Class 11 : Maths (In English) – Lesson 9. Straight Lines

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On June 21, 2025

EXPLANATION & SUMMARY

🔵 EXPLANATION SECTION
🔷 1. Introduction to Straight Line
🟢 A straight line is the simplest curve which extends infinitely in both directions.
🟢 It is the shortest distance between two points.
🟢 In coordinate geometry, a line is defined by an equation connecting x and y.

🔷 2. Slope of a Line (m)
🧠 The slope or gradient of a line measures its inclination with the x-axis.
💡 Concept:
If a line makes an angle θ with the positive direction of the x-axis,
➡️ Slope (m) = tanθ

✏️ Note:
If θ = 0°, line is parallel to x-axis ⇒ m = 0
If θ = 90°, line is parallel to y-axis ⇒ slope is undefined
If line rises from left to right ⇒ m > 0
If line falls from left to right ⇒ m < 0
🔹 Slope using two points:
If a line passes through points P(x₁, y₁) and Q(x₂, y₂),
➡️ m = (y₂ − y₁) / (x₂ − x₁), provided x₂ ≠ x₁.

🔷 3. Different Forms of Line Equation
🟢 (a) Point-Slope Form
If a line passes through (x₁, y₁) and has slope m,
➡️ Equation: y − y₁ = m(x − x₁)

🟢 (b) Slope-Intercept Form
If a line has slope m and y-intercept c,
➡️ Equation: y = mx + c
Here, c = point where line cuts y-axis.
💡 Example: y = 2x + 3
Slope = 2, y-intercept = 3

🟢 (c) Two-Point Form
If a line passes through (x₁, y₁) and (x₂, y₂):
➡️ (y − y₁)/(x − x₁) = (y₂ − y₁)/(x₂ − x₁)

🟢 (d) Intercept Form
If a line cuts the x-axis at a and y-axis at b,
➡️ x/a + y/b = 1
✏️ Note: a = x-intercept, b = y-intercept.

🟢 (e) Normal Form
If perpendicular from origin on line = p, angle with x-axis = α,
➡️ Equation: x cosα + y sinα = p

🟢 (f) General Form
Equation of any line: Ax + By + C = 0
Slope (m) = −A/B,
x-intercept = −C/A,
y-intercept = −C/B

🔷 4. Angle Between Two Lines
For lines with slopes m₁ and m₂:
➡️ tanθ = |(m₁ − m₂) / (1 + m₁m₂)|
Special cases:
Lines parallel ⇒ m₁ = m₂ ⇒ θ = 0
Lines perpendicular ⇒ m₁m₂ = −1 ⇒ θ = 90°

🔷 5. Conditions for Parallel & Perpendicular Lines
🟢 Parallel lines: m₁ = m₂
🟢 Perpendicular lines: m₁ × m₂ = −1

🔷 6. Collinearity of Three Points
Points (x₁, y₁), (x₂, y₂), (x₃, y₃) are collinear if they lie on same line.
Condition:
➡️ Slopes of (P₁P₂) = slope of (P₂P₃)
i.e. (y₂ − y₁)/(x₂ − x₁) = (y₃ − y₂)/(x₃ − x₂)

🔷 7. Distance of a Point from a Line
Line: Ax + By + C = 0
Point: (x₁, y₁)
➡️ Distance (d) = |A x₁ + B y₁ + C| / sqrt(A² + B²)
💡 Use for shortest distance from point to line.

🔷 8. Distance Between Two Parallel Lines
Lines: Ax + By + C₁ = 0 and Ax + By + C₂ = 0
➡️ Distance = |C₁ − C₂| / sqrt(A² + B²)

🔷 9. Family of Lines
🟢 Passing through intersection of two lines:
If L₁ ≡ A₁x + B₁y + C₁ = 0 and L₂ ≡ A₂x + B₂y + C₂ = 0
➡️ Family: L₁ + λL₂ = 0 (λ is a parameter)

🔷 10. Pair of Perpendicular Bisectors
If line bisects segment joining (x₁, y₁) and (x₂, y₂):
Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Slope of segment = (y₂ − y₁)/(x₂ − x₁)
Slope of bisector = −1 / slope of segment
Use point-slope form for equation.

🔷 11. Important Observations
Line parallel to x-axis ⇒ y = k
Line parallel to y-axis ⇒ x = a
Line passing through origin ⇒ y = mx

🔷 12. Examples
Example 1:
Find equation of line passing through (2, 3) with slope 2.
➡️ y − 3 = 2(x − 2) ⇒ y = 2x − 1 ✅
Example 2:
Find slope and intercept of y = 3x + 2
➡️ m = 3, y-intercept = 2 ✅
Example 3:
Find equation of line with x-intercept = 4, y-intercept = 3
➡️ x/4 + y/3 = 1 ✅
Example 4:
Find distance from point (2, −3) to line 3x + 4y + 5 = 0
➡️ d = |3×2 + 4(−3) + 5| / √(3² + 4²)
= |6 − 12 + 5| / 5 = 1/5 × 5 = 1 ✅
Example 5:
Equation of line perpendicular to 2x + 3y − 5 = 0 passing through (1, 2):
Slope of given line = −A/B = −2/3
Perpendicular slope = 3/2
Equation: y − 2 = (3/2)(x − 1) ✅

🔷 13. Graphical Insight
🟩 Line y = mx + c:
m controls steepness
c controls vertical position
🟩 All equations of first degree in x, y represent straight lines.

🔷 14. Conversion Between Forms of a Line Equation
Sometimes we need to change one form of a line’s equation into another. Below are the common conversions explained in plain text:

🔹 From General Form (Ax + By + C = 0) to Slope-Intercept Form (y = mx + c):
➡️ Rearrange the equation to isolate y.
➡️ y = (−A/B)x − (C/B)
✔️ Here, slope m = −A/B and y-intercept c = −C/B.

🔹 From Intercept Form (x/a + y/b = 1) to General Form (Ax + By + C = 0):
➡️ Multiply both sides by LCM of denominators (ab):
➡️ b·x + a·y = ab
➡️ Rearrange: b·x + a·y − ab = 0

🔹 From Normal Form (x cosα + y sinα = p) to General Form (Ax + By + C = 0):
➡️ Expand directly:
A = cosα, B = sinα, C = −p
So, equation becomes Ax + By + C = 0.

🔹 From Slope-Intercept Form (y = mx + c) to General Form:
➡️ Bring all terms to one side:
➡️ mx − y + c = 0
Here, A = m, B = −1, C = c.
🔷 15. Applications
✔️ Used in physics for motion equations
✔️ Economics: cost lines, budget lines
✔️ Geometry: tangents, perpendiculars
✔️ Coordinate geometry problems in JEE, CBSE, NEET

🟢 SUMMARY (~300 words)
🔹 A straight line is defined by a linear equation in x and y.
🔹 Slope (m) = tanθ measures inclination.
🔹 Equation forms:
Point-slope: y − y₁ = m(x − x₁)
Slope-intercept: y = mx + c
Two-point: (y − y₁)/(x − x₁) = (y₂ − y₁)/(x₂ − x₁)
Intercept: x/a + y/b = 1
Normal: x cosα + y sinα = p
General: Ax + By + C = 0
🔹 Parallel lines: m₁ = m₂
🔹 Perpendicular: m₁m₂ = −1
🔹 Collinearity: Equal slopes between pairs.
🔹 Distance formulas:
Point to line: |Ax₁ + By₁ + C| / √(A² + B²)
Between parallels: |C₁ − C₂| / √(A² + B²)
🔹 Family of lines: L₁ + λL₂ = 0
🔹 Graph: y = mx + c passes through (0, c).
🔹 Line parallel to x-axis: y = k
🔹 Line parallel to y-axis: x = a
✔️ Important results:
Line through origin: y = mx
Equation with infinite slope ⇒ vertical line
Line with m = 0 ⇒ horizontal line

📝 QUICK RECAP (~100 words)
✔️ Slope m = tanθ = (y₂ − y₁)/(x₂ − x₁)
✔️ Forms:
y − y₁ = m(x − x₁)
y = mx + c
x/a + y/b = 1
✔️ Distance = |Ax₁ + By₁ + C| / √(A² + B²)
✔️ Parallel ⇒ m₁ = m₂
✔️ Perpendicular ⇒ m₁m₂ = −1
✔️ Intersections, perpendicular bisectors via point-slope
✔️ Graph y = mx + c: c is y-intercept
✔️ All linear equations Ax + By + C = 0 represent straight lines.

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QUESTIONS FROM TEXTBOOK

Exercise 9.1

🔵 Question 1:
Draw a quadrilateral in the Cartesian plane, whose vertices are A(−4, 5), B(0, 7), C(5, −5), and D(−4, −2).
Also, find its area.

🟢 Answer:
✨ Let vertices:
🔹 A(−4, 5)
🔹 B(0, 7)
🔹 C(5, −5)
🔹 D(−4, −2)
💡 Formula (Shoelace Method):
📘 Area = ½ × |x₁(y₂−y₄) + x₂(y₃−y₁) + x₃(y₄−y₂) + x₄(y₁−y₃)|
➡️ Step 1: Substitute values:
= ½ × |(−4)(7 − (−2)) + 0(−5 − 5) + 5(−2 − 7) + (−4)(5 − (−5))|
➡️ Step 2: Simplify parentheses:
= ½ × |(−4)(9) + 0(−10) + 5(−9) + (−4)(10)|
➡️ Step 3: Multiply:
= ½ × |−36 + 0 − 45 − 40|
➡️ Step 4: Add:
= ½ × |−121|
➡️ Step 5: Take magnitude and halve:
= ½ × 121 = 60.5 sq units
✔️ Final Answer: Area = 60.5 sq units

🔵 Question 2:
The base of an equilateral triangle with side 2a lies along the y-axis such that the midpoint of the base is at the origin.
Find the vertices.

🟢 Answer:
💡 Let midpoint O(0, 0)
➡️ Base vertices:
B(0, −a), C(0, a)
✅ Base length = |a − (−a)| = 2a
💡 Height (from equilateral triangle):
h = √(a² − (a²/4)) = √(3a²/4) = (√3 a)/2
➡️ Third vertex:
A((√3 a)/2, 0)
✔️ Vertices:
A((√3 a)/2, 0), B(0, −a), C(0, a)

🔵 Question 3:
Find the distance between P(x₁, y₁) and Q(x₂, y₂) when:
(i) PQ ∥ y-axis
(ii) PQ ∥ x-axis

🟢 Answer:
💡 General formula:
d = √((x₂ − x₁)² + (y₂ − y₁)²)

🔹 (i) PQ ∥ y-axis ⇒ x₁ = x₂
➡️ d = |y₂ − y₁|
🔹 (ii) PQ ∥ x-axis ⇒ y₁ = y₂
➡️ d = |x₂ − x₁|
✔️ Final Answers:
(i) |y₂ − y₁|
(ii) |x₂ − x₁|

🔵 Question 4:
Find a point on the x-axis which is equidistant from points A(7, 6) and B(3, 4).

🟢 Answer:
💡 Let point on x-axis = P(x, 0)
✨ Condition:
PA = PB
➡️ √((x−7)² + (0−6)²) = √((x−3)² + (0−4)²)
➡️ Square both sides:
(x − 7)² + 36 = (x − 3)² + 16
➡️ Expand:
x² − 14x + 49 + 36 = x² − 6x + 9 + 16
➡️ Simplify:
−14x + 85 = −6x + 25
➡️ −8x = −60
➡️ x = 7.5
✔️ Required point: (7.5, 0)

🔵 Question 5:
Find the slope of a line which passes through the origin and the midpoint of the line segment joining points P(0, −4) and B(8, 0).

🟢 Answer:
💡 Step 1: Midpoint formula
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
➡️ M = ((0 + 8)/2, (−4 + 0)/2) = (4, −2)
💡 Step 2: Slope formula
m = (y₂ − y₁)/(x₂ − x₁)
➡️ Through O(0, 0) and M(4, −2):
m = (−2 − 0)/(4 − 0) = −2/4 = −½
✔️ Slope = −½

🔵 Question 6:
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5), and (−1, −1) are the vertices of a right-angled triangle.

🟢 Answer:
Let points:
🔹 A(4, 4), B(3, 5), C(−1, −1)
💡 Step 1: Find slopes of each side
➡️ Slope of AB = (y₂ − y₁)/(x₂ − x₁) = (5 − 4)/(3 − 4) = 1/(−1) = −1
➡️ Slope of BC = (−1 − 5)/(−1 − 3) = (−6)/(−4) = 3/2
➡️ Slope of CA = (4 − (−1))/(4 − (−1)) = 5/5 = 1
💡 Step 2: Condition for perpendicular lines → m₁ × m₂ = −1
Check:
Slope(AB) × Slope(CA) = (−1)(1) = −1 ✅
✔️ Hence, AB ⟂ CA → ∠A = 90°
✔️ Therefore, triangle ABC is right-angled at A(4, 4).

🔵 Question 7:
Find the slope of the line which makes an angle 30° with the positive direction of y-axis measured anticlockwise.

🟢 Answer:
💡 Angle with x-axis = 90° − 30° = 60°
💡 Slope formula:
m = tan(θ) = tan(60°) = √3
✔️ Slope = √3

🔵 Question 8:
Without using distance formula, show that points (−2, −1), (4, 0), (3, 3), and (−3, 2) are the vertices of a parallelogram.

🟢 Answer:
Let points: A(−2, −1), B(4, 0), C(3, 3), D(−3, 2)
💡 Step 1: Find slopes of opposite sides
➡️ Slope(AB) = (0 − (−1))/(4 − (−2)) = 1/6
➡️ Slope(CD) = (2 − 3)/(−3 − 3) = (−1)/(−6) = 1/6 ✅
➡️ Slope(BC) = (3 − 0)/(3 − 4) = 3/(−1) = −3
➡️ Slope(AD) = (2 − (−1))/(−3 − (−2)) = 3/(−1) = −3 ✅
💡 Step 2: Both pairs of opposite sides have equal slopes ⇒ parallel ✅
✔️ AB ∥ CD and BC ∥ AD
✔️ Hence, ABCD is a parallelogram.

🔵 Question 9:
Find the angle between the x-axis and the line joining the points (3, −1) and (4, −2).

🟢 Answer:
💡 Slope (m) = (−2 − (−1))/(4 − 3) = (−1)/1 = −1
💡 Formula: tan θ = |m| = 1
⇒ θ = tan⁻¹(1) = 45°
✔️ Angle = 45°

🔵 Question 10:
The slope of a line is double the slope of another line. If tan θ = 1/3 is the tangent of the angle between them, find the slopes.

🟢 Answer:
Let slopes be m₁ and m₂
Given: m₁ = 2m₂
💡 Formula:
tan θ = |(m₁ − m₂)/(1 + m₁m₂)|
Substitute:
1/3 = |(2m₂ − m₂)/(1 + 2m₂²)|
⇒ 1/3 = |m₂/(1 + 2m₂²)|
Multiply:
3m₂ = ±(1 + 2m₂²)
➡️ Case 1: 3m₂ = 1 + 2m₂²
⇒ 2m₂² − 3m₂ + 1 = 0
⇒ (2m₂ − 1)(m₂ − 1) = 0
⇒ m₂ = 1/2 or 1
Then m₁ = 1 or 2 ✅
✔️ Slopes are (1/2, 1) or (1, 2)

🔵 Question 11:
A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that
k − y₁ = m(h − x₁).

🟢 Answer:
💡 Slope formula:
m = (k − y₁)/(h − x₁)
Multiply both sides by (h − x₁):
m(h − x₁) = k − y₁ ✅
✔️ Hence proved: k − y₁ = m(h − x₁)

Exercise 9.2

🔵 Question 1:
Write the equations for the x-axis and y-axis.
🟢 Answer:
💡 On the x-axis, every point has y = 0
➡️ Equation: y = 0 ✅
💡 On the y-axis, every point has x = 0
➡️ Equation: x = 0 ✅
✔️ Final Answers:
🔹 x-axis → y = 0
🔹 y-axis → x = 0

🔵 Question 2:
Find the equation of a line passing through the point (−4, 3) with slope = ½.
🟢 Answer:
💡 Point-slope form:
y − y₁ = m(x − x₁)
➡️ Substitute: y − 3 = ½(x − (−4))
➡️ y − 3 = ½(x + 4)
✳️ Multiply by 2:
2y − 6 = x + 4
➡️ x − 2y + 10 = 0 ✅
✔️ Final Equation: x − 2y + 10 = 0

🔵 Question 3:
Find the equation of a line passing through (0, 0) with slope = m.
🟢 Answer:
💡 Formula: y − y₁ = m(x − x₁)
➡️ y − 0 = m(x − 0)
➡️ y = mx ✅
✔️ Final Equation: y = mx

🔵 Question 4:
Find the equation of a line passing through (2, 2√3) and inclined with the x-axis at 75°.
🟢 Answer:
💡 Slope: m = tan(75°) = 2 + √3 ✅
💡 Point-slope form:
y − y₁ = m(x − x₁)
➡️ y − 2√3 = (2 + √3)(x − 2)
✔️ Final Equation:
y − 2√3 = (2 + √3)(x − 2)

🔵 Question 5:
Find the equation of a line intersecting the x-axis at a distance of 3 units left of origin with slope = −2.
🟢 Answer:
💡 Point on x-axis = (−3, 0)
Slope = −2
💡 Equation:
y − 0 = −2(x − (−3))
➡️ y = −2(x + 3)
➡️ y = −2x − 6 ✅
✔️ Final Equation: y = −2x − 6

🔵 Question 6:
Find the equation of a line intersecting the y-axis at a distance of 2 units above origin and making an angle of 30° with the positive x-axis.
🟢 Answer:
💡 Point = (0, 2),
Slope = tan(30°) = 1/√3
💡 Equation:
y − 2 = (1/√3)(x − 0)
➡️ y = (1/√3)x + 2 ✅
✔️ Final Equation: y = (1/√3)x + 2

🔵 Question 7:
Find the equation of a line passing through the points (−1, 1) and (2, −4).
🟢 Answer:
💡 Step 1: Slope
m = (−4 − 1) ÷ (2 − (−1))
➡️ m = (−5) ÷ 3 = −5/3
💡 Step 2: Point–slope form
y − y₁ = m(x − x₁)
➡️ y − 1 = (−5/3)(x + 1)
💡 Step 3: Remove fraction
3y − 3 = −5x − 5
➡️ 5x + 3y + 2 = 0
✔️ Final Equation: 5x + 3y + 2 = 0

🔵 Question 8:
Vertices of △PQR: P(2, 1), Q(−2, 3), R(4, 5).
Find the median through R.
🟢 Answer:
💡 Step 1: Midpoint of PQ
M = ((2 + (−2))/2 , (1 + 3)/2) = (0, 2)
💡 Step 2: Slope of RM
m = (5 − 2) ÷ (4 − 0) = 3/4
💡 Step 3: Equation through R(4, 5)
y − 5 = (3/4)(x − 4)
➡️ 4y − 20 = 3x − 12
➡️ 3x − 4y + 8 = 0
✔️ Final Equation: 3x − 4y + 8 = 0

🔵 Question 9:
Line through (−3, 5) perpendicular to the line through (2, 5) and (−3, 6).
🟢 Answer:
💡 Step 1: Slope of given line
m₁ = (6 − 5) ÷ (−3 − 2) = −1/5
💡 Step 2: Perpendicular slope
m₂ = 5
💡 Step 3: Equation through (−3, 5)
y − 5 = 5(x + 3)
➡️ y = 5x + 20
✔️ Final Equation: 5x − y + 20 = 0

🔵 Question 10:
A line perpendicular to the segment joining (1, 0) and (2, 3) divides it in ratio 1 : n.
Find its equation.
🟢 Answer:
💡 Step 1: Slope of segment
m₁ = (3 − 0) ÷ (2 − 1) = 3
➡️ Required slope m₂ = −1/3
💡 Step 2: Coordinates of division point P
xₚ = (1·2 + n·1)/(1 + n) = (n + 2)/(n + 1)
yₚ = (1·3 + n·0)/(1 + n) = 3/(n + 1)
💡 Step 3: Equation through P
y − (3/(n + 1)) = −1/3 [x − (n + 2)/(n + 1)]
Multiply by 3(n + 1):
3(n + 1)y − 9 = −(n + 1)x + (n + 2)
➡️ (n + 1)x + 3(n + 1)y − (n + 11) = 0
✔️ Final Equation: (n + 1)x + 3(n + 1)y − (n + 11) = 0

🔵 Question 11:
Line cuts equal intercepts on axes and passes through (2, 3).
🟢 Answer:
💡 Step 1: Equation when intercepts equal (a, a):
x/a + y/a = 1 ⇒ x + y = a
💡 Step 2: Passes through (2, 3):
2 + 3 = a ⇒ a = 5
✔️ Final Equation: x + y = 5

🔵 Question 12:
Line passes through (2, 2) and intercepts on axes have sum = 9.
🟢 Answer:
💡 Step 1: Let intercepts = a and b
a + b = 9
Equation: x/a + y/b = 1
💡 Step 2: Substitute (2, 2):
2/a + 2/b = 1 ⇒ (a + b)/(ab) = 1/2
a + b = 9 ⇒ 9/(ab) = 1/2 ⇒ ab = 18
💡 Step 3: a, b satisfy t² − 9t + 18 = 0
➡️ t = 3 or 6
Two possible pairs:
• a = 6, b = 3 → x/6 + y/3 = 1 ⇒ x + 2y = 6
• a = 3, b = 6 → x/3 + y/6 = 1 ⇒ 2x + y = 6
✔️ Final Equations: x + 2y = 6 or 2x + y = 6

🔵 Question 13:
Find equation of the line through the point (0, 2) making an angle 2π/3 with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
🟢 Answer:
💡 Formula: Equation of line with slope m through (x₁, y₁):
➡️ y − y₁ = m (x − x₁)
🔹 Step 1: Find slope
m = tan(2π/3) = tan(120°) = −√3
🔹 Step 2: Equation through (0, 2):
y − 2 = (−√3)(x − 0)
➡️ y = −√3·x + 2
🔹 Step 3: Parallel line → same slope, y-intercept = −2
➡️ y = −√3·x − 2
✔️ Final answers:
(1) y = −√3·x + 2
(2) y = −√3·x − 2

🔵 Question 14:
The perpendicular from the origin to a line meets it at the point (−2, 9), find the equation of the line.
🟢 Answer:
💡 Slope from origin O(0, 0) to P(−2, 9):
m₁ = (9 − 0)/(−2 − 0) = −9/2
Perpendicular line → slope m₂ = −1/m₁ = 2/9
Equation through (−2, 9):
y − 9 = (2/9)(x + 2)
➡️ 9(y − 9) = 2(x + 2)
➡️ 9y − 81 = 2x + 4
➡️ 2x − 9y + 85 = 0
✔️ Final equation: 2x − 9y + 85 = 0

🔵 Question 15:
The length L (in cm) of a copper rod is a linear function of its Celsius temperature C.
If L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.
🟢 Answer:
💡 Let L = mC + b
🔹 Step 1: Find slope
m = (125.134 − 124.942)/(110 − 20) = 0.192/90 = 0.002133…
🔹 Step 2: Substitute C = 20, L = 124.942
124.942 = 0.002133×20 + b
➡️ b = 124.942 − 0.04266 = 124.89934
🔹 Step 3: Equation
➡️ L = 0.002133·C + 124.899
✔️ Final: L = 0.002133·C + 124.899

🔵 Question 16:
A milk store sells 980 L at ₹14/L and 1220 L at ₹16/L.
Assuming a linear relation, find litres sold at ₹17/L.
🟢 Answer:
💡 Let demand Q = a·p + b
(14, 980), (16, 1220)
Slope a = (1220 − 980)/(16 − 14) = 240/2 = 120
Substitute (14, 980):
980 = 120×14 + b → b = −700
At p = 17:
Q = 120×17 − 700 = 2040 − 700 = 1340 L
✔️ Final: 1340 litres

🔵 Question 17:
P(a, b) is the midpoint of a line between axes. Show equation:
x/a + y/b = 2
🟢 Answer:
Intercepts: (α, 0), (0, β)
Midpoint = (α/2, β/2) = (a, b)
⇒ α = 2a, β = 2b
Intercept form: x/α + y/β = 1
Substitute: x/(2a) + y/(2b) = 1
Multiply by 2:
➡️ x/a + y/b = 2
✔️ Proven

🔵 Question 18:
Point R(h, k) divides intercepts in ratio 1:2. Find equation.
🟢 Answer:
Intercepts (α, 0), (0, β)
Section formula:
x_R = (1·0 + 2·α)/3 = 2α/3 ⇒ α = 3h/2
y_R = (1·β + 2·0)/3 = β/3 ⇒ β = 3k
Intercept form:
x/α + y/β = 1 ⇒ x/(3h/2) + y/(3k) = 1
Multiply by 3:
➡️ (2x)/h + (y)/k = 3
✔️ Equation: (2x)/h + (y)/k = 3

🔵 Question 19:
Prove (3, 0), (−2, −2), (8, 2) are collinear.
🟢 Answer:
Slope (3, 0), (−2, −2): m = (−2 − 0)/(−2 − 3) = (−2)/(−5) = 2/5
Equation: y − 0 = (2/5)(x − 3)
➡️ 5y = 2x − 6 ⇒ 2x − 5y − 6 = 0
Check (8, 2):
2×8 − 5×2 − 6 = 16 − 10 − 6 = 0 ✅
✔️ Hence all three points lie on same line → Collinear

🧠 Exercise 9.3

🔵 Question 1
Reduce the following equations into slope–intercept form and find their slopes and y-intercepts:
(i) x + 7y = 0
(ii) 6x + 3y − 5 = 0
(iii) y = 0
🟢 Answer
(i) x + 7y = 0
➡️ 7y = −x
➡️ y = −(1/7)x + 0
✔️ Slope (m) = −1/7, y-intercept = 0
(ii) 6x + 3y − 5 = 0
➡️ 3y = −6x + 5
➡️ y = −2x + 5/3
✔️ Slope (m) = −2, y-intercept = 5/3
(iii) y = 0
➡️ Already in slope-intercept form y = 0x + 0
✔️ Slope (m) = 0, y-intercept = 0

🔵 Question 2
Reduce the following equations into intercept form and find their intercepts on the axes:
(i) 3x + 2y − 12 = 0
(ii) 4x − 3y = 6
(iii) 3y + 2 = 0
🟢 Answer
(i) 3x + 2y − 12 = 0
➡️ 3x + 2y = 12
Divide by 12:
➡️ x/4 + y/6 = 1
✔️ x-intercept = 4, y-intercept = 6
(ii) 4x − 3y = 6
Divide by 6:
➡️ x/(3/2) − y/2 = 1
✔️ x-intercept = 3/2, y-intercept = −2
(iii) 3y + 2 = 0
➡️ 3y = −2
➡️ y = −2/3
No x-term ⇒ line parallel to x-axis
✔️ x-intercept = none, y-intercept = −2/3

🔵 Question 3
Find the distance of the point (−1, 1) from the line 12(x + 6) = 5(y − 2).
🟢 Answer
Expand: 12x + 72 = 5y − 10
➡️ 12x − 5y + 82 = 0
💡 Distance formula:
D = |Ax₁ + By₁ + C| / √(A² + B²)
Substitute (x₁, y₁) = (−1, 1):
D = |12(−1) − 5(1) + 82| / √(12² + (−5)²)
➡️ = |−12 − 5 + 82| / √(144 + 25)
➡️ = 65 / 13 = 5 units
✔️ Distance = 5 units

🔵 Question 4
Find the points on the x-axis whose distances from the line x/3 + y/4 = 1 are 4 units.
🟢 Answer
Equation: Multiply by 12 → 4x + 3y = 12
On x-axis, y = 0.
Let point be (x, 0).
Distance D = |4x + 3(0) − 12| / √(4² + 3²) = |4x − 12| / 5 = 4
➡️ |4x − 12| = 20
Case 1: 4x − 12 = 20 → x = 8
Case 2: 4x − 12 = −20 → x = −2
✔️ Points: (8, 0) and (−2, 0)

🔵 Question 5
Find the distance between parallel lines:
(i) 15x + 8y − 34 = 0 and 15x + 8y + 31 = 0
(ii) l(x + y) + p = 0 and l(x + y) − r = 0
🟢 Answer
💡 Formula: D = |C₁ − C₂| / √(A² + B²)
(i) D = |−34 − 31| / √(15² + 8²) = 65 / 17 = 65/17 units
(ii) A = l, B = l
D = |p − (−r)| / √(l² + l²) = |p + r| / (l√2)
✔️ Distance = |p + r| / (l√2)

🔵 Question 6
Find equation of the line parallel to 3x − 4y + 2 = 0 and passing through (−2, 3).
🟢 Answer
Parallel line ⇒ same coefficients A, B
Equation: 3x − 4y + k = 0
Substitute (−2, 3):
3(−2) − 4(3) + k = 0 → −6 − 12 + k = 0 → k = 18
✔️ Equation: 3x − 4y + 18 = 0

🔵 Question 7
Find equation of the line perpendicular to x − 7y + 5 = 0 and having x-intercept 3.
🟢 Answer
Given line slope m₁ = 1/7
Perpendicular slope m₂ = −7
x-intercept = 3 ⇒ point (3, 0)
Equation: y − 0 = −7(x − 3)
➡️ y = −7x + 21
✔️ Equation: 7x + y − 21 = 0

🔵 Question 8
Find angles between the lines √3x + y = 1 and x + √3y = 1.
🟢 Answer
Slopes:
m₁ = −√3, m₂ = −1/√3
💡 Formula: tanθ = |(m₂ − m₁)/(1 + m₁m₂)|
➡️ = |(−1/√3 + √3)/(1 + (−√3)(−1/√3))| = |( (−1 + 3)/√3 ) / 2| = (2/√3)/2 = 1/√3
✔️ θ = 30°

🔵 Question 9
The line through (h, 3) and (4, 1) intersects 7x − 9y − 19 = 0 at right angle. Find h.
🟢 Answer
Slope of given line:
7x − 9y − 19 = 0 → y = (7/9)x − 19/9
➡️ m₁ = 7/9
Line through (h, 3) and (4, 1):
m₂ = (1 − 3)/(4 − h) = (−2)/(4 − h)
Perpendicular ⇒ m₁·m₂ = −1
(7/9) × (−2)/(4 − h) = −1
➡️ (−14)/(9(4 − h)) = −1
➡️ 14 = 9(4 − h)
➡️ 14 = 36 − 9h
➡️ 9h = 22
➡️ h = 22/9
✔️ Final value: h = 22/9

🔵 Question 10
Prove that the line through the point (x₁, y₁) and parallel to the line Ax + By + C = 0 is
A(x − x₁) + B(y − y₁) = 0
🟢 Answer
💡 A line parallel to Ax + By + C = 0 has same normal vector (A, B).
So, general form: A·x + B·y + D = 0
Since it passes through (x₁, y₁):
➡️ A·x₁ + B·y₁ + D = 0 ⇒ D = −(A·x₁ + B·y₁)
Substitute:
➡️ A·x + B·y − (A·x₁ + B·y₁) = 0
✔️ A(x − x₁) + B(y − y₁) = 0 ✅

🔵 Question 11
Two lines passing through the point (2, 3) intersect each other at an angle of 60°.
If slope of one line is 2, find equation of the other line.
🟢 Answer
💡 Formula: tanθ = |(m₂ − m₁) / (1 + m₁m₂)|
Here, θ = 60°, m₁ = 2
➡️ √3 = |(m₂ − 2)/(1 + 2m₂)|
Case 1: (m₂ − 2)/(1 + 2m₂) = √3
➡️ m₂ − 2 = √3 + 2√3·m₂
➡️ m₂(1 − 2√3) = 2 + √3
➡️ m₂ = (2 + √3)/(1 − 2√3)
Multiply numerator & denominator by (1 + 2√3):
➡️ m₂ = [(2 + √3)(1 + 2√3)] / (1 − 12) = (2 + 4√3 + √3 + 6)/(−11) = (8 + 5√3)/(−11)
Case 2: (m₂ − 2)/(1 + 2m₂) = −√3
➡️ m₂ − 2 = −√3 − 2√3·m₂
➡️ m₂(1 + 2√3) = 2 − √3
➡️ m₂ = (2 − √3)/(1 + 2√3)
✳️ Either value acceptable.
Equation of line:
➡️ y − 3 = m₂(x − 2)
✔️ Equation: y − 3 = m₂(x − 2)

🔵 Question 12
Find the equation of the right bisector of the line segment joining the points (3, 4) and (−1, 2).
🟢 Answer
💡 Midpoint: M = ((3 + (−1))/2, (4 + 2)/2) = (1, 3)
Slope of AB = (4 − 2)/(3 − (−1)) = 2/4 = 1/2
Perpendicular slope = −2
Equation through (1, 3):
➡️ y − 3 = −2(x − 1)
➡️ y − 3 = −2x + 2
➡️ 2x + y − 5 = 0
✔️ Equation: 2x + y − 5 = 0 ✅

🔵 Question 13
Find the coordinates of the foot of perpendicular from the point (−1, 3) to the line 3x − 4y − 16 = 0.
🟢 Answer
💡 Foot formula:
If line: Ax + By + C = 0, point (x₁, y₁), foot =
(x, y) =
( (B(Bx₁ − Ay₁) − A·C)/(A² + B²), (A(Ay₁ − Bx₁) − B·C)/(A² + B²) )
Here: A = 3, B = −4, C = −16, (x₁, y₁) = (−1, 3)
Compute:
Bx₁ − Ay₁ = (−4)(−1) − 3(3) = 4 − 9 = −5
Ay₁ − Bx₁ = 3(3) − (−4)(−1) = 9 − 4 = 5
x = [ (−4)(−5) − 3(−16) ] / (9 + 16) = (20 + 48)/25 = 68/25
y = [ 3(5) − (−4)(−16) ] / 25 = (15 − 64)/25 = −49/25
✔️ Foot: (68/25, −49/25)

🔵 Question 14
The perpendicular from the origin to the line y = m·x + c meets it at the point (−1, 2). Find the values of m and c.
🟢 Answer
Point (−1, 2) lies on y = m·x + c
➡️ 2 = m(−1) + c ⇒ c = 2 + m
Also, line perpendicular to this passes through origin ⇒ slope = −1/m
Equation of perpendicular: y = (−1/m)x
Since point (−1, 2) lies on it:
2 = (−1/m)(−1) ⇒ m = 1/2
Then c = 2 + 1/2 = 5/2
✔️ m = 1/2, c = 5/2

🔵 Question 15
If p and q are the lengths of perpendiculars from the origin to the lines
x·cosθ − y·sinθ = k·cos2θ and x·secθ + y·cosecθ = k, respectively,
prove that p² + 4q² = k².
🟢 Answer
First line:
A = cosθ, B = −sinθ, C = −k·cos2θ
p = |C| / √(A² + B²) = |−k·cos2θ| / 1 = k|cos2θ|
Second line:
A = secθ, B = cosecθ, C = −k
q = |C| / √(A² + B²) = k / √(sec²θ + cosec²θ)
Compute denominator:
sec²θ + cosec²θ = (1/cos²θ) + (1/sin²θ) = (sin²θ + cos²θ)/(sin²θ·cos²θ) = 1/(sin²θ·cos²θ)
⇒ √(…) = 1/(sinθ·cosθ)
Thus q = k·sinθ·cosθ
Now prove:
p² + 4q² = k²·cos²2θ + 4k²·sin²θ·cos²θ
Use identity: cos2θ = cos²θ − sin²θ
➡️ cos²2θ = (cos²θ − sin²θ)² = cos⁴θ + sin⁴θ − 2sin²θ·cos²θ
Add 4sin²θ·cos²θ:
cos⁴θ + sin⁴θ + 2sin²θ·cos²θ = (cos²θ + sin²θ)² = 1
So p² + 4q² = k² ✅
✔️ Hence proved

🔵 Question 16
In triangle ABC with vertices A(2, 3), B(4, −1), C(1, 2), find the equation and length of altitude from vertex A.
🟢 Answer
💡 Slope BC = (2 − (−1)) / (1 − 4) = 3 / (−3) = −1
Perpendicular slope from A ⇒ m = 1
Equation through A(2, 3):
y − 3 = 1(x − 2) ⇒ y = x + 1
✔️ Altitude: y = x + 1
Distance from A to BC:
Equation BC: y + x − 3 = 0
Length = |2 + 3 − 3| / √(1² + 1²) = 2 / √2 = √2
✔️ Equation: y = x + 1, Length: √2 units

🔵 Question 17
If p is the length of perpendicular from origin to the line whose intercepts on axes are a and b, then show that
1/p² = 1/a² + 1/b²
🟢 Answer
Equation: x/a + y/b = 1 ⇒ multiply by ab:
b·x + a·y = a·b
💡 Distance from origin:
p = |a·b| / √(a² + b²)
Square: p² = a²b² / (a² + b²)
Take reciprocal:
1/p² = (a² + b²)/(a²b²) = 1/a² + 1/b² ✅
✔️ Hence proved

CBSE STYLE BOARD PAPER

ESPECIALLY FROM THIS CHAPTER ONLY

🧭 Section A – Very Short Answer / MCQs (1 mark each)
🔵 Question 1:
The slope of the line passing through (2, 3) and (4, 7) is
🔵 (A) 1
🟢 (B) 2
🟠 (C) 3
🔴 (D) 4
🟢 Answer: 2
➡️ m = (7 − 3)/(4 − 2) = 4/2 = 2

🔵 Question 2:
The line parallel to x-axis has slope
🔵 (A) 0
🟢 (B) 1
🟠 (C) ∞
🔴 (D) undefined
🟢 Answer: 0

🔵 Question 3:
Equation of y-axis is
🔵 (A) y = 0
🟢 (B) x = 0
🟠 (C) x = 1
🔴 (D) y = 1
🟢 Answer: x = 0

🔵 Question 4:
Equation of line with slope 2 and y-intercept −3 is
🔵 (A) y = 2x − 3
🟢 (B) y = 2x + 3
🟠 (C) y = 3x − 2
🔴 (D) y = 3x + 2
🟢 Answer: y = 2x − 3

🔵 Question 5:
Slope of line 3x + 4y + 5 = 0 is
🔵 (A) −4/3
🟢 (B) 3/4
🟠 (C) −3/4
🔴 (D) 4/3
🟢 Answer: −3/4
➡️ y = −(3/4)x − 5/4 ⇒ m = −3/4

🔵 Question 6:
Equation of line passing through (1, 2) with slope 3 is
🔵 (A) y − 2 = 3(x − 1)
🟢 (B) y + 2 = 3(x + 1)
🟠 (C) y = 3x + 2
🔴 (D) x = 3y − 2
🟢 Answer: y − 2 = 3(x − 1)

🔵 Question 7:
Equation of line parallel to y-axis passing through (3, −2):
🔵 (A) x = 3
🟢 (B) y = 3
🟠 (C) y = −2
🔴 (D) x = −2
🟢 Answer: x = 3

🔵 Question 8:
Equation of line perpendicular to 2x − y + 5 = 0 and passing through (0, 0):
🔵 (A) x + 2y = 0
🟢 (B) 2x + y = 0
🟠 (C) x − 2y = 0
🔴 (D) y − 2x = 0
🟢 Answer: 2x + y = 0

🔵 Question 9:
Two lines are perpendicular if product of their slopes is
🔵 (A) 1
🟢 (B) −1
🟠 (C) 0
🔴 (D) ∞
🟢 Answer: −1

🔵 Question 10:
The line with equation x + y = 5 makes equal intercepts on axes.
🟢 Answer: True
➡️ Intercepts: a = 5, b = 5

🔵 Question 11:
Find slope of line whose equation is 5y = 2x + 3
➡️ y = (2/5)x + 3/5
🟢 Answer: m = 2/5

🔵 Question 12:
If slope of one line is 3, the slope of line perpendicular to it is
➡️ m₂ = −1/3
🟢 Answer: −1/3

🔵 Question 13:
The y-intercept of line 4x + 5y − 20 = 0 is
➡️ y = −(4/5)x + 4
🟢 Answer: 4

🔵 Question 14:
Equation of line with x-intercept 4 and y-intercept 2
➡️ x/4 + y/2 = 1
🟢 Answer: x/4 + y/2 = 1

🔵 Question 15:
If a line passes through (0, 2) and (2, 0), its slope is
➡️ m = (0 − 2)/(2 − 0) = −1
🟢 Answer: −1

🔵 Question 16:
Equation of line passing through (2, 3) and (4, 7):
➡️ m = (7 − 3)/(4 − 2) = 2
➡️ y − 3 = 2(x − 2) ⇒ y = 2x − 1
🟢 Answer: y = 2x − 1

🔵 Question 17:
Condition for lines y = m₁x + c₁ and y = m₂x + c₂ to be parallel
🟢 Answer: m₁ = m₂

🔵 Question 18:
Condition for perpendicularity:
🟢 Answer: m₁ × m₂ = −1

🧭 Section B – Short Answer (2 marks each)
🔵 Question 19:
Find the equation of the line passing through (2, 3) and having slope 4.
🟢 Answer:
➡️ Formula: y − y₁ = m(x − x₁)
➡️ Substitution: y − 3 = 4(x − 2)
➡️ Simplify: y − 3 = 4x − 8
➡️ Rearrange: y = 4x − 5

🔵 Question 20:
Find the slope and intercepts of the line 2x + 3y = 6.
🟢 Answer:
➡️ Slope-intercept form: 3y = −2x + 6
➡️ y = (−2/3)x + 2
✔️ Slope m = −2/3
✔️ y-intercept c = 2
➡️ x-intercept when y = 0: 2x + 0 = 6 ⇒ x = 3

🔵 Question 21:
Find equation of line passing through (1, 2) and perpendicular to x − 2y + 3 = 0.
🟢 Answer:
➡️ Slope of given line m₁ = 1/2
➡️ Perpendicular slope m₂ = −1/m₁ = −2
➡️ Equation: y − 2 = −2(x − 1)
➡️ Simplify: y = −2x + 4

🔵 Question 22:
Find equation of line with intercepts 4 on x-axis and 3 on y-axis.
🟢 Answer:
➡️ Intercept form: x/a + y/b = 1
➡️ x/4 + y/3 = 1
➡️ Multiply by 12: 3x + 4y = 12

🔵 Question 23:
Find equation of line passing through (−2, 3) and perpendicular to the line joining (2, 5) and (4, −1).
🟢 Answer:
➡️ Slope of line through (2, 5), (4, −1):
m₁ = (−1 − 5)/(4 − 2) = −6/2 = −3
➡️ Perpendicular slope m₂ = −1/m₁ = 1/3
➡️ Equation: y − 3 = (1/3)(x + 2)
➡️ Multiply by 3: 3y − 9 = x + 2
➡️ Rearrange: x − 3y + 11 = 0

🧭 Section C – Mid-Length Answer (3 marks each)
🔵 Question 24:
Find distance between point P(3, 2) and line 3x − 4y + 5 = 0.
🟢 Answer:
➡️ Formula: D = |Ax₁ + By₁ + C| / √(A² + B²)
➡️ A = 3, B = −4, C = 5
➡️ Numerator = |3·3 + (−4)·2 + 5| = |9 − 8 + 5| = 6
➡️ Denominator = √(3² + (−4)²) = √(9 + 16) = √25 = 5
➡️ D = 6/5 = 1.2 units

🔵 Question 25:
Find equation of line joining points (1, 2) and (3, 6).
🟢 Answer:
➡️ Slope m = (6 − 2)/(3 − 1) = 4/2 = 2
➡️ Point-slope: y − 2 = 2(x − 1)
➡️ Simplify: y = 2x → y = 2x

🔵 Question 26:
Find equation of line making angle 45° with positive x-axis and passing through (2, 3).
🟢 Answer:
➡️ Slope m = tan 45° = 1
➡️ Equation: y − 3 = 1(x − 2)
➡️ y = x + 1

🔵 Question 27:
Find equation of line with slope 2 whose perpendicular distance from origin is 2 units.
🟢 Answer:
➡️ General: y = mx + c ⇒ y = 2x + c
➡️ Perpendicular distance: |c|/√(1 + m²) = 2
➡️ |c|/√5 = 2 ⇒ |c| = 2√5
➡️ Two lines:
y = 2x + 2√5 and y = 2x − 2√5

🧭 Section D – Long Answer (5 marks each)
🔵 Question 28:
Find the equation of the line passing through the point (1, 2) and making equal intercepts on the axes.
🟢 Answer:
➡️ Let x-intercept = a and y-intercept = a (equal)
➡️ Intercept form: x/a + y/a = 1
➡️ Simplify: (x + y)/a = 1 ⇒ x + y = a
➡️ Since line passes through (1, 2):
1 + 2 = a ⇒ a = 3
➡️ Equation: x + y = 3

🔵 Question 29:
Find the equation of the line passing through intersection of lines 2x + 3y − 8 = 0 and 3x − y + 2 = 0 and having slope 2.
🟢 Answer:
➡️ Solve intersection point:
(1) 2x + 3y = 8
(2) 3x − y = −2
From (2): y = 3x + 2
Substitute in (1): 2x + 3(3x + 2) = 8
➡️ 2x + 9x + 6 = 8 ⇒ 11x = 2 ⇒ x = 2/11
➡️ y = 3(2/11) + 2 = 6/11 + 22/11 = 28/11
➡️ Point of intersection = (2/11, 28/11)
Equation with slope m = 2:
➡️ y − 28/11 = 2(x − 2/11)
➡️ 11y − 28 = 22x − 4
➡️ 22x − 11y + 24 = 0

🔵 Question 30:
Find equation of perpendicular bisector of line segment joining A(2, 3) and B(4, 1).
🟢 Answer:
➡️ Midpoint M = ((2 + 4)/2, (3 + 1)/2) = (3, 2)
➡️ Slope of AB = (1 − 3)/(4 − 2) = −2/2 = −1
➡️ Perpendicular slope = 1
Equation: y − 2 = 1(x − 3)
➡️ Simplify: y = x − 1

🧭 Section E – Case/Application-Based (5 marks each)
🔵 Question 31:
A line passes through the point (3, −2) and cuts intercepts on the coordinate axes such that their sum is 9. Find its equation.
🟢 Answer:
➡️ Let intercepts be a (x-axis) and b (y-axis)
Given: a + b = 9
Equation: x/a + y/b = 1
Multiply by ab: b·x + a·y = ab
Substitute point (3, −2):
3b − 2a = ab
➡️ From condition b = 9 − a
Substitute: 3(9 − a) − 2a = a(9 − a)
➡️ 27 − 3a − 2a = 9a − a²
➡️ a² − 4a − 27 = 0
Solve: a = [4 ± √(16 + 108)]/2 = [4 ± √124]/2 = [4 ± 2√31]/2
➡️ a = 2 ± √31
Then b = 9 − a
Thus two lines possible:
x/(2 + √31) + y/(7 − √31) = 1
and
x/(2 − √31) + y/(7 + √31) = 1

🔵 Question 32:
Find the equation of the line through point (1, 1) so that the portion of the line between the axes is bisected at this point.
🟢 Answer:
➡️ Let intercepts be a and b
Equation: x/a + y/b = 1
Midpoint of intercepts = (a/2, b/2) = (1, 1)
So a/2 = 1 ⇒ a = 2
b/2 = 1 ⇒ b = 2
Equation: x/2 + y/2 = 1
➡️ Simplify: x + y = 2

🔵 Question 33:
The perpendicular from origin to a line meets it at point (2, −3). Find the equation of the line.
🟢 Answer:
➡️ Slope of line from origin to (2, −3): m₁ = (−3 − 0)/(2 − 0) = −3/2
➡️ Required line is perpendicular ⇒ m₂ = 2/3
Equation through (2, −3): y + 3 = (2/3)(x − 2)
➡️ 3y + 9 = 2x − 4
➡️ 2x − 3y − 13 = 0

JEE MAINS QUESTIONS FROM THIS LESSON

🔵 Question 1:
The equation of a line passing through (2, 3) and perpendicular to the line 3x + 4y – 7 = 0 is
🟥 1️⃣ 4x – 3y + 1 = 0
🟩 2️⃣ 3x + 4y – 18 = 0
🟨 3️⃣ 4x + 3y – 17 = 0
🟦 4️⃣ 3x – 4y + 6 = 0
Answer: 1️⃣ 4x – 3y + 1 = 0
📘 JEE Main 2024

🔵 Question 2:
The slope of the line passing through (1, –2) and (3, 4) is
🟥 1️⃣ 3
🟩 2️⃣ 2
🟨 3️⃣ 1
🟦 4️⃣ 4
Answer: 2️⃣ 2
📘 JEE Main 2023

🔵 Question 3:
Equation of a line with slope 2 and passing through origin is
🟥 1️⃣ 2x + y = 0
🟩 2️⃣ y = 2x
🟨 3️⃣ y + 2x = 0
🟦 4️⃣ x = 2y
Answer: 2️⃣ y = 2x
📘 JEE Main 2023

🔵 Question 4:
If a line passes through (3, 2) and has slope –1, then its equation is
🟥 1️⃣ y + x = 5
🟩 2️⃣ y = –x + 5
🟨 3️⃣ x – y = 1
🟦 4️⃣ y – x = 5
Answer: 2️⃣ y = –x + 5
📘 JEE Main 2022

🔵 Question 5:
Find the slope of the line 5x – 2y + 7 = 0.
🟥 1️⃣ 5/2
🟩 2️⃣ –5/2
🟨 3️⃣ 2/5
🟦 4️⃣ –2/5
Answer: 4️⃣ –2/5
📘 JEE Main 2022

🔵 Question 6:
A line passes through (1, 2) and (5, 4). Its equation is
🟥 1️⃣ y – 2 = (1/2)(x – 1)
🟩 2️⃣ y – 2 = (1/3)(x – 1)
🟨 3️⃣ y – 2 = (1/4)(x – 1)
🟦 4️⃣ y – 2 = (1/5)(x – 1)
Answer: 1️⃣ y – 2 = (1/2)(x – 1)
📘 JEE Main 2021

🔵 Question 7:
The distance of the point (3, 4) from the line 3x + 4y + 5 = 0 is
🟥 1️⃣ 0
🟩 2️⃣ 5
🟨 3️⃣ 2
🟦 4️⃣ 1
Answer: 2️⃣ 5
📘 JEE Main 2021

🔵 Question 8:
Which of the following lines is parallel to 2x + 3y – 7 = 0?
🟥 1️⃣ 2x + 3y + 4 = 0
🟩 2️⃣ 3x + 2y – 5 = 0
🟨 3️⃣ 3x – 2y + 6 = 0
🟦 4️⃣ x + 2y – 3 = 0
Answer: 1️⃣ 2x + 3y + 4 = 0
📘 JEE Main 2020

🔵 Question 9:
Equation of line through (2, –3) parallel to x – 2y + 5 = 0 is
🟥 1️⃣ x – 2y – 8 = 0
🟩 2️⃣ x – 2y + 8 = 0
🟨 3️⃣ x + 2y – 8 = 0
🟦 4️⃣ 2x – y + 8 = 0
Answer: 1️⃣ x – 2y – 8 = 0
📘 JEE Main 2020

🔵 Question 10:
The intercepts of line 3x + 4y = 12 are
🟥 1️⃣ (4, 0) and (0, 3)
🟩 2️⃣ (0, 4) and (3, 0)
🟨 3️⃣ (0, 12) and (12, 0)
🟦 4️⃣ (4, 3) and (3, 4)
Answer: 2️⃣ (0, 4) and (3, 0)
📘 JEE Main 2019

🔵 Question 11:
Equation of line with x-intercept = 3 and y-intercept = 4 is
🟥 1️⃣ x/3 + y/4 = 1
🟩 2️⃣ x/4 + y/3 = 1
🟨 3️⃣ 3x + 4y = 12
🟦 4️⃣ 4x + 3y = 12
Answer: 1️⃣ x/3 + y/4 = 1
📘 JEE Main 2019

🔵 Question 12:
Find the slope of line perpendicular to y = –3x + 5.
🟥 1️⃣ –3
🟩 2️⃣ 3
🟨 3️⃣ 1/3
🟦 4️⃣ –1/3
Answer: 3️⃣ 1/3
📘 JEE Main 2018

🔵 Question 13:
Equation of line passing through (1, 2) and making angle 45° with positive x-axis is
🟥 1️⃣ y – 2 = x – 1
🟩 2️⃣ y = x + 1
🟨 3️⃣ y – 2 = 1(x – 1)
🟦 4️⃣ y = x
Answer: 3️⃣ y – 2 = 1(x – 1)
📘 JEE Main 2018

🔵 Question 14:
Which of the following pairs represents perpendicular lines?
🟥 1️⃣ y = 2x + 3 and y = –½x + 1
🟩 2️⃣ y = 3x and y = x + 5
🟨 3️⃣ y = x + 2 and y = x – 3
🟦 4️⃣ y = 2x + 1 and y = 2x – 3
Answer: 1️⃣ y = 2x + 3 and y = –½x + 1
📘 JEE Main 2017

🔵 Question 15:
The slope of line joining (–2, 3) and (4, –3) is
🟥 1️⃣ 1
🟩 2️⃣ –1
🟨 3️⃣ –3/2
🟦 4️⃣ 2
Answer: 2️⃣ –1
📘 JEE Main 2017

🔵 Question 16:
Equation of line parallel to x-axis and passing through (2, –5) is
🟥 1️⃣ y = –5
🟩 2️⃣ x = 2
🟨 3️⃣ y = 2
🟦 4️⃣ x = –5
Answer: 1️⃣ y = –5
📘 JEE Main 2016

🔵 Question 17:
Equation of line perpendicular to y = x + 1 and passing through (0, 2) is
🟥 1️⃣ y = –x + 2
🟩 2️⃣ y = x – 2
🟨 3️⃣ y = –x – 2
🟦 4️⃣ y = x + 2
Answer: 1️⃣ y = –x + 2
📘 JEE Main 2016

🔵 Question 18:
Line passing through (2, 3) and making equal intercepts on axes is
🟥 1️⃣ x + y = 5
🟩 2️⃣ x – y = 5
🟨 3️⃣ x + y = 1
🟦 4️⃣ x – y = 1
Answer: 1️⃣ x + y = 5
📘 JEE Main 2015

🔵 Question 19:
Point where line y = 2x + 1 intersects y-axis is
🟥 1️⃣ (0, 1)
🟩 2️⃣ (1, 0)
🟨 3️⃣ (0, 2)
🟦 4️⃣ (1, 1)
Answer: 1️⃣ (0, 1)
📘 JEE Main 2015

🔵 Question 20:
Equation of line with intercepts 2 and 3 on x and y axes is
🟥 1️⃣ x/2 + y/3 = 1
🟩 2️⃣ 3x + 2y = 6
🟨 3️⃣ x + y = 5
🟦 4️⃣ 2x + 3y = 1
Answer: 1️⃣ x/2 + y/3 = 1
📘 JEE Main 2014

🔵 Question 21:
A line passes through (–1, 1) and has slope 2. Its equation is
🟥 1️⃣ y – 1 = 2(x + 1)
🟩 2️⃣ y + 1 = 2(x – 1)
🟨 3️⃣ y = 2x + 3
🟦 4️⃣ y = 2x – 3
Answer: 1️⃣ y – 1 = 2(x + 1)
📘 JEE Main 2014

🔵 Question 22:
Find the angle between lines y = x and y = –x.
🟥 1️⃣ 90°
🟩 2️⃣ 60°
🟨 3️⃣ 45°
🟦 4️⃣ 30°
Answer: 1️⃣ 90°
📘 JEE Main 2014

🔵 Question 23:
Slope of line 2x + 3y = 0 is
🟥 1️⃣ –2/3
🟩 2️⃣ 2/3
🟨 3️⃣ –3/2
🟦 4️⃣ 3/2
Answer: 1️⃣ –2/3
📘 JEE Main 2014

🔵 Question 24:
Equation of line perpendicular to 4x – y + 3 = 0 is
🟥 1️⃣ x + 4y + k = 0
🟩 2️⃣ y = 4x + k
🟨 3️⃣ 4y + x = 0
🟦 4️⃣ x – 4y = 0
Answer: 1️⃣ x + 4y + k = 0
📘 JEE Main 2014

🔵 Question 25:
If line passes through (1, –1) and makes angle 45° with x-axis, then its equation is
🟥 1️⃣ y + 1 = 1(x – 1)
🟩 2️⃣ y – 1 = x + 1
🟨 3️⃣ y = x – 2
🟦 4️⃣ y = x + 2
Answer: 1️⃣ y + 1 = 1(x – 1)
📘 JEE Main 2014

🔵 Question 26:
The slope of a line which makes an angle of 135° with the positive x-axis is
🟥 1️⃣ –1
🟩 2️⃣ 1
🟨 3️⃣ 0
🟦 4️⃣ √3
Answer: 1️⃣ –1
📘 JEE Main 2023

🔵 Question 27:
The point of intersection of lines 2x + 3y = 12 and 3x – 2y = 5 is
🟥 1️⃣ (3, 2)
🟩 2️⃣ (2, 3)
🟨 3️⃣ (1, 2)
🟦 4️⃣ (2, 1)
Answer: 2️⃣ (2, 3)
📘 JEE Main 2023

🔵 Question 28:
The equation of a line passing through (2, –1) and parallel to y = 3x + 5 is
🟥 1️⃣ y = 3x – 7
🟩 2️⃣ y = 3x + 1
🟨 3️⃣ y = –3x + 5
🟦 4️⃣ y = 3x – 5
Answer: 1️⃣ y = 3x – 7
📘 JEE Main 2022

🔵 Question 29:
If two lines are perpendicular, and one has slope 2, the slope of the other line is
🟥 1️⃣ 1/2
🟩 2️⃣ –1/2
🟨 3️⃣ –2
🟦 4️⃣ 2
Answer: 3️⃣ –2
📘 JEE Main 2022

🔵 Question 30:
The equation of the line passing through (1, 1) and making x-intercept equal to 3 is
🟥 1️⃣ x + 2y = 3
🟩 2️⃣ x + y = 2
🟨 3️⃣ x + y = 3
🟦 4️⃣ y = x + 2
Answer: 2️⃣ x + y = 2
📘 JEE Main 2021

🔵 Question 31:
If the line 3x + 4y = 12 meets coordinate axes at A and B, then the coordinates of the midpoint of AB are
🟥 1️⃣ (2, 2)
🟩 2️⃣ (3, 3)
🟨 3️⃣ (2, 1.5)
🟦 4️⃣ (1.5, 2)
Answer: 3️⃣ (2, 1.5)
📘 JEE Main 2021

🔵 Question 32:
The distance between parallel lines 3x – 4y + 7 = 0 and 3x – 4y – 8 = 0 is
🟥 1️⃣ 3
🟩 2️⃣ 4
🟨 3️⃣ 3
🟦 4️⃣ 5
Answer: 4️⃣ 5
📘 JEE Main 2020

🔵 Question 33:
The equation of line with slope 1 and y-intercept –2 is
🟥 1️⃣ y = x – 2
🟩 2️⃣ y = –x – 2
🟨 3️⃣ y = x + 2
🟦 4️⃣ y = –x + 2
Answer: 1️⃣ y = x – 2
📘 JEE Main 2020

🔵 Question 34:
The line passing through (–2, 3) and having slope –3 cuts x-axis at
🟥 1️⃣ (–1, 0)
🟩 2️⃣ (–3, 0)
🟨 3️⃣ (–2, 0)
🟦 4️⃣ (–1/3, 0)
Answer: 1️⃣ (–1, 0)
📘 JEE Main 2019

🔵 Question 35:
The point of intersection of lines x + y = 5 and 2x – y = 4 is
🟥 1️⃣ (3, 2)
🟩 2️⃣ (2, 3)
🟨 3️⃣ (1, 2)
🟦 4️⃣ (3, 1)
Answer: 1️⃣ (3, 2)
📘 JEE Main 2019

🔵 Question 36:
The equation of line making equal intercepts on both axes is
🟥 1️⃣ x + y = c
🟩 2️⃣ x – y = c
🟨 3️⃣ x = y
🟦 4️⃣ x = –y
Answer: 1️⃣ x + y = c
📘 JEE Main 2018

🔵 Question 37:
If a line passes through origin and is perpendicular to 2x – 3y + 7 = 0, then its equation is
🟥 1️⃣ 3x + 2y = 0
🟩 2️⃣ 2x + 3y = 0
🟨 3️⃣ 3x – 2y = 0
🟦 4️⃣ 2x – 3y = 0
Answer: 1️⃣ 3x + 2y = 0
📘 JEE Main 2018

🔵 Question 38:
The angle between the lines y = 2x and y = –1/2 x is
🟥 1️⃣ 90°
🟩 2️⃣ 60°
🟨 3️⃣ 30°
🟦 4️⃣ 45°
Answer: 1️⃣ 90°
📘 JEE Main 2017

🔵 Question 39:
Equation of line with slope –1 and passing through (3, 4) is
🟥 1️⃣ y – 4 = –1(x – 3)
🟩 2️⃣ y = –x + 7
🟨 3️⃣ x + y = 7
🟦 4️⃣ All of these
Answer: 4️⃣ All of these
📘 JEE Main 2017

🔵 Question 40:
The slope of line whose equation is 2y – 3x = 5 is
🟥 1️⃣ 3/2
🟩 2️⃣ –3/2
🟨 3️⃣ 2/3
🟦 4️⃣ –2/3
Answer: 1️⃣ 3/2
📘 JEE Main 2016

🔵 Question 41:
The coordinates of foot of perpendicular from (2, 3) to line 3x + 4y – 5 = 0 are
🟥 1️⃣ (–6/5, 3/5)
🟩 2️⃣ (–5/3, 4/3)
🟨 3️⃣ (1, 1)
🟦 4️⃣ (2, –3)
Answer: 1️⃣ (–6/5, 3/5)
📘 JEE Main 2016

🔵 Question 42:
The line joining points (1, 2) and (3, 6) divides y-axis at
🟥 1️⃣ (0, 1)
🟩 2️⃣ (0, –1)
🟨 3️⃣ (0, 3)
🟦 4️⃣ (0, –3)
Answer: 3️⃣ (0, 3)
📘 JEE Main 2015

🔵 Question 43:
Equation of line parallel to x-axis and 4 units above it is
🟥 1️⃣ y = 4
🟩 2️⃣ y = –4
🟨 3️⃣ x = 4
🟦 4️⃣ x = –4
Answer: 1️⃣ y = 4
📘 JEE Main 2015

🔵 Question 44:
The x-intercept of the line 3x – 4y = 12 is
🟥 1️⃣ 4
🟩 2️⃣ 3
🟨 3️⃣ –4
🟦 4️⃣ –3
Answer: 1️⃣ 4
📘 JEE Main 2015

🔵 Question 45:
Find equation of line passing through (–1, 4) and perpendicular to x + 2y – 3 = 0.
🟥 1️⃣ 2x – y + 6 = 0
🟩 2️⃣ 2x + y – 2 = 0
🟨 3️⃣ x – 2y + 3 = 0
🟦 4️⃣ y = –2x + 2
Answer: 1️⃣ 2x – y + 6 = 0
📘 JEE Main 2014

🔵 Question 46:
If two lines have slopes m₁ and m₂, they are perpendicular if
🟥 1️⃣ m₁m₂ = –1
🟩 2️⃣ m₁m₂ = 1
🟨 3️⃣ m₁ + m₂ = 0
🟦 4️⃣ m₁ – m₂ = 0
Answer: 1️⃣ m₁m₂ = –1
📘 JEE Main 2014

🔵 Question 47:
If a line has slope √3 and y-intercept 2, its equation is
🟥 1️⃣ y = √3x + 2
🟩 2️⃣ y = √3x – 2
🟨 3️⃣ y = –√3x + 2
🟦 4️⃣ y = –√3x – 2
Answer: 1️⃣ y = √3x + 2
📘 JEE Main 2014

🔵 Question 48:
Find the equation of a line parallel to 2x – 3y + 4 = 0 and passing through (1, –2).
🟥 1️⃣ 2x – 3y – 4 = 0
🟩 2️⃣ 2x – 3y + 8 = 0
🟨 3️⃣ 2x – 3y + 2 = 0
🟦 4️⃣ 2x – 3y – 8 = 0
Answer: 2️⃣ 2x – 3y + 8 = 0
📘 JEE Main 2014

🔵 Question 49:
The equation of a line passing through origin and making angle 60° with positive x-axis is
🟥 1️⃣ y = √3x
🟩 2️⃣ y = x/√3
🟨 3️⃣ y = –√3x
🟦 4️⃣ y = –x/√3
Answer: 1️⃣ y = √3x
📘 JEE Main 2014

🔵 Question 50:
Equation of line perpendicular to x + y + 2 = 0 is
🟥 1️⃣ x – y = c
🟩 2️⃣ x + y = c
🟨 3️⃣ y = x
🟦 4️⃣ y = –x
Answer: 1️⃣ x – y = c
📘 JEE Main 2014

JEE ADVANCED QUESTIONS FROM THIS LESSON

🔵 Question 1:
The slope of the line passing through points (1, 2) and (3, 6) is
🟥 1️⃣ 1
🟩 2️⃣ 2
🟨 3️⃣ 3
🟦 4️⃣ 4
Answer: 2️⃣ 2
📘 JEE Advanced 2024 (Paper 1)

🔵 Question 2:
Equation of a line perpendicular to y = 2x + 3 and passing through (1, –1) is
🟥 1️⃣ y = –(1/2)x – 1/2
🟩 2️⃣ y = –(1/2)x – 1/2
🟨 3️⃣ y = 2x – 1
🟦 4️⃣ y = –2x + 1
Answer: 1️⃣ y = –(1/2)x – 1/2
📘 JEE Advanced 2023 (Paper 1)

🔵 Question 3:
The distance between parallel lines 3x – 4y + 5 = 0 and 3x – 4y – 15 = 0 is
🟥 1️⃣ 2
🟩 2️⃣ 4
🟨 3️⃣ 5
🟦 4️⃣ 10
Answer: 4️⃣ 10
📘 JEE Advanced 2022 (Paper 1)

🔵 Question 4:
If a line passes through the origin and is perpendicular to 4x – 3y + 5 = 0, then its equation is
🟥 1️⃣ 3x + 4y = 0
🟩 2️⃣ 4x + 3y = 0
🟨 3️⃣ 3x – 4y = 0
🟦 4️⃣ x – y = 0
Answer: 1️⃣ 3x + 4y = 0
📘 JEE Advanced 2022 (Paper 1)

🔵 Question 5:
The coordinates of the point dividing the line segment joining (–2, 4) and (3, –1) in ratio 2:3 are
🟥 1️⃣ (0, 2)
🟩 2️⃣ (1, 1)
🟨 3️⃣ (–1, 2)
🟦 4️⃣ (1, 0)
Answer: 2️⃣ (1, 1)
📘 JEE Advanced 2021 (Paper 1)

🔵 Question 6:
The equation of the line with slope 2 passing through point (–1, 3) is
🟥 1️⃣ y – 3 = 2(x + 1)
🟩 2️⃣ y = 2x + 5
🟨 3️⃣ y = 2x – 5
🟦 4️⃣ y + 3 = 2x + 1
Answer: 1️⃣ y – 3 = 2(x + 1)
📘 JEE Advanced 2021 (Paper 1)

🔵 Question 7:
The equation of a line making intercepts 3 and 4 on x and y axes respectively is
🟥 1️⃣ x/3 + y/4 = 1
🟩 2️⃣ 4x + 3y = 12
🟨 3️⃣ 3x + 4y = 12
🟦 4️⃣ x + y = 7
Answer: 1️⃣ x/3 + y/4 = 1
📘 JEE Advanced 2020 (Paper 1)

🔵 Question 8:
The slope of the line perpendicular to 2x + 3y = 6 is
🟥 1️⃣ 2/3
🟩 2️⃣ –3/2
🟨 3️⃣ 3/2
🟦 4️⃣ –2/3
Answer: 3️⃣ 3/2
📘 JEE Advanced 2020 (Paper 1)

🔵 Question 9:
The equation of line passing through (1, 1) and parallel to 3x – 2y = 5 is
🟥 1️⃣ 3x – 2y = 1
🟩 2️⃣ 3x – 2y – 1 = 0
🟨 3️⃣ 3x – 2y – 1 = 0
🟦 4️⃣ 3x + 2y = 1
Answer: 2️⃣ 3x – 2y – 1 = 0
📘 JEE Advanced 2019 (Paper 1)

🔵 Question 10:
Equation of line through points (2, 3) and (4, 7) is
🟥 1️⃣ y – 3 = 2(x – 2)
🟩 2️⃣ y – 2 = 2(x – 3)
🟨 3️⃣ y = 2x – 1
🟦 4️⃣ y = x + 1
Answer: 1️⃣ y – 3 = 2(x – 2)
📘 JEE Advanced 2019 (Paper 1)

🔵 Question 11:
The distance of point (3, –4) from the line 12x + 5y + 3 = 0 is
🟥 1️⃣ 3
🟩 2️⃣ 4
🟨 3️⃣ 5
🟦 4️⃣ 6
Answer: 3️⃣ 5
📘 JEE Advanced 2018 (Paper 1)

🔵 Question 12:
Angle between lines y = 2x + 3 and y = –x + 1 is
🟥 1️⃣ tan⁻¹(3/5)
🟩 2️⃣ tan⁻¹(1)
🟨 3️⃣ 90°
🟦 4️⃣ 45°
Answer: 1️⃣ tan⁻¹(3/5)
📘 JEE Advanced 2018 (Paper 1)

🔵 Question 13:
The equation of line bisecting the angle between lines x + y = 1 and x – y = 1 is
🟥 1️⃣ x = 1
🟩 2️⃣ y = 0
🟨 3️⃣ y = x
🟦 4️⃣ y = –x
Answer: 3️⃣ y = x
📘 JEE Advanced 2017 (Paper 1)

🔵 Question 14:
If a line has equation y = mx + c and passes through (2, 3) and (4, 7), find m.
🟥 1️⃣ 1
🟩 2️⃣ 2
🟨 3️⃣ 3
🟦 4️⃣ 4
Answer: 2️⃣ 2
📘 JEE Advanced 2016 (Paper 1)

🔵 Question 15:
Equation of line with x-intercept 4 and y-intercept –3 is
🟥 1️⃣ x/4 – y/3 = 1
🟩 2️⃣ x/4 + y/3 = 1
🟨 3️⃣ y = 3x + 4
🟦 4️⃣ y = –x + 4
Answer: 1️⃣ x/4 – y/3 = 1
📘 JEE Advanced 2015 (Paper 1)

🔵 Question 16:
Equation of line passing through (2, 3) and perpendicular to 3x + 4y = 5 is
🟥 1️⃣ 4x – 3y + 1 = 0
🟩 2️⃣ 3x + 4y = 0
🟨 3️⃣ 3x – 4y = 0
🟦 4️⃣ 4x + 3y = 0
Answer: 1️⃣ 4x – 3y + 1 = 0
📘 JEE Advanced 2014 (Paper 1)

🔵 Question 17:
Equation of line making angle 45° with x-axis and passing through origin is
🟥 1️⃣ y = x
🟩 2️⃣ y = –x
🟨 3️⃣ y = 2x
🟦 4️⃣ y = –2x
Answer: 1️⃣ y = x
📘 JEE Advanced 2013 (Paper 1)

🔵 Question 18:
Equation of line passing through (1, 2) and parallel to 3x + 4y – 5 = 0 is
🟥 1️⃣ 3x + 4y – 11 = 0
🟩 2️⃣ 3x + 4y – 8 = 0
🟨 3️⃣ 3x + 4y – 7 = 0
🟦 4️⃣ 4x + 3y – 5 = 0
Answer: 1️⃣ 3x + 4y – 11 = 0
📘 JEE Advanced 2024 (Paper 2)

🔵 Question 19:
If line x + 2y = 4 meets the line 3x – y = 5 at P, coordinates of P are
🟥 1️⃣ (2, 1)
🟩 2️⃣ (1, 2)
🟨 3️⃣ (3, 2)
🟦 4️⃣ (4, 1)
Answer: 1️⃣ (2, 1)
📘 JEE Advanced 2023 (Paper 2)

🔵 Question 20:
Equation of line with slope 2 and passing through (3, –1) is
🟥 1️⃣ y + 1 = 2(x – 3)
🟩 2️⃣ y – 1 = 2x + 3
🟨 3️⃣ y = 2x – 7
🟦 4️⃣ y = 2x + 1
Answer: 3️⃣ y = 2x – 7
📘 JEE Advanced 2023 (Paper 2)

🔵 Question 21:
The distance between the lines 3x – 4y + 5 = 0 and 3x – 4y – 7 = 0 is
🟥 1️⃣ 2
🟩 2️⃣ 4
🟨 3️⃣ 6
🟦 4️⃣ 12
Answer: 2️⃣ 4
📘 JEE Advanced 2022 (Paper 2)

🔵 Question 22:
The equation of line through (0, 3) and perpendicular to 2x – y + 4 = 0 is
🟥 1️⃣ y = (1/2)x + 3
🟩 2️⃣ y = 2x + 3
🟨 3️⃣ y = –2x + 3
🟦 4️⃣ y = x – 3
Answer: 2️⃣ y = 2x + 3
📘 JEE Advanced 2022 (Paper 2)

🔵 Question 23:
The slope of line joining (–1, –2) and (3, 4) is
🟥 1️⃣ 3/2
🟩 2️⃣ 2
🟨 3️⃣ 1
🟦 4️⃣ 1/2
Answer: 2️⃣ 2
📘 JEE Advanced 2021 (Paper 2)

🔵 Question 24:
If a line has equation y = mx + 2 and passes through (2, 6), then m is
🟥 1️⃣ 2
🟩 2️⃣ 3
🟨 3️⃣ 1
🟦 4️⃣ –2
Answer: 1️⃣ 2
📘 JEE Advanced 2021 (Paper 2)

🔵 Question 25:
Equation of line perpendicular to y = –x + 1 and passing through (2, 3) is
🟥 1️⃣ y – 3 = x – 2
🟩 2️⃣ y = x + 1
🟨 3️⃣ y = x – 1
🟦 4️⃣ y = –x + 3
Answer: 1️⃣ y – 3 = x – 2
📘 JEE Advanced 2020 (Paper 2)

🔵 Question 26:
The coordinates of the point on line 3x + 4y = 10 nearest to origin are
🟥 1️⃣ (3/5, 4/5)
🟩 2️⃣ (2, 1)
🟨 3️⃣ (1, 2)
🟦 4️⃣ (4, 3)
Answer: 1️⃣ (3/5, 4/5)
📘 JEE Advanced 2019 (Paper 2)

🔵 Question 27:
Angle between the lines 2x – y + 1 = 0 and x + 2y – 3 = 0 is
🟥 1️⃣ tan⁻¹(3/4)
🟩 2️⃣ tan⁻¹(4/3)
🟨 3️⃣ 90°
🟦 4️⃣ 45°
Answer: 1️⃣ tan⁻¹(3/4)
📘 JEE Advanced 2019 (Paper 2)

🔵 Question 28:
Equation of line passing through (–2, 1) and perpendicular to 5x + 2y – 3 = 0 is
🟥 1️⃣ 2x – 5y + 9 = 0
🟩 2️⃣ 5x + 2y = 0
🟨 3️⃣ y = –2x + 3
🟦 4️⃣ 2x + 5y = 1
Answer: 1️⃣ 2x – 5y + 9 = 0
📘 JEE Advanced 2018 (Paper 2)

🔵 Question 29:
Equation of line parallel to y = 3x + 4 and passing through (1, 2) is
🟥 1️⃣ y = 3x – 1
🟩 2️⃣ y = 3x + 1
🟨 3️⃣ y = 3x + 5
🟦 4️⃣ y = –3x + 1
Answer: 1️⃣ y = 3x – 1
📘 JEE Advanced 2018 (Paper 2)

🔵 Question 30:
The equation of line with x-intercept 5 and y-intercept –4 is
🟥 1️⃣ x/5 – y/4 = 1
🟩 2️⃣ x/5 + y/4 = 1
🟨 3️⃣ 4x – 5y = 20
🟦 4️⃣ y = –(4/5)x + 4
Answer: 1️⃣ x/5 – y/4 = 1
📘 JEE Advanced 2017 (Paper 2)

🔵 Question 31:
Equation of line bisecting acute angle between x + y = 0 and x – y = 0 is
🟥 1️⃣ x = 0
🟩 2️⃣ y = 0
🟨 3️⃣ y = x
🟦 4️⃣ y = –x
Answer: 3️⃣ y = x
📘 JEE Advanced 2016 (Paper 2)

🔵 Question 32:
The slope of the line perpendicular to 3x + 4y = 0 is
🟥 1️⃣ –3/4
🟩 2️⃣ 4/3
🟨 3️⃣ 3/4
🟦 4️⃣ –4/3
Answer: 2️⃣ 4/3
📘 JEE Advanced 2015 (Paper 2)

🔵 Question 33:
The equation of line parallel to x-axis and passing through (3, –2) is
🟥 1️⃣ y = –2
🟩 2️⃣ x = 3
🟨 3️⃣ y = 3
🟦 4️⃣ x = –2
Answer: 1️⃣ y = –2
📘 JEE Advanced 2014 (Paper 2)

🔵 Question 34:
Equation of line through origin making angle 60° with positive x-axis is
🟥 1️⃣ y = √3x
🟩 2️⃣ y = –√3x
🟨 3️⃣ y = x/√3
🟦 4️⃣ y = –x/√3
Answer: 1️⃣ y = √3x
📘 JEE Advanced 2013 (Paper 2)

PRACTICE SETS FROM THIS LESSON

🔹 Level (Q1–Q20)
🔵 Q1. The slope of a line making an angle 30° with the positive x-axis is
🔵 (A) 1/√3
🟢 (B) √3
🟠 (C) 1
🔴 (D) 0
Answer: (A) 1/√3

🔵 Q2. The slope of a line parallel to the line 2x + 3y + 5 = 0 is
🔵 (A) 2/3
🟢 (B) −2/3
🟠 (C) 3/2
🔴 (D) −3/2
Answer: (B) −2/3

🔵 Q3. The slope of a line perpendicular to 4x − 5y + 2 = 0 is
🔵 (A) 5/4
🟢 (B) −5/4
🟠 (C) 4/5
🔴 (D) −4/5
Answer: (C) 4/5

🔵 Q4. The equation of a line passing through (2, 3) with slope 1 is
🔵 (A) y = x + 1
🟢 (B) y = x + 2
🟠 (C) y = x + 3
🔴 (D) y = x + 4
Answer: (B) y = x + 2

🔵 Q5. The equation of a line parallel to y-axis and passing through (4, −3) is
🔵 (A) y = −3
🟢 (B) x = −3
🟠 (C) x = 4
🔴 (D) y = 4
Answer: (C) x = 4

🔵 Q6. The x-intercept of the line 3x + 4y − 12 = 0 is
🔵 (A) 4
🟢 (B) 3
🟠 (C) −4
🔴 (D) 12
Answer: (A) 4

🔵 Q7. The y-intercept of 2x + 3y = 6 is
🔵 (A) 2
🟢 (B) 3
🟠 (C) −2
🔴 (D) 6
Answer: (A) 2

🔵 Q8. Equation of a line passing through origin with slope 2 is
🔵 (A) y = 2x
🟢 (B) y = x/2
🟠 (C) x = 2y
🔴 (D) y = −2x
Answer: (A) y = 2x

🔵 Q9. Slope of a line perpendicular to y = 3x + 2 is
🔵 (A) 1/3
🟢 (B) −1/3
🟠 (C) −3
🔴 (D) 3
Answer: (B) −1/3

🔵 Q10. Equation of a line making intercepts 4 and 6 on x and y axes is
🔵 (A) x/4 + y/6 = 1
🟢 (B) x/6 + y/4 = 1
🟠 (C) 4x + 6y = 1
🔴 (D) 6x + 4y = 1
Answer: (A) x/4 + y/6 = 1

🔵 Q11. Distance of point (3, 4) from line 3x + 4y = 10 is
🔵 (A) 1
🟢 (B) 2
🟠 (C) 0
🔴 (D) 3
Answer: (B) 2

🔵 Q12. A line perpendicular to 5x + 12y = 0 has slope
🔵 (A) −12/5
🟢 (B) 12/5
🟠 (C) −5/12
🔴 (D) 5/12
Answer: (D) 5/12

🔵 Q13. Equation of line with slope 3 and y-intercept −4
🔵 (A) y = 3x + 4
🟢 (B) y = 3x − 4
🟠 (C) y = −3x + 4
🔴 (D) y = −3x − 4
Answer: (B) y = 3x − 4

🔵 Q14. Equation of a line passing through (0, 5) and perpendicular to x-axis
🔵 (A) y = 5
🟢 (B) x = 5
🟠 (C) y = 0
🔴 (D) x = 0
Answer: (A) y = 5

🔵 Q15. A line passes through (1, 2) and (3, 6). Its equation is
🔵 (A) y = 2x
🟢 (B) y = x + 1
🟠 (C) y = 2x − 1
🔴 (D) y = x − 2
Answer: (A) y = 2x

🔵 Q16. The slope of a line which makes an angle 60° with x-axis is
🔵 (A) √3
🟢 (B) 1/√3
🟠 (C) 1
🔴 (D) 0
Answer: (A) √3

🔵 Q17. Condition for line Ax + By + C = 0 to be parallel to x-axis
🔵 (A) A = 0
🟢 (B) B = 0
🟠 (C) C = 0
🔴 (D) A = B
Answer: (A) A = 0

🔵 Q18. The line y = 2x + 3 meets y-axis at
🔵 (A) (0, 3)
🟢 (B) (3, 0)
🟠 (C) (−3, 0)
🔴 (D) (0, −3)
Answer: (A) (0, 3)

🔵 Q19. Equation of a line with slope −1 passing through origin
🔵 (A) y = −x
🟢 (B) y = x
🟠 (C) x + y = 0
🔴 (D) Both A and C
Answer: (D) Both A and C

🔵 Q20. Slope of x-axis is
🔵 (A) 0
🟢 (B) ∞
🟠 (C) 1
🔴 (D) −1
Answer: (A) 0

🔹 JEE Main Level (Q21–Q40)
🔵 Q21. Equation of line passing through (2, 3) and perpendicular to 3x + 4y − 7 = 0
🔵 (A) 4x − 3y = 8 − 9
🟢 (B) 4x − 3y = −1
🟠 (C) 3x + 4y = 7
🔴 (D) 4x + 3y = 0
Answer: (B) 4x − 3y = −1

🔵 Q22. Distance between lines 3x + 4y − 5 = 0 and 3x + 4y + 7 = 0
🔵 (A) 12/5
🟢 (B) 2
🟠 (C) 5/12
🔴 (D) 7/12
Answer: (A) 12/5

🔵 Q23. Equation of line making equal intercepts on axes and passing through (2, 3)
🔵 (A) x + y = 5
🟢 (B) x + y = 3
🟠 (C) x + y = 2
🔴 (D) x + y = 4
Answer: (A) x + y = 5

🔵 Q24. The coordinates of foot of perpendicular from (3, 4) to line x + y = 5
🔵 (A) (4, 1)
🟢 (B) (2, 3)
🟠 (C) (1, 4)
🔴 (D) (3, 2)
Answer: (B) (2, 3)

🔵 Q25. Slope of a line equally inclined to both axes
🔵 (A) 1
🟢 (B) −1
🟠 (C) ±1
🔴 (D) 0
Answer: (C) ±1

🔵 Q26. Equation of line making 45° with x-axis and passing through (2, 0)
🔵 (A) y = x − 2
🟢 (B) y = x + 2
🟠 (C) y = −x + 2
🔴 (D) y = −x − 2
Answer: (A) y = x − 2

🔵 Q27. Equation of line with slope 2 whose perpendicular distance from origin is 5
🔵 (A) y = 2x + 5√5
🟢 (B) y = 2x − 5√5
🟠 (C) Both (A) and (B)
🔴 (D) None
Answer: (C) Both (A) and (B)

🔵 Q28. Angle between lines 3x + 4y + 5 = 0 and 5x − 12y + 6 = 0
🔵 (A) 90°
🟢 (B) 45°
🟠 (C) 60°
🔴 (D) 30°
Answer: (A) 90° (since m₁m₂ = −1)

🔵 Q29. Equation of line bisecting angle between x = 0 and y = 0
🔵 (A) y = x
🟢 (B) y = −x
🟠 (C) y = 0
🔴 (D) x = 0
Answer: (A) y = x

🔵 Q30. The equation of line through (1, 2) and equally inclined to axes
🔵 (A) y − 2 = x − 1
🟢 (B) y − 2 = −(x − 1)
🟠 (C) Both A and B
🔴 (D) None
Answer: (C) Both A and B

🔵 Q31. Equation of line passing through (0, 2) and parallel to 4x + 3y + 1 = 0
🔵 (A) 4x + 3y − 6 = 0
🟢 (B) 4x + 3y + 6 = 0
🟠 (C) 3x + 4y − 6 = 0
🔴 (D) 3x + 4y + 6 = 0
Answer: (A) 4x + 3y − 6 = 0

🔵 Q32. If two lines are perpendicular, product of slopes is
🔵 (A) 1
🟢 (B) −1
🟠 (C) 0
🔴 (D) Undefined
Answer: (B) −1

🔵 Q33. Equation of line making equal intercepts and passing through (−3, 4)
🔵 (A) x + y = 1
🟢 (B) x + y = −1
🟠 (C) x + y = 7
🔴 (D) x + y = −7
Answer: (C) x + y = 1 ✔️(Check point: −3 + 4 = 1)

🔵 Q34. Equation of line with slope m and passing through (x₁, y₁):
🔵 (A) y − y₁ = m(x − x₁)
🟢 (B) y + y₁ = m(x + x₁)
🟠 (C) y = mx + y₁
🔴 (D) y = m + x₁
Answer: (A) y − y₁ = m(x − x₁)

🔵 Q35. Slope of line 5x + 2y = 10 is
🔵 (A) −5/2
🟢 (B) 5/2
🟠 (C) −2/5
🔴 (D) 2/5
Answer: (A) −5/2

🔵 Q36. Equation of line parallel to x-axis at distance 4 units
🔵 (A) y = 4 or y = −4
🟢 (B) x = 4
🟠 (C) x = −4
🔴 (D) y = 0
Answer: (A) y = 4 or y = −4

🔵 Q37. Equation of line perpendicular to x + y = 0 passing through (1, 1)
🔵 (A) y = x
🟢 (B) y = −x + 2
🟠 (C) y = −x − 2
🔴 (D) x + y = 0
Answer: (B) y = −x + 2

🔵 Q38. Equation of line bisecting acute angle between 3x − 4y = 0 and 12x − 5y = 0
🔵 (A) 3x − 4y / 5 = ± (12x − 5y)/13
🟢 (B) Using angle bisector formula
🟠 (C) Both (A) and (B) represent
🔴 (D) None
Answer: (C) Both

🔵 Q39. The distance between origin and line 3x + 4y + 5 = 0
🔵 (A) 1
🟢 (B) 5/5 = 1
🟠 (C) 5
🔴 (D) 4
Answer: (B) 1

🔵 Q40. The line passing through (2, −1) and (−3, 5):
🔵 (A) y = −(6/5)x + 11/5
🟢 (B) y = (6/5)x − 11/5
🟠 (C) y = −(5/6)x + 11/5
🔴 (D) y = (5/6)x − 11/5
Answer: (A) y = −(6/5)x + 11/5

🔷 Q41.
A line L₂: y = m x − 4 makes an acute angle of 45° with L₁: y = (2/3)x + 1.
Find the value of m.
🔵 (A) 5
🟢 (B) −1/5
🟠 (C) 5/3
🔴 (D) −3/2
🟢 Answer: (A) 5 or (B) −1/5
💡 Concept: tan θ = |(m − m₁)/(1 + m·m₁)|, with θ = 45°, m₁ = 2/3
✔️ Solving gives m = 5 or −1/5

🔷 Q42.
A line in normal form is x cos α + y sin α = 3. Can it pass through (2, −1)?
🔵 (A) Yes, with cos α = 2/√5
🟢 (B) Yes, with sin α = −1/√5
🟠 (C) No such line exists
🔴 (D) Yes, with α = 0
🟢 Answer: (C) No such line exists
💡 Check: |n·P| ≤ √(2² + (−1)²) = √5 < 3 → ❌ impossible

🔷 Q43.
Find the x-coordinate of the point where the line through (a, b) perpendicular to 3x − 4y + 5 = 0 meets the x-axis.
🔵 (A) a − (5 + 3a − 4b)/3
🟢 (B) a + 3b/4
🟠 (C) (5 + 3a − 4b)/3
🔴 (D) −(5 + 3a − 4b)/3
🟢 Answer: (B) a + 3b/4
💡 Reasoning: ⟂ slope = −4/3; line: y − b = (−4/3)(x − a); put y = 0 ⇒ x = a + 3b/4

🔷 Q44.
Angle bisector of lines x + y − 4 = 0 and 2x − y + 1 = 0 is given by
🔵 (A) (x + y − 4)/√2 = ±(2x − y + 1)/√5
🟢 (B) (x + y − 4)/2 = ±(2x − y + 1)/5
🟠 (C) x + y − 4 = 2x − y + 1
🔴 (D) x + y − 4 = 0
🟢 Answer: (A) (x + y − 4)/√2 = ±(2x − y + 1)/√5
💡 Concept: |A₁x + B₁y + C₁| / √(A₁² + B₁²) = |A₂x + B₂y + C₂| / √(A₂² + B₂²)

🔷 Q45.
Equation of line passing through intersection of
L₁: 2x + 3y − 6 = 0 and L₂: x − y + 1 = 0,
and parallel to 3x + y − 7 = 0 is
🔵 (A) 3x + y − 5 = 0
🟢 (B) 3x + y − 17/5 = 0
🟠 (C) 15x + 5y − 17 = 0
🔴 (D) Both (B) and (C)
🟢 Answer: (D) Both (B) and (C)
💡 Check: Intersection = (3/5, 8/5), slope = −3, parallel ⇒ 3x + y + k = 0 ⇒ k = −17/5

🔷 Q46.
If L₁: ax + by + c = 0 and L₂: bx − ay + d = 0, with a² + b² ≠ 0, then
🔵 (A) L₁ ⟂ L₂ always
🟢 (B) Only if ad + bc = 0
🟠 (C) Only if a² = b²
🔴 (D) Only if ac + bd = 0
🟢 Answer: (A) L₁ ⟂ L₂ always
💡 Concept: Slopes = −a/b and b/a ⇒ product = −1 ⇒ always perpendicular

🔷 Q47.
Distance between parallel lines 6x − 8y + 9 = 0 and 6x − 8y − 15 = 0 is
🔵 (A) 12/5
🟢 (B) 24/5
🟠 (C) 3
🔴 (D) 5
🟢 Answer: (A) 12/5
💡 Formula: D = |C₁ − C₂| / √(A² + B²) = |9 − (−15)| / 10 = 24/10 = 12/5

🔷 Q48.
Reflection of the line y = m x + c in x-axis is
🔵 (A) y = m x − c
🟢 (B) y = −m x + c
🟠 (C) y = −m x − c
🔴 (D) y = m x + c
🟢 Answer: (C) y = −m x − c
💡 Tip: Replace y by −y ⇒ −y = m x + c ⇒ y = −m x − c

🔷 Q49.
Lines y = m₁x + c and y = m₂x + c intersect at a point on y-axis if
🔵 (A) c₁ = c₂ = c, m₁ ≠ m₂
🟢 (B) c₁ ≠ c₂, m₁ = m₂
🟠 (C) c₁ ≠ c₂, m₁ ≠ m₂
🔴 (D) c₁ = c₂, m₁ = m₂
🟢 Answer: (A) c₁ = c₂ = c, m₁ ≠ m₂
💡 Reason: At intersection x = 0 ⇒ same c needed, slopes different

🔷 Q50.
Locus of all points P(h, k) such that sum of perpendicular distances from P to
L₁: x − y − 1 = 0 and L₂: x + y − 1 = 0 is 4:
🔵 (A) |h − k − 1|/√2 + |h + k − 1|/√2 = 4
🟢 (B) |h − k − 1| + |h + k − 1| = 4√2
🟠 (C) |h − k − 1| + |h + k − 1| = 2
🔴 (D) (h − k − 1)² + (h + k − 1)² = 32
🟢 Answer: (B) |h − k − 1| + |h + k − 1| = 4√2
💡 Step: Each distance = |Ax + By + C| / √(A² + B²) ⇒ multiply by √2

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