Straight Lines: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Straight Lines  
(January 2026) 
Q1: Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on 
the line x + 2v2y = 4. If the co-ordinates of the vertex A are (a, ß), then the greatest integer 
less than or equal to |a + v2ß| is 
A: 5 
B: 4 
C: 2 
D: 3 
Answer: B 
Explanation: 
 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Straight Lines  
(January 2026) 
Q1: Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on 
the line x + 2v2y = 4. If the co-ordinates of the vertex A are (a, ß), then the greatest integer 
less than or equal to |a + v2ß| is 
A: 5 
B: 4 
C: 2 
D: 3 
Answer: B 
Explanation: 
 
Orthocenter (O) : (0, 0). Since it's an equilateral triangle, this is also the Centroid. 
Side BC Line : x + 2v2y = 4. 
Vertex A : (a, ß). 
Find the distance from the Centroid to side BC 
The side BC lies on the line x + 2v2y - 4 = 0. The distance from the origin G(0, 0) to this line is 
the length of the "lower" part of the median (r). 
Using the point-to-line distance formula: 
 
Find the coordinates of Vertex A(a, ß) 
The centroid G divides the segment AD (where D is the foot of the altitude on BC) in the ratio 2 :  
1. Therefore, the distance from A to G is 2r. 
AG = 2 × 4/3 = 8/3 
The line AG is perpendicular to side BC. The slope of BC is m = -1/2v2, so the slope of the line 
AG (the altitude) is the negative reciprocal: 
 
Since A lies on the line passing through the origin with slope 2v2, its coordinates can be written 
as: 
a = k,    ß = 2v2k 
Because A and the side BC are on opposite sides of the origin (the centroid), and the constant 
term in our line equation is negative, A must satisfy the condition that its position yields an 
opposite sign. Substituting these into the distance formula AG = 8/3: 
 
Calculate |a + v2ß| 
Substitute a = k and ß = 2v2k : 
|a + v2ß| = |k + v2(2v2k)| 
= |k + 4k| = |5k| 
= 5 × 8/9 = 40/9 ˜ 4.44 
greatest integer less than or equal to |a + v2ß| = [4.44] = 4. 
 
Q2: Let the angles made with the positive x-axis by two straight lines drawn from the 
point P(2, 3) and meeting the line x + y = 6 at a distance v(2/3) from the point P be ? 1 and 
? 2. Then the value of (? 1 + ? 2) is: 
A: p/2 
B: p/3 
C: p/12 
D: p/6 
Answer: A 
Explanation: 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Straight Lines  
(January 2026) 
Q1: Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on 
the line x + 2v2y = 4. If the co-ordinates of the vertex A are (a, ß), then the greatest integer 
less than or equal to |a + v2ß| is 
A: 5 
B: 4 
C: 2 
D: 3 
Answer: B 
Explanation: 
 
Orthocenter (O) : (0, 0). Since it's an equilateral triangle, this is also the Centroid. 
Side BC Line : x + 2v2y = 4. 
Vertex A : (a, ß). 
Find the distance from the Centroid to side BC 
The side BC lies on the line x + 2v2y - 4 = 0. The distance from the origin G(0, 0) to this line is 
the length of the "lower" part of the median (r). 
Using the point-to-line distance formula: 
 
Find the coordinates of Vertex A(a, ß) 
The centroid G divides the segment AD (where D is the foot of the altitude on BC) in the ratio 2 :  
1. Therefore, the distance from A to G is 2r. 
AG = 2 × 4/3 = 8/3 
The line AG is perpendicular to side BC. The slope of BC is m = -1/2v2, so the slope of the line 
AG (the altitude) is the negative reciprocal: 
 
Since A lies on the line passing through the origin with slope 2v2, its coordinates can be written 
as: 
a = k,    ß = 2v2k 
Because A and the side BC are on opposite sides of the origin (the centroid), and the constant 
term in our line equation is negative, A must satisfy the condition that its position yields an 
opposite sign. Substituting these into the distance formula AG = 8/3: 
 
Calculate |a + v2ß| 
Substitute a = k and ß = 2v2k : 
|a + v2ß| = |k + v2(2v2k)| 
= |k + 4k| = |5k| 
= 5 × 8/9 = 40/9 ˜ 4.44 
greatest integer less than or equal to |a + v2ß| = [4.44] = 4. 
 
Q2: Let the angles made with the positive x-axis by two straight lines drawn from the 
point P(2, 3) and meeting the line x + y = 6 at a distance v(2/3) from the point P be ? 1 and 
? 2. Then the value of (? 1 + ? 2) is: 
A: p/2 
B: p/3 
C: p/12 
D: p/6 
Answer: A 
Explanation: 
Let any point Q(t, 6 - t) on line x + y = 6 such that. 
distance PQ = v(2/3) ? PQ² = 2/3. 
using distance formula for points P(2, 3) and Q(t, 6 - t). 
(2 - t)² + (3 - 6 + t)² = ? 
(2 - t)² + (t - 3)² = ? 
4 + t² - 4t + t² + 9 - 6t = ? 
2t² - 10t + 13 = ? 
6t² - 30t + 39 - 2 = 0 
6t² - 30t + 37 = 0 
let t
1
 & t
2
 are roots of this equation. 
t
1
 + t
2
 = 30/6 = 5 and t 1t 2 = 37/6. 
So, there are two points Q
1
(t
1
, 6 - t
1
) & Q
2
(t
2
, 6 - t
2
). 
It is given that angle made with positive x-axis by straight lines from P(2, 3) to Q 1 & Q
2
 is ? 1 & ?
2
 
respectively. 
This means tan ?
1
 = slope of line PQ
1
. 
 
 
Q3: Let A(1, 0), B(2, -1) and C(7/3, 4/3) be three points. If the equation of the bisector of 
the angle ABC is ax + ßy = 5, then the value of a² + ß² is 
A: 5 
B: 10 
C: 8 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Straight Lines  
(January 2026) 
Q1: Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on 
the line x + 2v2y = 4. If the co-ordinates of the vertex A are (a, ß), then the greatest integer 
less than or equal to |a + v2ß| is 
A: 5 
B: 4 
C: 2 
D: 3 
Answer: B 
Explanation: 
 
Orthocenter (O) : (0, 0). Since it's an equilateral triangle, this is also the Centroid. 
Side BC Line : x + 2v2y = 4. 
Vertex A : (a, ß). 
Find the distance from the Centroid to side BC 
The side BC lies on the line x + 2v2y - 4 = 0. The distance from the origin G(0, 0) to this line is 
the length of the "lower" part of the median (r). 
Using the point-to-line distance formula: 
 
Find the coordinates of Vertex A(a, ß) 
The centroid G divides the segment AD (where D is the foot of the altitude on BC) in the ratio 2 :  
1. Therefore, the distance from A to G is 2r. 
AG = 2 × 4/3 = 8/3 
The line AG is perpendicular to side BC. The slope of BC is m = -1/2v2, so the slope of the line 
AG (the altitude) is the negative reciprocal: 
 
Since A lies on the line passing through the origin with slope 2v2, its coordinates can be written 
as: 
a = k,    ß = 2v2k 
Because A and the side BC are on opposite sides of the origin (the centroid), and the constant 
term in our line equation is negative, A must satisfy the condition that its position yields an 
opposite sign. Substituting these into the distance formula AG = 8/3: 
 
Calculate |a + v2ß| 
Substitute a = k and ß = 2v2k : 
|a + v2ß| = |k + v2(2v2k)| 
= |k + 4k| = |5k| 
= 5 × 8/9 = 40/9 ˜ 4.44 
greatest integer less than or equal to |a + v2ß| = [4.44] = 4. 
 
Q2: Let the angles made with the positive x-axis by two straight lines drawn from the 
point P(2, 3) and meeting the line x + y = 6 at a distance v(2/3) from the point P be ? 1 and 
? 2. Then the value of (? 1 + ? 2) is: 
A: p/2 
B: p/3 
C: p/12 
D: p/6 
Answer: A 
Explanation: 
Let any point Q(t, 6 - t) on line x + y = 6 such that. 
distance PQ = v(2/3) ? PQ² = 2/3. 
using distance formula for points P(2, 3) and Q(t, 6 - t). 
(2 - t)² + (3 - 6 + t)² = ? 
(2 - t)² + (t - 3)² = ? 
4 + t² - 4t + t² + 9 - 6t = ? 
2t² - 10t + 13 = ? 
6t² - 30t + 39 - 2 = 0 
6t² - 30t + 37 = 0 
let t
1
 & t
2
 are roots of this equation. 
t
1
 + t
2
 = 30/6 = 5 and t 1t 2 = 37/6. 
So, there are two points Q
1
(t
1
, 6 - t
1
) & Q
2
(t
2
, 6 - t
2
). 
It is given that angle made with positive x-axis by straight lines from P(2, 3) to Q 1 & Q
2
 is ? 1 & ?
2
 
respectively. 
This means tan ?
1
 = slope of line PQ
1
. 
 
 
Q3: Let A(1, 0), B(2, -1) and C(7/3, 4/3) be three points. If the equation of the bisector of 
the angle ABC is ax + ßy = 5, then the value of a² + ß² is 
A: 5 
B: 10 
C: 8 
D: 13 
Answer: B 
Explanation: 
 
solution : A(1, 0), B(2, -1) and C(7/3, 4/3) 
equation of line AB using two point form 
y - 0 = (0 - (-1))/(1 - 2) (x - 1)  
y = -x + 1  
L
1
 : x + y - 1 = 0 
similarly equation of line BC. 
y + 1 = (4/3 - (-1))/(7/3 - 2) (x - 2)  
y + 1 = (7/3)/(1/3) (x - 2)  
y = 7x - 14 - 1  
L 2 : 7x - y - 15 = 0 
from graph (2, 0) lies in the region of ?ABC. 
calculating L
1
(2, 0) · L
2
(2, 0) 
(2 + 0 - 1) · (7 × 2 - 0 - 15) = 1 · (-1) = -1 
equation of angular bisector of ?ABC is given as 
 
since L
1
(2, 0) · L
2
(2, 0) is negative so equation of ?ABC bisector in the region (2, 0) can be 
find using negative sign from equation (1) 
5x + 5y - 5 = -(7x - y - 15)  
5x + 5y - 5 + 7x - y - 15 = 0  
12x + 4y - 20 = 0 
3x + y = 5 comparing with bisector ax + ßy = 5. 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Straight Lines  
(January 2026) 
Q1: Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on 
the line x + 2v2y = 4. If the co-ordinates of the vertex A are (a, ß), then the greatest integer 
less than or equal to |a + v2ß| is 
A: 5 
B: 4 
C: 2 
D: 3 
Answer: B 
Explanation: 
 
Orthocenter (O) : (0, 0). Since it's an equilateral triangle, this is also the Centroid. 
Side BC Line : x + 2v2y = 4. 
Vertex A : (a, ß). 
Find the distance from the Centroid to side BC 
The side BC lies on the line x + 2v2y - 4 = 0. The distance from the origin G(0, 0) to this line is 
the length of the "lower" part of the median (r). 
Using the point-to-line distance formula: 
 
Find the coordinates of Vertex A(a, ß) 
The centroid G divides the segment AD (where D is the foot of the altitude on BC) in the ratio 2 :  
1. Therefore, the distance from A to G is 2r. 
AG = 2 × 4/3 = 8/3 
The line AG is perpendicular to side BC. The slope of BC is m = -1/2v2, so the slope of the line 
AG (the altitude) is the negative reciprocal: 
 
Since A lies on the line passing through the origin with slope 2v2, its coordinates can be written 
as: 
a = k,    ß = 2v2k 
Because A and the side BC are on opposite sides of the origin (the centroid), and the constant 
term in our line equation is negative, A must satisfy the condition that its position yields an 
opposite sign. Substituting these into the distance formula AG = 8/3: 
 
Calculate |a + v2ß| 
Substitute a = k and ß = 2v2k : 
|a + v2ß| = |k + v2(2v2k)| 
= |k + 4k| = |5k| 
= 5 × 8/9 = 40/9 ˜ 4.44 
greatest integer less than or equal to |a + v2ß| = [4.44] = 4. 
 
Q2: Let the angles made with the positive x-axis by two straight lines drawn from the 
point P(2, 3) and meeting the line x + y = 6 at a distance v(2/3) from the point P be ? 1 and 
? 2. Then the value of (? 1 + ? 2) is: 
A: p/2 
B: p/3 
C: p/12 
D: p/6 
Answer: A 
Explanation: 
Let any point Q(t, 6 - t) on line x + y = 6 such that. 
distance PQ = v(2/3) ? PQ² = 2/3. 
using distance formula for points P(2, 3) and Q(t, 6 - t). 
(2 - t)² + (3 - 6 + t)² = ? 
(2 - t)² + (t - 3)² = ? 
4 + t² - 4t + t² + 9 - 6t = ? 
2t² - 10t + 13 = ? 
6t² - 30t + 39 - 2 = 0 
6t² - 30t + 37 = 0 
let t
1
 & t
2
 are roots of this equation. 
t
1
 + t
2
 = 30/6 = 5 and t 1t 2 = 37/6. 
So, there are two points Q
1
(t
1
, 6 - t
1
) & Q
2
(t
2
, 6 - t
2
). 
It is given that angle made with positive x-axis by straight lines from P(2, 3) to Q 1 & Q
2
 is ? 1 & ?
2
 
respectively. 
This means tan ?
1
 = slope of line PQ
1
. 
 
 
Q3: Let A(1, 0), B(2, -1) and C(7/3, 4/3) be three points. If the equation of the bisector of 
the angle ABC is ax + ßy = 5, then the value of a² + ß² is 
A: 5 
B: 10 
C: 8 
D: 13 
Answer: B 
Explanation: 
 
solution : A(1, 0), B(2, -1) and C(7/3, 4/3) 
equation of line AB using two point form 
y - 0 = (0 - (-1))/(1 - 2) (x - 1)  
y = -x + 1  
L
1
 : x + y - 1 = 0 
similarly equation of line BC. 
y + 1 = (4/3 - (-1))/(7/3 - 2) (x - 2)  
y + 1 = (7/3)/(1/3) (x - 2)  
y = 7x - 14 - 1  
L 2 : 7x - y - 15 = 0 
from graph (2, 0) lies in the region of ?ABC. 
calculating L
1
(2, 0) · L
2
(2, 0) 
(2 + 0 - 1) · (7 × 2 - 0 - 15) = 1 · (-1) = -1 
equation of angular bisector of ?ABC is given as 
 
since L
1
(2, 0) · L
2
(2, 0) is negative so equation of ?ABC bisector in the region (2, 0) can be 
find using negative sign from equation (1) 
5x + 5y - 5 = -(7x - y - 15)  
5x + 5y - 5 + 7x - y - 15 = 0  
12x + 4y - 20 = 0 
3x + y = 5 comparing with bisector ax + ßy = 5. 
a = 3 and ß = 1 
a² + ß² = 9 + 1 = 10. 
 
Q4: Let A(1, 2) and C(-3, -6) be two diagonally opposite vertices of a rhombus, whose 
sides AD and BC are parallel to the line 7x - y = 14. If B(a, ß) and D(?, d) are the other two 
vertices, then |a + ß + ? + d| is equal to: 
A: 3 
B: 6 
C: 1 
D: 9 
Answer: B 
Explanation: 
Let A = (1, 2) and C = (-3, -6) be diagonally opposite vertices of a rhombus. 
In a rhombus, the diagonals bisect each other at a common point M.. 
Calculation of Midpoint M : 
 
M = (-1, -2)