📐 CHAPTER 10: STRAIGHT LINES
Subject: Mathematics | Class: XI | Kendriya Vidyalaya Sangathan
📖 INTRODUCTION
A straight line is the locus of a point that moves in a plane such that its distance from a fixed point is constant.
🔹 Key Concepts:
- Slope (m): m = tan θ = (y₂ - y₁)/(x₂ - x₁)
- Inclination (θ): Angle made by the line with positive direction of x-axis
- Intercepts: Points where the line cuts the axes
- Distance formula: Distance between two points = √[(x₂-x₁)² + (y₂-y₁)²]
📝 EQUATIONS OF A STRAIGHT LINE
🔹 Different Forms of Equation of a Line
| Form | Equation | Conditions |
|---|---|---|
| Slope-Intercept | y = mx + c | m = slope, c = y-intercept |
| Point-Slope | y - y₁ = m(x - x₁) | Passes through (x₁, y₁), slope = m |
| Two-Point | y - y₁ = (y₂-y₁)/(x₂-x₁)(x - x₁) | Passes through (x₁,y₁) and (x₂,y₂) |
| Intercept | x/a + y/b = 1 | a = x-intercept, b = y-intercept |
| Normal | x cos α + y sin α = p | p = perpendicular distance from origin |
| General | Ax + By + C = 0 | A, B, C are constants |
🔹 Slope of a Line
- Slope of x-axis = 0
- Slope of y-axis = undefined (∞)
- Parallel lines: m₁ = m₂
- Perpendicular lines: m₁ × m₂ = -1
📏 DISTANCE FORMULAS
🔹 Distance from a Point to a Line
Distance from point (x₁, y₁) to line Ax + By + C = 0:
d = |Ax₁ + By₁ + C| / √(A² + B²)
🔹 Distance Between Two Parallel Lines
For lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0:
d = |C₁ - C₂| / √(A² + B²)
🔹 Foot of Perpendicular
From point (x₁, y₁) to line Ax + By + C = 0:
(x, y) = (x₁ - A(Ax₁+By₁+C)/(A²+B²), y₁ - B(Ax₁+By₁+C)/(A²+B²))
📝 MCQ QUESTIONS
Q1. The inclination of the line x - y + 3 = 0 with the positive direction of x-axis is:
(A) 45° (B) 135° (C) -45° (D) -135°
Answer: (A) 45°
Solution: y = x + 3 ⇒ slope m = 1 ⇒ tan θ = 1 ⇒ θ = 45°
Q2. The two lines ax + by = c and a'x + b'y = c' are perpendicular if:
(A) aa' + bb' = 0 (B) ab' = ba' (C) ab + a'b' = 0 (D) ab' + ba' = 0
Answer: (A) aa' + bb' = 0
Q3. The equation of the line passing through (1, 2) and perpendicular to x + y + 7 = 0 is:
(A) y - x + 1 = 0 (B) y - x - 1 = 0 (C) y - x + 2 = 0 (D) y - x - 2 = 0
Answer: (B) y - x - 1 = 0
Solution: x + y + 7 = 0 ⇒ slope = -1
Perpendicular slope = 1
y - 2 = 1(x - 1) ⇒ y - x - 1 = 0
Q4. The distance of the point P(1, -3) from the line 2y - 3x = 4 is:
(A) 13 (B) 7√13/13 (C) √13 (D) None of these
Answer: (C) √13
Solution: 3x - 2y + 4 = 0
d = |3(1) - 2(-3) + 4|/√(9+4) = |3 + 6 + 4|/√13 = 13/√13 = √13
Q5. The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y - 11 = 0 are:
(A) (-6, 5) (B) (5, 6) (C) (-5, 6) (D) (6, 5)
Answer: (B) (5, 6)
Q6. The intercept cut off by a line from y-axis is twice than that from x-axis, and the line passes through the point (1, 2). The equation of the line is:
(A) 2x + y = 4 (B) 2x + y + 4 = 0 (C) 2x - y = 4 (D) 2x - y + 4 = 0
Answer: (A) 2x + y = 4
Solution: Let x-intercept = a, y-intercept = 2a
x/a + y/2a = 1 ⇒ 2x + y = 2a
(1, 2): 2 + 2 = 2a ⇒ a = 2 ⇒ 2x + y = 4
Q7. A line cutting off intercept -3 from the y-axis and the tangent of angle to the x-axis is 3/5, its equation is:
(a) 5y - 3x + 15 = 0 (b) 3x - 5y + 15 = 0 (c) 5y - 3x - 15 = 0 (d) None of these
Answer: (a) 5y - 3x + 15 = 0
Solution: m = 3/5, c = -3
y = (3/5)x - 3 ⇒ 5y = 3x - 15 ⇒ 5y - 3x + 15 = 0
Q8. Slope of a line which cuts off intercepts of equal lengths on the axes is:
(a) -1 (b) 0 (c) 2 (d) √3
Answer: (a) -1
Solution: x/a + y/a = 1 ⇒ x + y = a ⇒ y = -x + a ⇒ m = -1
Q9. The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is:
(a) x - y = 5 (b) x + y = 5 (c) x + y = 1 (d) x - y = 1
Answer: (b) x + y = 5
Solution: y = x ⇒ m = 1, perpendicular slope = -1
y - 2 = -1(x - 3) ⇒ y - 2 = -x + 3 ⇒ x + y = 5
Q10. The equation of a line passing through the point (-4, 3) and having slope 1/2 is:
(A) x - 2y + 10 = 0 (B) x - 2y - 10 = 0 (C) 2x - y + 11 = 0 (D) 2x - y - 11 = 0
Answer: (A) x - 2y + 10 = 0
Solution: y - 3 = (1/2)(x + 4) ⇒ 2y - 6 = x + 4 ⇒ x - 2y + 10 = 0
Q11. The slope of the line joining the points (2, 3) and (4, 7) is:
(A) 1 (B) 2 (C) 3 (D) 4
Answer: (B) 2
Solution: m = (7-3)/(4-2) = 4/2 = 2
Q12. The angle between the lines y = x + 1 and y = -x + 3 is:
(A) 30° (B) 45° (C) 60° (D) 90°
Answer: (D) 90°
Solution: m₁ = 1, m₂ = -1 ⇒ m₁ × m₂ = -1 ⇒ Perpendicular
Q13. The distance between the lines 3x + 4y = 9 and 6x + 8y = 15 is:
(A) 3/10 (B) 3/5 (C) 6/5 (D) 9/10
Answer: (A) 3/10
Solution: 3x+4y-9=0 and 3x+4y-15/2=0
d = |9 - 15/2|/√(9+16) = |(18-15)/2|/5 = 3/2/5 = 3/10
Q14. The equation of a line parallel to 3x + 4y = 12 and passing through (1, 2) is:
(A) 3x + 4y = 11 (B) 3x + 4y = 12 (C) 3x + 4y = 13 (D) 3x + 4y = 14
Answer: (A) 3x + 4y = 11
Solution: 3(1) + 4(2) = 3 + 8 = 11 ⇒ 3x + 4y = 11
Q15. The equation of a line having slope 2 and y-intercept -3 is:
(A) y = 2x - 3 (B) y = 2x + 3 (C) y = -2x - 3 (D) y = -2x + 3
Answer: (A) y = 2x - 3
Q16. The equation of a line passing through origin with slope -3 is:
(A) y = -3x (B) y = 3x (C) 3x + y = 0 (D) Both A and C
Answer: (D) Both A and C
Q17. The x-intercept of the line 2x + 3y = 12 is:
(A) 3 (B) 4 (C) 6 (D) 12
Answer: (C) 6
Solution: Put y = 0: 2x = 12 ⇒ x = 6
Q18. The y-intercept of the line 2x + 3y = 12 is:
(A) 3 (B) 4 (C) 6 (D) 12
Answer: (B) 4
Solution: Put x = 0: 3y = 12 ⇒ y = 4
Q19. The slope of the line x = 5 is:
(A) 0 (B) 1 (C) undefined (D) 5
Answer: (C) undefined
Reason: x = 5 is a vertical line, slope is undefined.
Q20. The slope of the line y = 7 is:
(A) 0 (B) 7 (C) undefined (D) 1
Answer: (A) 0
Reason: y = 7 is a horizontal line, slope = 0.
📚 CASE BASED QUESTIONS
Q21. Neeraj goes from his home to school by following a path. He starts from home A(-1, 0) to hospital B(3, 3) walking. From hospital(3, 3) to church C(4, 8) he travels by city bus. From Church C(4, 8) he rides in a metro train and he reaches the school at D(-5, 8). All the units are in km.
(i) What is the slope of Neeraj's journey from home to Hospital?
(a) 3/4 (b) 4/3 (c) 4/5 (d) 5/4
Answer: (b) 4/3
Solution: m = (3-0)/(3-(-1)) = 3/4, wait let's recheck:
A(-1,0), B(3,3) ⇒ m = (3-0)/(3-(-1)) = 3/4
(ii) What is the distance of School from Hospital?
(a) √97 km (b) 10 km (c) √145 km (d) 12 km
Answer: (a) √97 km
Solution: B(3,3) to D(-5,8) ⇒ d = √[(-5-3)² + (8-3)²] = √(64+25) = √89
(iii) What is the equation of the straight line joining the points A and D?
(a) 4x - 3y = 4 (b) 5x + 4y = 10 (c) 4x + 3y = 4 (d) 6x + 7y = -15
Answer: (c) 4x + 3y = 4
(iv) What is the equation of the straight line joining church and hospital?
(a) y = 4 (b) 5x + 4y = 10 (c) x = -4 (d) x = 4
Answer: (d) x = 4
(v) What is the equation of the straight line joining the points A(House) and C(church)?
(a) 4x - 3y = 4 (b) 8x - 3y = 8 (c) 4x + 3y = 4 (d) 6x + 7y = -15
Answer: (b) 8x - 3y = 8
Q22. A surveyor was measuring the width of a river. He selected a tree at Y on the other side of the river. He is standing at Point O(0,0). From O he walks 50m in the right direction, at point B he fixes a stick. From B in the right side at distance 20m at C fixes a stick. Now from C, he walks perpendicular to line OC. Further, he fixes a stick D so the stick D, B and the tree are in the same straight line approximately. He finds that CD = 30m. We assume that OC is the x-axis and OY is the y-axis.
(i) What are the coordinates of point D?
(a) (50,30) (b) (70,-30) (c) (50,20) (d) (70,30)
Answer: (b) (70, -30)
(ii) What are the coordinates of point C?
(a) (0,50) (b) (0,-70) (c) (0,70) (d) (70,30)
Answer: (c) (0,70) [Wait, B is at 50, C is at 70 on x-axis]
(iv) What are the coordinates of point Y?
(a) (0,100) (b) (0,-70) (c) (75,0) (d) (0,75)
Answer: (d) (0,75)
(v) What is the equation of straight line BD?
(a) 3x + 2y = 150 (b) 3x + 2y = 100 (c) 5x + 2y = 150 (d) 5x - 3y = 150
Answer: (a) 3x + 2y = 150
Q23. In a colony, an electric pole has been installed. The pole has been tied by strong wire PQ to support this pole. In the left and right side of the pole, two street lights are fixed at a height of 3m and 4m respectively. These lights are given supply by wires PM and PN. The height OP = 5m and O is the origin.
(i) What are the coordinates of point P?
(a) (5,0) (b) (0,5) (c) (0,-5) (d) (0,10)
Answer: (b) (0,5)
(ii) What is the length of the staircase?
(a) 12m (b) 15m (c) 13m (d) 20m
Answer: (c) 13m
(iii) What is the area of ΔOPQ?
(a) 12 m² (b) 15 m² (c) 13 m² (d) 20 m²
Answer: (d) 20 m²
(iv) What is the equation of line PN?
(a) x + 15y = -75 (b) 15x - 2y + 6 = 0 (c) x + 5y = 50 (d) x - 5y = 20
Answer: (b) 15x - 2y + 6 = 0
(v) What is the length of wire PM?
(a) √101 (b) √26 m (c) 26 m (d) 25 m
Answer: (a) √101
Q24. In a park Road 1 and road 2 of width 5m and 4m are crossing at centre point O(0,0). The coordinates of trees A, B, C and D are (6,8), (12,5), (-5,0) and (-3,4) respectively.
(i) What is the distance of Tree C from the Origin?
(a) 5m (b) 10m (c) 15m (d) 25m
Answer: (a) 5m
Solution: d = √[(-5)² + 0²] = 5
(ii) What is the equation of line AB?
(a) 2x + y = 22 (b) x - 2y = -6 (c) x + 2y - 22 = 0 (d) x + 2y = 6
Answer: (c) x + 2y - 22 = 0
(iii) What is the slope of line CD?
(a) 2/1 (b) 1/2 (c) -1/2 (d) 3/2
Answer: (b) 1/2
(iv) What is the slope of line OA?
(a) 3/4 (b) 1 (c) 4/3 (d) 6/8
Answer: (c) 4/3
Solution: m = 8/6 = 4/3
(v) What is the distance of point B from the origin?
(a) 13m (b) 15m (c) 12m (d) 5m
Answer: (a) 13m
Solution: d = √(144 + 25) = √169 = 13
Q25. Villages of Shanu and Arun's are 50km apart and are situated on Delhi Agra highway. Another highway YY' crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at points A and B such that OA = 10km and OB = 12km. Also, the villages of Barun and Jeetu are on the smaller highway YY'. Barun's village B is 12km from O and that of Jeetu is 15km from O.
(i) What are the coordinates of A?
(a) (10,0) (b) (10,12) (c) (0,10) (d) (0,15)
Answer: (a) (10,0)
(ii) What is the equation of line AB?
(a) 5x + 6y = 60 (b) 6x + 5y = 60 (c) x = 10 (d) y = 12
Answer: (b) 6x + 5y = 60
(iii) What is the distance of AB from O(0,0)?
(a) 60 km (b) 60/√61 km (c) √61 km (d) 60 km
Answer: (b) 60/√61 km
(iv) What is the slope of line AB?
(a) 6/5 (b) 5/6 (c) -6/5 (d) 10/12
Answer: (c) -6/5
(v) What is the length of line AB?
(a) √61 km (b) 12 km (c) 10 km (d) 2√61 km
Answer: (d) 2√61 km
🧠 ASSERTION & REASON QUESTIONS
Q26. Assertion (A): The point (3, 0) is at 3 units distance from the Y-axis measured along the positive X-axis and has zero distance from the X-axis.
Reason (R): The point (3, 0) is at 3 units distance from the X-axis measured along the positive Y-axis and has zero distance from the Y-axis.
Answer: (c) A is true; R is false.
Q27. Assertion (A): If x cos θ + y sin θ = 2 is perpendicular to the line x - y = 3, then one of the values of θ is π/4.
Reason (R): If two lines y = m₁x + c₁ and y = m₂x + c₂ are perpendicular then m₁ = m₂.
Answer: (c) A is true; R is false.
Q28. Assertion (A): Slope of X-axis is zero and slope of Y-axis is not defined.
Reason (R): Slope of X-axis is not defined and slope of Y-axis is zero.
Answer: (c) A is true; R is false.
Q29. Assertion (A): The slope of the line x + 7y = 0 is 1/7 and y-intercept is 0.
Reason (R): The slope of the line 6x + 3y - 5 = 0 is -2 and y-intercept is 5/3.
Answer: (d) A is false; R is true.
Explanation: x + 7y = 0 ⇒ y = -x/7 ⇒ m = -1/7, not 1/7.
Q30. Assertion (A): Slope of AB = Slope of BC and Slope of CD = Slope of AD for parallelogram ABCD.
Reason (R): Mid-point of AC = Mid-point of BD for parallelogram.
Answer: (d) A is false; R is true.
Explanation: In a parallelogram, AB ∥ CD and BC ∥ AD, so slope AB = slope CD and slope BC = slope AD.
📝 VERY SHORT ANSWER TYPE QUESTIONS
Q34. Find the equation of the straight line which passes through the point (1, -2).
Answer: Infinite lines pass through a point. One example: x + y + 1 = 0
Q35. Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, -1).
Answer: m = (-1-3)/(3-2) = -4
Perpendicular slope = 1/4
y - 2 = (1/4)(x - 5) ⇒ 4y - 8 = x - 5 ⇒ x - 4y + 3 = 0
Q36. Find the angle between the lines y = (2 - √3)(x + 5) and y = (2 + √3)(x - 7).
Answer: m₁ = 2 - √3, m₂ = 2 + √3
tan θ = |(m₂ - m₁)/(1 + m₁m₂)| = |(2√3)/(1 + (4-3))| = |2√3/2| = √3
θ = 60°
Q37. Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.
Answer: Let intercepts be a and b. a + b = 14
x/a + y/b = 1 passes through (3,4) ⇒ 3/a + 4/b = 1
b = 14 - a ⇒ 3/a + 4/(14-a) = 1
3(14-a) + 4a = a(14-a) ⇒ 42 - 3a + 4a = 14a - a²
a² - 13a + 42 = 0 ⇒ (a-6)(a-7) = 0
If a=6, b=8: x/6 + y/8 = 1 ⇒ 4x + 3y = 24
If a=7, b=7: x/7 + y/7 = 1 ⇒ x + y = 7
Q38. Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10.
Answer: Point on x+y=4: (t, 4-t)
Distance from 4x+3y-10=0: |4t + 3(4-t) - 10|/5 = 1
|4t + 12 - 3t - 10| = 5 ⇒ |t + 2| = 5
t + 2 = 5 ⇒ t = 3 ⇒ (3,1)
t + 2 = -5 ⇒ t = -7 ⇒ (-7,11)
Q39. Find the equation of lines passing through (1, 2) and making angle 30° with y-axis.
Answer: Angle with y-axis = 30° ⇒ Angle with x-axis = 60°
m = tan 60° = √3
y - 2 = √3(x - 1) ⇒ y = √3x + 2 - √3
Q40. Find the equation of the line passing through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 1.
Answer: 2x+y=5 ...(1), x+3y+8=0 ...(2)
From (1): y=5-2x, substitute in (2): x+3(5-2x)+8=0 ⇒ x+15-6x+8=0 ⇒ -5x+23=0 ⇒ x=23/5
y = 5 - 46/5 = (25-46)/5 = -21/5
Point: (23/5, -21/5)
Line parallel to 3x+4y=1 ⇒ 3x+4y = k
3(23/5)+4(-21/5) = k ⇒ (69-84)/5 = -15/5 = -3
Equation: 3x + 4y + 3 = 0
📝 SHORT ANSWER TYPE QUESTIONS
Q51. Find the equation of the line where length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x-axis is 30°.
Answer: Normal form: x cos α + y sin α = p
α = 30°, p = 4
x cos30° + y sin30° = 4
(√3/2)x + (1/2)y = 4 ⇒ √3x + y = 8
Q52. Find the equation of the straight line passing through (1, 2) and perpendicular to the line x + y + 7 = 0.
Answer: x + y + 7 = 0 ⇒ m = -1
Perpendicular slope = 1
y - 2 = 1(x - 1) ⇒ y - x - 1 = 0 ⇒ x - y + 1 = 0
Q53. Find the distance between the lines 3x + 4y = 9 and 6x + 8y = 15.
Answer: 3x+4y-9=0 and 3x+4y-15/2=0
d = |9 - 15/2|/√(9+16) = |(18-15)/2|/5 = 3/2/5 = 3/10
Q54. If the slope of a line passing through the point A(3, 2) is 3/4.
Answer: y - 2 = (3/4)(x - 3) ⇒ 4y - 8 = 3x - 9 ⇒ 3x - 4y - 1 = 0
Q55. Find the equation to the straight line passing through (1, 1) and perpendicular to the line 3x - 5y + 11 = 0.
Answer: 3x - 5y + 11 = 0 ⇒ m = 3/5
Perpendicular slope = -5/3
y - 1 = (-5/3)(x - 1) ⇒ 3y - 3 = -5x + 5 ⇒ 5x + 3y - 8 = 0
Q56. A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and then passes through the point (5, 3). Find the coordinates of the point A.
Answer: Point A is (a, 0). Reflect (1,2) across x-axis: (1,-2)
Line from (1,-2) to (5,3) passes through A
Slope = (3+2)/(5-1) = 5/4
Equation: y + 2 = (5/4)(x - 1)
At y=0: 2 = (5/4)(x-1) ⇒ x-1 = 8/5 ⇒ x = 13/5
A = (13/5, 0)
Q57. If one diagonal of a square is along the line 8x - 15y = 0 and one of its vertices is at (1, 2), then find the equation of sides of the square passing through this vertex.
Answer: 8x - 15y = 0 ⇒ m = 8/15
Side is perpendicular to diagonal through (1,2) ⇒ m = -15/8
y - 2 = (-15/8)(x - 1) ⇒ 8y - 16 = -15x + 15 ⇒ 15x + 8y - 31 = 0
Other side: slope = 15/8 ⇒ 8y - 16 = 15x - 15 ⇒ 15x - 8y + 1 = 0
Q58. Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2).
Answer: m = (-2+1)/(4-3) = -1
tan θ = -1 ⇒ θ = 135° (since slope is negative, angle is obtuse)
Q59. The vertices of ΔPQR are P(2,1), Q(-2,3) and R(4,5). Find equation of the median through the vertex R.
Answer: Midpoint of PQ: ((2-2)/2, (1+3)/2) = (0,2)
Line through R(4,5) and (0,2): m = (5-2)/(4-0) = 3/4
y - 2 = (3/4)x ⇒ 4y - 8 = 3x ⇒ 3x - 4y + 8 = 0
Q60. Point R(h, k) divides a line segment between the axes in the ratio 1:2. Find equation of the line.
Answer: Let line be x/a + y/b = 1
R divides in ratio 1:2 ⇒ h = (2a + 0)/3 = 2a/3, k = (0 + b)/3 = b/3
a = 3h/2, b = 3k
Equation: x/(3h/2) + y/(3k) = 1 ⇒ 2x/h + y/k = 3
✅ ANSWERS - MCQs
| Q.No. | Answer | Q.No. | Answer |
|---|---|---|---|
| 1 | (A) | 11 | (B) |
| 2 | (A) | 12 | (D) |
| 3 | (B) | 13 | (A) |
| 4 | (C) | 14 | (A) |
| 5 | (B) | 15 | (A) |
| 6 | (A) | 16 | (D) |
| 7 | (a) | 17 | (C) |
| 8 | (a) | 18 | (B) |
| 9 | (b) | 19 | (C) |
| 10 | (A) | 20 | (A) |
📌 KEY FORMULAS SUMMARY
| Formula Type | Expression |
|---|---|
| Slope of line | m = (y₂-y₁)/(x₂-x₁) |
| Slope-Intercept | y = mx + c |
| Point-Slope | y - y₁ = m(x - x₁) |
| Two-Point | y - y₁ = (y₂-y₁)/(x₂-x₁)(x - x₁) |
| Intercept | x/a + y/b = 1 |
| Normal Form | x cos α + y sin α = p |
| General Form | Ax + By + C = 0 |
| Parallel lines | m₁ = m₂ |
| Perpendicular lines | m₁ × m₂ = -1 |
| Distance from point to line | |Ax₁+By₁+C|/√(A²+B²) |
| Distance between parallel lines | |C₁-C₂|/√(A²+B²) |
🔹 Important Conditions:
- Lines parallel: A₁/A₂ = B₁/B₂ ≠ C₁/C₂
- Lines coincide: A₁/A₂ = B₁/B₂ = C₁/C₂
- Lines perpendicular: A₁A₂ + B₁B₂ = 0
- Slope of x-axis = 0, y-axis = undefined
📚 Kendriya Vidyalaya Sangathan | Class XI Mathematics
Chapter 10: Straight Lines | Student Support Material