Straight Lines — Class 11 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Straight Lines" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Straight Lines" — 8 important questions with detailed answers for CBSE board exam…
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NCERT Solutions for Class 11 Mathematics — Straight Lines. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Straight Lines — All 8 Questions Solved
Q1. Find the equation of the line passing through points (2, 3) and (4, 7).
Given points: P₁(2, 3) and P₂(4, 7)
Step 1: Find the slope of the line:
m = (y₂ - y₁)/(x₂ - x₁) = (7 - 3)/(4 - 2) = 4/2 = 2
Step 2: Use the point-slope form:
y - y₁ = m(x - x₁)
Using point (2, 3):
y - 3 = 2(x - 2)
y - 3 = 2x - 4
y = 2x - 1
Step 3: Write in standard form:
2x - y - 1 = 0
Final Answer: The equation is y = 2x - 1 or 2x - y - 1 = 0
Q2. Find the distance from the point (3, 4) to the line 3x + 4y - 5 = 0.
Given point P(3, 4) and line 3x + 4y - 5 = 0
Line in form ax + by + c = 0 where a = 3, b = 4, c = -5
Step 1: Use the distance formula:
Distance = |ax₀ + by₀ + c|/√(a² + b²)
Step 2: Substitute the values:
Distance = |3(3) + 4(4) + (-5)|/√(3² + 4²)
Distance = |9 + 16 - 5|/√(9 + 16)
Distance = |20|/√25
Distance = 20/5
Distance = 4 units
Final Answer: The distance is 4 units
Q3. Find the angle between the lines 2x + y - 3 = 0 and x - 2y + 4 = 0.
Line 1: 2x + y - 3 = 0, so m₁ = -2/1 = -2
Line 2: x - 2y + 4 = 0, so 2y = x + 4, m₂ = 1/2
Step 1: Use the angle formula:
tan θ = |(m₁ - m₂)/(1 + m₁m₂)|
Step 2: Calculate m₁m₂:
m₁m₂ = (-2)(1/2) = -1
Step 3: Substitute in the formula:
tan θ = |(-2 - 1/2)/(1 + (-1))|
tan θ = |(-5/2)/0|
Step 4: Since denominator is 0 (m₁m₂ = -1), the lines are perpendicular.
Angle θ = 90°
Final Answer: The lines are perpendicular; angle = 90°
Q4. A line passes through (1, 2) and makes an angle of 45° with the positive x-axis. Find its equation.
Given: Point (1, 2) and angle with x-axis = 45°
Step 1: Find the slope using the angle:
tan 45° = m
m = 1
Step 2: Use point-slope form:
y - y₁ = m(x - x₁)
y - 2 = 1(x - 1)
y - 2 = x - 1
y = x + 1
Step 3: Write in standard form:
x - y + 1 = 0
Final Answer: The equation is y = x + 1 or x - y + 1 = 0
Q5. Find the equation of the line which is perpendicular to x + 2y = 5 and passes through (0, 0).
Given line: x + 2y = 5
Required line passes through origin (0, 0) and is perpendicular to the given line.
Step 1: Find the slope of the given line:
x + 2y = 5
2y = -x + 5
y = -x/2 + 5/2
Slope m₁ = -1/2
Step 2: Find the slope of perpendicular line:
For perpendicular lines: m₁ × m₂ = -1
(-1/2) × m₂ = -1
m₂ = 2
Step 3: Use point-slope form with (0, 0):
y - 0 = 2(x - 0)
y = 2x
Step 4: Write in standard form:
2x - y = 0
Final Answer: The equation is y = 2x or 2x - y = 0
Q6. Find the equation of the line parallel to 2x - 3y + 7 = 0 and passing through (3, 1).
Given line: 2x - 3y + 7 = 0
Required line is parallel and passes through (3, 1).
Step 1: Find the slope of the given line:
2x - 3y + 7 = 0
-3y = -2x - 7
y = (2/3)x + 7/3
Slope m = 2/3
Step 2: Parallel lines have equal slopes:
Required slope = 2/3
Step 3: Use point-slope form:
y - 1 = (2/3)(x - 3)
3(y - 1) = 2(x - 3)
3y - 3 = 2x - 6
2x - 3y - 3 = 0
Final Answer: The equation is 2x - 3y - 3 = 0
Q7. Find the coordinates of the point where the lines 2x + 3y = 5 and 3x - 2y = 4 intersect.
Given equations:
2x + 3y = 5 ... (1)
3x - 2y = 4 ... (2)
Step 1: Multiply equation (1) by 2:
4x + 6y = 10 ... (3)
Step 2: Multiply equation (2) by 3:
9x - 6y = 12 ... (4)
Step 3: Add equations (3) and (4):
4x + 6y + 9x - 6y = 10 + 12
13x = 22
x = 22/13
Step 4: Substitute x in equation (1):
2(22/13) + 3y = 5
44/13 + 3y = 5
3y = 5 - 44/13
3y = (65 - 44)/13
3y = 21/13
y = 7/13
Step 5: Verify in equation (2):
3(22/13) - 2(7/13) = 66/13 - 14/13 = 52/13 = 4 ✓
Final Answer: Intersection point is (22/13, 7/13)
Q8. If the line ax + by + c = 0 passes through (2, -1) and (3, 1), find the relationship between a, b, and c.
Given: Line ax + by + c = 0 passes through (2, -1) and (3, 1)
Step 1: Since (2, -1) lies on the line:
a(2) + b(-1) + c = 0
2a - b + c = 0 ... (1)
Step 2: Since (3, 1) lies on the line:
a(3) + b(1) + c = 0
3a + b + c = 0 ... (2)
Step 3: Subtract equation (1) from equation (2):
(3a + b + c) - (2a - b + c) = 0
3a + b + c - 2a + b - c = 0
a + 2b = 0
a = -2b
Step 4: Substitute in equation (1):
2(-2b) - b + c = 0
-4b - b + c = 0
c = 5b
Step 5: The relationship:
a : b : c = -2b : b : 5b = -2 : 1 : 5
Final Answer: The relationship is a = -2b and c = 5b, or a : b : c = -2 : 1 : 5
Straight Lines — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. The distance between the points (2, 3) and (4, 1) is:
- A. √8
- B. √12
- C. √10
- D. √16
Answer: A. √8 — Using the distance formula, d = √[(x2-x1)² + (y2-y1)²], we get d = √[(4-2)² + (1-3)²] = √[2² + (-2)²] = √[4+4] = √8.
MCQ 2. What is the equation of a line with slope 3 and y-intercept -5?
- A. y = 3x + 5
- B. y = 3x - 5
- C. y = -3x + 5
- D. y = -3x - 5
Answer: B. y = 3x - 5 — The slope-intercept form of a line is y = mx + c, where m is the slope and c is the y-intercept. Substituting m=3 and c=-5 gives y = 3x - 5.
MCQ 3. The midpoint of the line segment joining (5, -3) and (7, 1) is:
- A. (6, -2)
- B. (6, -1)
- C. (12, -2)
- D. (1, -2)
Answer: B. (6, -1) — Using the midpoint formula ((x1+x2)/2, (y1+y2)/2), we get ((5+7)/2, (-3+1)/2) = (12/2, -2/2) = (6, -1).
MCQ 4. The slope of the line represented by the equation 3x + 4y = 7 is:
- A. 3/4
- B. -3/4
- C. 4/3
- D. -4/3
Answer: B. -3/4 — Rewrite the equation in slope-intercept form (y = mx + c): 4y = -3x + 7, so y = (-3/4)x + 7/4. The slope 'm' is -3/4.
MCQ 5. What is the angle of inclination of a line with a slope of 1?
- A. 30°
- B. 45°
- C. 60°
- D. 90°
Answer: B. 45° — The slope 'm' is equal to tan(θ), where θ is the angle of inclination. If m=1, then tan(θ)=1, which means θ = 45°.
MCQ 6. What is the y-intercept of the line 5x - 2y = 10?
- A. 5
- B. -5
- C. 2
- D. -2
Answer: B. -5 — To find the y-intercept, set x = 0 in the equation: 5(0) - 2y = 10. This gives -2y = 10, so y = -5.
MCQ 7. Find the coordinates of the point that divides the line segment joining (1, -3) and (4, 6) in the ratio 2:1 internally.
- A. (3, 3)
- B. (2, 0)
- C. (3, 0)
- D. (2, 3)
Answer: A. (3, 3) — Using the section formula ( (mx2+nx1)/(m+n), (my2+ny1)/(m+n) ), with m=2, n=1, (x1,y1)=(1,-3), (x2,y2)=(4,6), the x-coordinate is (2*4 + 1*1)/(2+1) = 9/3 = 3, and the y-coordinate is (2*6 + 1*(-3))/(2+1) = 9/3 = 3. Thus, the point is (3, 3).
MCQ 8. The equation of the line passing through (1, 2) and (3, 4) is:
- A. x + y = 3
- B. x - y = -1
- C. y - x = 1
- D. x + y = 5
Answer: C. y - x = 1 — First, find the slope m = (4-2)/(3-1) = 2/2 = 1. Using the point-slope form y - y1 = m(x - x1) with (1,2), we get y - 2 = 1(x - 1), which simplifies to y - x = 1.
MCQ 9. The distance of the point (3, 4) from the line 3x + 4y - 5 = 0 is:
- A. 5
- B. 2
- C. 4
- D. 3
Answer: C. 4 — Using the formula d = |Ax0 + By0 + C| / √(A² + B²), we have d = |3(3) + 4(4) - 5| / √(3² + 4²) = |9 + 16 - 5| / √(9 + 16) = |20| / √25 = 20 / 5 = 4.
MCQ 10. The area of the triangle with vertices (0,0), (a,0), and (0,b) is:
- A. ab
- B. (1/2)ab
- C. a+b
- D. √(a²+b²)
Answer: B. (1/2)ab — For a right-angled triangle with vertices at the origin and on the axes, the base is 'a' units and the height is 'b' units. The area is (1/2) * base * height = (1/2) * a * b.
MCQ 11. A line makes x-intercept 3 and y-intercept 4. Its equation is:
- A. 4x + 3y = 12
- B. 3x + 4y = 12
- C. x/3 + y/4 = 0
- D. x/4 + y/3 = 1
Answer: A. 4x + 3y = 12 — The intercept form of a line is x/a + y/b = 1, where 'a' is the x-intercept and 'b' is the y-intercept. Substituting a=3, b=4 gives x/3 + y/4 = 1. Multiplying by 12 (LCM of 3 and 4) gives 4x + 3y = 12.
MCQ 12. The equation of a line passing through the intersection of x + y = 1 and 2x - y = 5 and parallel to the x-axis is:
- A. y = -1
- B. y = 1
- C. x = 2
- D. x = 1
Answer: A. y = -1 — First, find the intersection point of x+y=1 and 2x-y=5. Adding the two equations: 3x = 6 => x=2. Substitute x=2 into x+y=1 => y=-1. The intersection point is (2,-1). A line parallel to the x-axis has the equation y = constant. Since it passes through (2,-1), its equation must be y = -1.
MCQ 13. The angle between the lines x - y√3 = 5 and x√3 + y = 7 is:
- A. 30°
- B. 45°
- C. 60°
- D. 90°
Answer: D. 90° — The slope of the first line (m1) from y = (1/√3)x - 5/√3 is 1/√3. The slope of the second line (m2) from y = -√3x + 7 is -√3. Since m1 * m2 = (1/√3) * (-√3) = -1, the lines are perpendicular, and the angle between them is 90°.
MCQ 14. The distance between the parallel lines 3x + 4y + 5 = 0 and 3x + 4y - 5 = 0 is:
- A. 1
- B. 2
- C. 5
- D. 10
Answer: B. 2 — The distance between two parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is |C1 - C2| / √(A² + B²). Here, d = |5 - (-5)| / √(3² + 4²) = |10| / √(9 + 16) = 10 / √25 = 10 / 5 = 2.
MCQ 15. If the centroid of a triangle with vertices (x, 1), (1, y), and (-2, 4) is (0, 0), then the values of x and y are:
- A. x = 1, y = -5
- B. x = -1, y = -5
- C. x = 1, y = 5
- D. x = -1, y = 5
Answer: A. x = 1, y = -5 — The centroid (G) coordinates are ((x1+x2+x3)/3, (y1+y2+y3)/3). Given G=(0,0), we have: (x+1-2)/3 = 0 => x-1 = 0 => x=1. And (1+y+4)/3 = 0 => y+5 = 0 => y=-5. So, x=1, y=-5.
MCQ 16. Find the equation of the line perpendicular to 2x - 3y + 5 = 0 and passing through the point (-1, 2).
- A. 3x + 2y - 1 = 0
- B. 3x - 2y + 7 = 0
- C. 2x + 3y - 4 = 0
- D. 2x - 3y + 8 = 0
Answer: A. 3x + 2y - 1 = 0 — The slope of the given line 2x - 3y + 5 = 0 is m1 = 2/3. The slope of a line perpendicular to it (m2) will be -1/m1 = -3/2. Using the point-slope form y - y1 = m(x - x1) with (-1, 2) and m = -3/2: y - 2 = (-3/2)(x - (-1)) => 2(y - 2) = -3(x + 1) => 2y - 4 = -3x - 3 => 3x + 2y - 1 = 0.
MCQ 17. For what value of 'p' are the lines (p+1)x - y + 7 = 0 and 3x - 2y + 5 = 0 parallel?
- A. 1/2
- B. -1/2
- C. 5/2
- D. -5/2
Answer: A. 1/2 — For parallel lines, their slopes must be equal. Slope of first line m1 = - (p+1) / (-1) = p+1. Slope of second line m2 = -3 / (-2) = 3/2. Equating slopes: p+1 = 3/2 => p = 3/2 - 1 => p = 1/2.
MCQ 18. The orthocentre of the triangle with vertices (0,0), (2,0) and (0,3) is:
- A. (0,0)
- B. (2,0)
- C. (0,3)
- D. (2,3)
Answer: A. (0,0) — The given vertices (0,0), (2,0), and (0,3) form a right-angled triangle with the right angle at the origin (0,0). For a right-angled triangle, the orthocentre is the vertex at the right angle. Therefore, the orthocentre is (0,0).
MCQ 19. Find the equation of the line passing through (1, 1) and making an angle of 60° with the positive x-axis.
- A. y - 1 = √3(x - 1)
- B. y - 1 = (1/√3)(x - 1)
- C. y - 1 = -√3(x - 1)
- D. y - 1 = -(1/√3)(x - 1)
Answer: A. y - 1 = √3(x - 1) — The slope 'm' of the line is tan(θ). Given θ = 60°, m = tan(60°) = √3. Using the point-slope form y - y1 = m(x - x1) with (1, 1) and m = √3, the equation is y - 1 = √3(x - 1).
MCQ 20. Find the image of the point (2, 3) with respect to the line y = -x.
- A. (-3, -2)
- B. (3, 2)
- C. (-2, -3)
- D. (-3, 2)
Answer: A. (-3, -2) — For reflection across y = -x, the coordinates (x,y) transform to (-y, -x). So, the image of (2, 3) is (-3, -2).
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.