometric condition as equation in x, y
| 2. Angle between lines: tan(theta) = | (m2-m1)/(1+m1*m2) | |
|---|---|---|
| 10. Distance of point from line: d = | Ax1+By1+C | /sqrt(A^2+B^2) |
| 11. Distance between parallel lines: d = | C1-C2 | /sqrt(A^2+B^2) |