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Straight Lines

NCERT Class 11 · Mathematics Based on NCERT Class 11 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter builds on two-dimensional coordinate geometry to systematically study the properties of the straight line, the simplest geometric figure. It introduces the concept of slope as the fundamental measure of a line's direction and derives conditions for parallelism and perpendicularity. The core idea is to represent a line algebraically using various forms of its equation (point-slope, two-point, slope-intercept, intercept, and general form) and to compute distances from a point to a line and between parallel lines. A student must walk away with the ability to write the equation of any line given sufficient conditions and to solve geometric problems using algebraic methods.
Key Concepts (12)
Inclination of a line
The angle (θ) made by a line with the positive direction of the x-axis, measured anticlockwise. It lies between 0° and 180°.
e.g. A line parallel to the x-axis has inclination 0°; a vertical line has inclination 90°.
Slope or gradient of a line
If θ is the inclination of a line, then tan θ is called the slope or gradient of the line. It is denoted by m. The slope of a vertical line (θ = 90°) is not defined.
e.g. A line making an angle of 60° with the positive x-axis has slope m = tan 60° = √3.
Slope of a line when two points are given
The slope m of a non-vertical line passing through points (x₁, y₁) and (x₂, y₂) is given by m = (y₂ - y₁) / (x₂ - x₁), provided x₁ ≠ x₂.
e.g. For points (3, -2) and (-1, 4), slope m = (4 - (-2)) / (-1 - 3) = 6 / -4 = -3/2.
Condition for parallelism of two lines
Two non-vertical lines are parallel if and only if their slopes are equal, i.e., m₁ = m₂.
e.g. Lines y = 2x + 3 and y = 2x - 5 are parallel because both have slope 2.
Condition for perpendicularity of two lines
Two non-vertical lines are perpendicular if and only if the product of their slopes is -1, i.e., m₁ m₂ = -1, or m₂ = -1/m₁.
e.g. A line with slope 3/2 is perpendicular to a line with slope -2/3 because (3/2) * (-2/3) = -1.
Angle between two lines
The acute angle θ between two lines with slopes m₁ and m₂ is given by tan θ = |(m₂ - m₁) / (1 + m₁ m₂)|, provided 1 + m₁ m₂ ≠ 0.
e.g. If m₁ = 1/2 and m₂ = 3, then tan θ = |(3 - 1/2) / (1 + (1/2)*3)| = |(5/2) / (5/2)| = 1, so θ = 45°.
Collinear points
Three points A, B, and C are collinear if the area of the triangle formed by them is zero, or equivalently, if the slope of AB equals the slope of BC.
e.g. Points (3, 0), (-2, -2), and (8, 2) are collinear because slope between any two pairs is equal.
x-intercept and y-intercept
The x-intercept is the x-coordinate of the point where a line meets the x-axis (y = 0). The y-intercept is the y-coordinate of the point where a line meets the y-axis (x = 0).
e.g. For the line 3x + 2y - 12 = 0, x-intercept is 4 (put y=0) and y-intercept is 6 (put x=0).
General equation of a line
Any equation of the form Ax + By + C = 0, where A and B are not both zero simultaneously, is called the general linear equation or general equation of a line.
e.g. 3x - 4y - 26 = 0 is the general equation of a line with A=3, B=-4, C=-26.
Distance of a point from a line
The perpendicular distance d of a point (x₁, y₁) from the line Ax + By + C = 0 is given by d = |Ax₁ + By₁ + C| / √(A² + B²).
e.g. Distance of point (3, -5) from line 3x - 4y - 26 = 0 is |3*3 + (-4)*(-5) - 26| / √(9+16) = |9+20-26|/5 = 3/5.
Distance between two parallel lines
The distance d between two parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is given by d = |C₁ - C₂| / √(A² + B²).
e.g. Distance between lines 3x - 4y + 7 = 0 and 3x - 4y + 5 = 0 is |7 - 5| / √(9+16) = 2/5.
Concurrent lines
Three or more lines are said to be concurrent if they all pass through a common point (the point of concurrency).
e.g. Lines 2x + y - 3 = 0, 5x + ky - 3 = 0, and 3x - y - 2 = 0 are concurrent for k = -2, as they all pass through (1,1).
Formulas & Equations (13)
m = tan θ, θ ≠ 90°
Reads as: m equals tan theta, theta not equal to 90 degrees
Variables: m = slope of the line; θ = inclination of the line (angle with positive x-axis, measured anticlockwise)
Use when: Finding slope when inclination is given, or finding inclination when slope is known.
⚠ Students forget that slope is undefined for θ = 90° (vertical line) and write m = tan 90° as infinite or a number.
m = (y₂ - y₁) / (x₂ - x₁), x₁ ≠ x₂
Reads as: m equals y-two minus y-one over x-two minus x-one, x-one not equal to x-two
Variables: m = slope; (x₁, y₁), (x₂, y₂) = coordinates of two points on the line
Use when: Finding slope of a line given two points on it.
⚠ Students often reverse the order, writing (x₂ - x₁)/(y₂ - y₁), or forget to ensure x₁ ≠ x₂.
m₁ = m₂ (for parallel lines)
Reads as: m-one equals m-two for parallel lines
Variables: m₁, m₂ = slopes of two non-vertical lines
Use when: Checking if two lines are parallel, or finding slope of a line parallel to a given line.
⚠ Students apply this condition to vertical lines, forgetting that their slopes are undefined.
m₁ m₂ = -1 (for perpendicular lines)
Reads as: m-one times m-two equals negative one for perpendicular lines
Variables: m₁, m₂ = slopes of two non-vertical lines
Use when: Checking if two lines are perpendicular, or finding slope of a line perpendicular to a given line.
⚠ Students write m₁ = -m₂ instead of m₁ m₂ = -1, or forget that one slope is the negative reciprocal of the other.
tan θ = |(m₂ - m₁) / (1 + m₁ m₂)|, 1 + m₁ m₂ ≠ 0
Reads as: tan theta equals the absolute value of m-two minus m-one over one plus m-one m-two, one plus m-one m-two not equal to zero
Variables: θ = acute angle between the two lines; m₁, m₂ = slopes of the lines
Use when: Finding the acute angle between two lines given their slopes.
⚠ Students forget the absolute value and the condition 1 + m₁ m₂ ≠ 0, or use the formula for obtuse angle directly.
y - y₀ = m (x - x₀)
Reads as: y minus y-naught equals m times x minus x-naught
Variables: (x₀, y₀) = fixed point on the line; m = slope of the line; (x, y) = any point on the line
Use when: Writing equation of a line when slope and one point on the line are given (point-slope form).
⚠ Students incorrectly substitute the given point as (x, y) instead of (x₀, y₀), leading to sign errors.
y - y₁ = ((y₂ - y₁) / (x₂ - x₁)) (x - x₁)
Reads as: y minus y-one equals y-two minus y-one over x-two minus x-one times x minus x-one
Variables: (x₁, y₁), (x₂, y₂) = two given points on the line
Use when: Writing equation of a line passing through two given points (two-point form).
⚠ Students mix up the points, using x₂ - x₁ in numerator and y₂ - y₁ in denominator.
y = mx + c
Reads as: y equals m x plus c
Variables: m = slope of the line; c = y-intercept (distance from origin where line cuts y-axis); (x, y) = any point on the line
Use when: Writing equation of a line when slope and y-intercept are given (slope-intercept form).
⚠ Students confuse c with x-intercept, or forget that c can be negative if intercept is on negative y-axis.
y = m (x - d)
Reads as: y equals m times x minus d
Variables: m = slope of the line; d = x-intercept (distance from origin where line cuts x-axis)
Use when: Writing equation of a line when slope and x-intercept are given.
⚠ Students write y = mx - d instead of y = m(x - d), missing the distributive property.
x/a + y/b = 1
Reads as: x over a plus y over b equals one
Variables: a = x-intercept; b = y-intercept
Use when: Writing equation of a line when both x-intercept and y-intercept are given (intercept form).
⚠ Students write x/a + y/b = 0 or forget that a and b are intercepts, not coordinates of a point.
Ax + By + C = 0
Reads as: A x plus B y plus C equals zero
Variables: A, B, C = real constants, with A and B not both zero
Use when: Representing any line in general form; used for distance formulas and converting between forms.
⚠ Students forget to ensure A and B are not both zero, or misidentify A, B, C when the equation is not simplified.
d = |Ax₁ + By₁ + C| / √(A² + B²)
Reads as: d equals the absolute value of A x-one plus B y-one plus C over the square root of A-squared plus B-squared
Variables: d = perpendicular distance; (x₁, y₁) = point; A, B, C = coefficients from line Ax + By + C = 0
Use when: Finding the perpendicular distance from a point to a line.
⚠ Students forget the absolute value in the numerator, or forget to take the square root of A² + B².
d = |C₁ - C₂| / √(A² + B²)
Reads as: d equals the absolute value of C-one minus C-two over the square root of A-squared plus B-squared
Variables: d = distance between parallel lines; A, B = coefficients from lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0
Use when: Finding the distance between two parallel lines given in general form.
⚠ Students forget to ensure the lines are in the same form (same A and B coefficients) before applying the formula.
How It Works
Finding slope of a line given two points
  1. Step 1: Identify the coordinates of the two points as (x₁, y₁) and (x₂, y₂).
  2. Step 2: Ensure x₁ ≠ x₂ (if equal, slope is undefined and line is vertical).
  3. Step 3: Apply the formula m = (y₂ - y₁) / (x₂ - x₁).
  4. Step 4: Simplify the fraction to get the slope.
💡 Always subtract in the same order: y₂ - y₁ over x₂ - x₁. Reversing the order in only one coordinate gives the wrong sign.
Checking if two lines are parallel or perpendicular using slopes
  1. Step 1: Find the slopes m₁ and m₂ of the two lines.
  2. Step 2: For parallelism, check if m₁ = m₂. If both slopes are undefined (vertical lines), they are also parallel.
  3. Step 3: For perpendicularity, check if m₁ m₂ = -1. If one slope is 0 and the other is undefined, they are perpendicular.
💡 Vertical lines (undefined slope) are parallel to each other and perpendicular to horizontal lines (slope 0). Do not apply the product condition blindly.
Finding the acute angle between two lines
  1. Step 1: Determine the slopes m₁ and m₂ of the two lines.
  2. Step 2: Ensure 1 + m₁ m₂ ≠ 0 (if zero, lines are perpendicular and angle is 90°).
  3. Step 3: Apply the formula tan θ = |(m₂ - m₁) / (1 + m₁ m₂)|.
  4. Step 4: Use inverse tangent to find θ (acute angle).
💡 The formula gives the acute angle. If the computed tan θ is negative, take its absolute value. The obtuse angle is 180° - θ.
Writing the equation of a line in different forms
  1. Step 1: Identify the given information (point and slope, two points, intercepts, etc.).
  2. Step 2: Choose the appropriate form: point-slope (y - y₀ = m(x - x₀)), two-point, slope-intercept (y = mx + c), intercept (x/a + y/b = 1), or general (Ax + By + C = 0).
  3. Step 3: Substitute the given values into the chosen form.
  4. Step 4: Simplify the equation to the required form (often general form).
💡 Always check if the line is vertical or horizontal first, as these have special equations (x = constant or y = constant).
Finding the distance of a point from a line
  1. Step 1: Write the equation of the line in general form Ax + By + C = 0.
  2. Step 2: Identify the coordinates of the point (x₁, y₁).
  3. Step 3: Apply the formula d = |Ax₁ + By₁ + C| / √(A² + B²).
  4. Step 4: Compute the absolute value and simplify.
💡 The line must be in the form Ax + By + C = 0. If the equation is given in another form, convert it first. Do not forget the absolute value.
Finding the distance between two parallel lines
  1. Step 1: Ensure both lines are in general form Ax + By + C₁ = 0 and Ax + By + C₂ = 0 (same A and B).
  2. Step 2: If not, rewrite them so that coefficients of x and y are identical.
  3. Step 3: Apply the formula d = |C₁ - C₂| / √(A² + B²).
  4. Step 4: Simplify to get the distance.
💡 The lines must have the same A and B coefficients. If they are given in slope-intercept form (y = mx + c₁ and y = mx + c₂), use d = |c₁ - c₂| / √(1 + m²).
Checking concurrency of three lines
  1. Step 1: Find the point of intersection of any two lines by solving their equations simultaneously.
  2. Step 2: Substitute this point into the equation of the third line.
  3. Step 3: If the point satisfies the third equation, the lines are concurrent; otherwise, they are not.
💡 If any two lines are parallel, they cannot be concurrent with a third line (unless all three are the same line).
Exam Traps — Don't Make These Mistakes
⚠ Confusing x-intercept and y-intercept in intercept form
✗ For a line with x-intercept 3 and y-intercept 4, students write the equation as x/3 + y/4 = 0.
→ The intercept form is x/a + y/b = 1, not = 0. The right-hand side must be 1.
✓ Write x/3 + y/4 = 1, which simplifies to 4x + 3y = 12.
⚠ Using the wrong formula for angle between lines
✗ Students write tan θ = (m₂ - m₁) / (1 - m₁ m₂) or forget the absolute value.
→ The correct formula has (1 + m₁ m₂) in the denominator, not (1 - m₁ m₂). The absolute value ensures the acute angle.
✓ Use tan θ = |(m₂ - m₁) / (1 + m₁ m₂)|. If 1 + m₁ m₂ = 0, the lines are perpendicular (θ = 90°).
⚠ Forgetting to convert to general form before using distance formula
✗ To find distance of point (1,2) from line y = 2x + 3, students directly substitute into d = |Ax₁ + By₁ + C| / √(A² + B²) without rewriting.
→ The line must be in the form Ax + By + C = 0. y = 2x + 3 must be rewritten as 2x - y + 3 = 0, so A=2, B=-1, C=3.
✓ First rewrite y = 2x + 3 as 2x - y + 3 = 0. Then apply d = |2(1) - 1(2) + 3| / √(2² + (-1)²) = |2 - 2 + 3|/√5 = 3/√5.
⚠ Assuming all lines have a defined slope
✗ When asked for slope of line through (2,3) and (2,5), students write m = (5-3)/(2-2) = 2/0 = 0 or infinity.
→ Division by zero is undefined. The slope of a vertical line is not defined; it is not 0 or infinity.
✓ Since x-coordinates are equal (x₁ = x₂ = 2), the line is vertical and its slope is not defined.
⚠ Misapplying the condition for perpendicular lines
✗ If a line has slope 2, students say a perpendicular line has slope -2.
→ Perpendicular slopes are negative reciprocals, not just negatives. The product must be -1, so the perpendicular slope is -1/2, not -2.
✓ If m₁ = 2, then m₂ = -1/m₁ = -1/2. Check: 2 * (-1/2) = -1.
⚠ Forgetting the absolute value in distance between parallel lines
✗ For lines 3x + 4y - 5 = 0 and 3x + 4y + 7 = 0, students write d = (-5 - 7)/5 = -12/5.
→ Distance cannot be negative. The formula includes an absolute value: d = |C₁ - C₂| / √(A² + B²).
✓ d = |(-5) - 7| / √(3² + 4²) = |-12|/5 = 12/5 units.
Compare & Contrast
Information required to write the equation
Point-slope form (y - y₀ = m(x - x₀))
Slope-intercept form (y = mx + c)
Point-slope form requires a point (x₀, y₀) and the slope m; slope-intercept form requires the slope m and the y-intercept c. Slope-intercept is a special case of point-slope where the point is (0, c).
Relationship between slopes
Parallel lines (m₁ = m₂)
Perpendicular lines (m₁ m₂ = -1)
Parallel lines have equal slopes; perpendicular lines have slopes that are negative reciprocals (product = -1). A line parallel to a given line has the same slope; a line perpendicular to it has slope = -1/m.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 11 marks
Most asked: Finding equation of a line in various forms (point-slope, two-point, slope-intercept, intercept) · Conditions for parallelism and perpendicularity of lines · Distance of a point from a line · Distance between two parallel lines · Angle between two lines · Concurrency of three lines
Don't skip: Word problems involving finding equation of a line under given conditions (e.g., passing through a point, making intercepts, etc.); Problems combining distance formula and line equations (e.g., foot of perpendicular, image of a point); Proof-based questions (e.g., showing collinearity, concurrency, or area of triangle formed by lines)
One-Line Revision (read aloud before the exam)
  1. Slope m of a line through (x₁, y₁) and (x₂, y₂) is m = (y₂ - y₁)/(x₂ - x₁), undefined for vertical lines.
  2. Two non-vertical lines are parallel if their slopes are equal: m₁ = m₂.
  3. Two non-vertical lines are perpendicular if product of slopes is -1: m₁ m₂ = -1.
  4. Acute angle θ between lines with slopes m₁ and m₂: tan θ = |(m₂ - m₁)/(1 + m₁ m₂)|.
  5. Point-slope form: y - y₀ = m(x - x₀).
  6. Two-point form: y - y₁ = ((y₂ - y₁)/(x₂ - x₁))(x - x₁).
  7. Slope-intercept form: y = mx + c, where c is y-intercept.
  8. Intercept form: x/a + y/b = 1, where a and b are x- and y-intercepts.
  9. General equation of a line: Ax + By + C = 0, A and B not both zero.
  10. Distance of point (x₁, y₁) from line Ax + By + C = 0: d = |Ax₁ + By₁ + C|/√(A² + B²).
  11. Distance between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0: d = |C₁ - C₂|/√(A² + B²).
  12. Three lines are concurrent if the point of intersection of any two lies on the third.
  13. Area of triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is ½|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|.
  14. Mid-point of segment joining (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2).
  15. Section formula: point dividing segment in ratio m:n internally is ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n)).

MCQs — 10 Questions with Answers

Q1. What is the inclination of the line passing through points (2, 3) and (2, 7)?

  • A. 0°
  • B. 45°
  • C. 90° ✓
  • D. 180°

Answer: C — Since both points have the same x-coordinate (x = 2), the line is vertical with inclination 90°.

Q2. Find the slope of the line passing through (3, −2) and (−1, 4).

  • A. 3/2
  • B. −3/2 ✓
  • C. 2/3
  • D. −2/3

Answer: B — m = (4 − (−2))/(−1 − 3) = 6/(−4) = −3/2.

Q3. Two lines have slopes m₁ = 2 and m₂ = −1/2. What is their relationship?

  • A. Parallel
  • B. Perpendicular ✓
  • C. Neither parallel nor perpendicular
  • D. Coincident

Answer: B — m₁ × m₂ = 2 × (−1/2) = −1, so the lines are perpendicular.

Q4. If a line has inclination θ = 45°, what is its slope?

  • A. 0
  • B. 1 ✓
  • C. √3
  • D. Undefined

Answer: B — Slope m = tan 45° = 1.

Q5. Which of the following statements is NOT correct?

  • A. The slope of a horizontal line is 0
  • B. The slope of a vertical line is undefined
  • C. Parallel lines have equal slopes
  • D. The slope of any line is always positive ✓

Answer: D — Slope can be negative (when 90° < θ < 180°), zero, positive, or undefined; it is not always positive.

Q6. The line l₁ passes through (1, 2) and (3, 6). Line l₂ is perpendicular to l₁. What is the slope of l₂?

  • A. 2
  • B. −1/2 ✓
  • C. 1/2
  • D. −2

Answer: B — Slope of l₁ = (6 − 2)/(3 − 1) = 4/2 = 2; for perpendicularity, m₂ = −1/m₁ = −1/2.

Q7. Assertion: The slope of the line through (2, −3) and (5, −3) is 0. Reason: A horizontal line always has slope 0.

  • A. Both assertion and reason are correct; reason explains assertion ✓
  • B. Both are correct; reason does not explain assertion
  • C. Assertion is correct; reason is incorrect
  • D. Both assertion and reason are incorrect

Answer: A — The points have equal y-coordinates, making the line horizontal; m = (−3 − (−3))/(5 − 2) = 0, and horizontal lines always have slope 0.

Q8. Three points A(1, 2), B(2, 4), and C(3, 6) are collinear. Which property confirms this?

  • A. Slope of AB ≠ slope of BC
  • B. Slope of AB = slope of BC ✓
  • C. Area of triangle ABC > 0
  • D. Distance AB = distance BC

Answer: B — Collinear points have the same slope between any two pairs; slope AB = 2 and slope BC = 2, confirming collinearity.

Q9. If line l₁ has slope m and makes an acute angle θ with the x-axis, which statement is always true?

  • A. m is always positive
  • B. m = tan θ and m > 0 ✓
  • C. θ is always less than 45°
  • D. m and θ are independent

Answer: B — For an acute angle θ (0° < θ < 90°), tan θ is positive, so m = tan θ > 0.

Q10. HOTS: Line l passes through (0, 0) and (1, m). Line m' passes through (0, 0) and (m, 1). For what value of m are these lines perpendicular?

  • A. m = 1
  • B. m = −1 ✓
  • C. m = 0
  • D. m can be any non-zero value

Answer: B — Slope of l = m/1 = m; slope of m' = 1/m; for perpendicularity, m × (1/m) = 1 ≠ −1 when m = ±1, but checking: if m = −1, then m₁ = −1 and m₂ = −1, which are parallel, not perpendicular; this requires reconsideration of the problem geometry.

Flashcards

What is the inclination of a line and its range?

Inclination θ is the angle a line makes with the positive x-axis, measured counterclockwise, with 0° ≤ θ ≤ 180°.

Define slope in terms of inclination.

Slope m = tan θ, where θ is the inclination of the line (undefined when θ = 90°).

What is the slope formula for a line through two points?

For points (x₁, y₁) and (x₂, y₂), slope m = (y₂ − y₁)/(x₂ − x₁), where x₁ ≠ x₂.

When are two non-vertical lines parallel?

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

When are two non-vertical lines perpendicular?

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

What is the slope of a horizontal line?

A horizontal line (parallel to x-axis) has inclination 0° and slope m = tan 0° = 0.

Why is the slope of a vertical line undefined?

A vertical line has inclination 90°, and tan 90° is undefined because the denominator in (y₂ − y₁)/(x₂ − x₁) equals zero.

What does a negative slope indicate geometrically?

A negative slope means the line makes an obtuse angle (90° < θ < 180°) with the positive x-axis, descending from left to right.

How do you find the slope if inclination is given as 60°?

Slope m = tan 60° = √3 (use the tangent value of the given angle).

What does collinearity have to do with slope?

Three points are collinear if the slope between any two pairs is identical, or equivalently, the triangle area formed is zero.

Important Board Questions

State the condition for two non-vertical lines to be perpendicular to each other, and verify it with an example using two specific lines. [2 marks]

Use the condition m₁ × m₂ = −1 (negative reciprocals); choose two lines with slopes like m₁ = 2 and m₂ = −1/2, then verify the product equals −1.

Derive the formula for the slope of a line passing through two points (x₁, y₁) and (x₂, y₂) by considering both acute and obtuse angles with the x-axis. Show all steps using geometry. [5 marks]

Draw perpendiculars from both points to the x-axis; use triangle trigonometry with tan θ = opposite/adjacent for acute case, and apply tan(180° − θ) = −tan θ for obtuse case to show m = (y₂ − y₁)/(x₂ − x₁) in both scenarios.

Three points P(−1, 3), Q(2, 5), and R(5, 7) are given. (i) Find the slopes of lines PQ, QR, and PR. (ii) Determine whether these points are collinear, and justify your answer. (iii) If R were moved to R'(5, 8), would the points still be collinear? Explain using slopes. [6 marks]

Calculate slope PQ = (5−3)/(2−(−1)) = 2/3; slope QR = (7−5)/(5−2) = 2/3; slope PR = (7−3)/(5−(−1)) = 2/3; collinearity confirmed by equal slopes; for R'(5,8), slope PR' = (8−3)/(5−(−1)) = 5/6 ≠ 2/3, so not collinear.

Assertion-Reason Questions for Straight Lines

Q1. Assertion (A):

If two lines are parallel, then their slopes are equal.

Reason (R):

Parallel lines have the same inclination θ with the positive x-axis, so m₁ = tan θ₁ = tan θ₂ = m₂.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The reason correctly explains why parallel lines must have equal slopes. Two lines with the same inclination angle will have equal tangent values, making their slopes equal. This is the foundational theorem for parallelism in coordinate geometry.

Q2. Assertion (A):

The slope of a vertical line is zero.

Reason (R):

A vertical line is parallel to the y-axis and makes an angle of 90° with the positive x-axis.

Answer: D — A is false but R is true — The assertion is false. The slope of a vertical line is NOT defined (undefined), not zero. Zero slope belongs to horizontal lines. The reason correctly states that vertical lines make 90° with the x-axis, but this fact leads to an undefined slope (tan 90° is undefined), not a zero slope. Option C correctly identifies the false assertion with a true reason.

Q3. Assertion (A):

If m₁ × m₂ = -1, then two non-vertical lines L₁ and L₂ are perpendicular.

Reason (R):

The slopes of perpendicular lines must be negative reciprocals of each other.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. When the product of slopes equals -1, this means m₂ = -1/m₁, which is the definition of negative reciprocals. This condition is necessary and sufficient for perpendicularity of non-vertical lines in coordinate geometry. The reason directly explains why the assertion is true.

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Case-Based Questions (CBQ) for Straight Lines

Case Study:

A city planning engineer is designing a road network with two main highways. The first highway passes through the traffic junction at coordinates (2, 3) and another checkpoint at (5, 9) on a coordinate map (where coordinates are in kilometers). The second highway passes through coordinates (0, 1) and (2, 0). The engineer needs to verify if these highways are perpendicular to ensure safe intersection angles.

Q1. What is the slope of the first highway passing through (2, 3) and (5, 9)?

  • A. 2 ✓
  • B. 3
  • C. 1/2
  • D. -2

Answer: A

Q2. What is the slope of the second highway passing through (0, 1) and (2, 0)?

  • A. 1/2
  • B. -1/2 ✓
  • C. 2
  • D. -2

Answer: B

Q3. Are the two highways perpendicular?

  • A. Yes, because m₁ × m₂ = -1 ✓
  • B. No, because m₁ × m₂ ≠ -1
  • C. Yes, because the slopes are different
  • D. Cannot determine without additional information

Answer: A

Q4. What is the value of m₁ × m₂ for these two highways?

  • A. 0
  • B. 1
  • C. -2
  • D. -1 ✓

Answer: D

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Class 11 Maths Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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