JEE Main Previous Year Questions (2025): Straight Lines | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


JEE Main Previous Year Questions 
(2025): Straight Lines 
 
Q1: Let the distance between two parallel lines be 5 units and a point ?? lie between 
the lines at a unit distance from one of them. An equilateral triangle ?????? is formed 
such that ?? lies on one of the parallel lines, while ?? lies on the other. Then ( ???? )
?? is 
equal to ____ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 28 
Solution: 
We set up a coordinate system so that the two parallel lines are given by 
?? = 0  and  ?? = 5, 
since their distance is 5 units. Choose point 
?? = ( 0,1) 
so that the distance from ?? to the line ?? = 0 is 1 unit (and its distance to the line ?? = 5 is 4 
units). 
Let point 
?? = ( ?? , 0) 
be on the line ?? = 0, and let point 
?? = ( ?? , 5) 
be on the line ?? = 5. Since triangle ?????? is equilateral with side length ?? , we require: 
???? = ???? = ???? = ?? . 
A convenient method is to "rotate" ?? about ?? by an angle of 60
°
 to obtain ?? . In complex-
number (or vector) terms, if we translate so that ?? is at the origin, then the rotation is given by 
?? ?? 60
°
= cos 60
°
+ ?? sin 60
°
=
1
2
+ ?? v 3
2
. 
Thus, writing ?? in vector form relative to ?? , we have 
?? - ?? = ( ?? , -1) . 
Rotating this by 60
°
 gives 
?? - ?? = ( ?? cos 60
°
- ( -1) sin 60
°
, ?? sin 60
°
+ ( -1) cos 60
°
) . 
Substituting the values cos 60
°
=
1
2
 and sin 60
°
=
v 3
2
, we obtain 
?? - ?? = (
?? 2
+
v 3
2
,
?? v 3
2
-
1
2
) , 
so  ?? = (
?? +v 3
2
, 1 +
?? v 3
2
-
1
2
)= (
?? +v 3
2
,
?? v 3+1
2
) . 
Since ?? lies on ?? = 5, its ?? -coordinate must equal 5 : 
?? v 3+1
2
= 5. 
Solve for ?? : 
Page 2


JEE Main Previous Year Questions 
(2025): Straight Lines 
 
Q1: Let the distance between two parallel lines be 5 units and a point ?? lie between 
the lines at a unit distance from one of them. An equilateral triangle ?????? is formed 
such that ?? lies on one of the parallel lines, while ?? lies on the other. Then ( ???? )
?? is 
equal to ____ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 28 
Solution: 
We set up a coordinate system so that the two parallel lines are given by 
?? = 0  and  ?? = 5, 
since their distance is 5 units. Choose point 
?? = ( 0,1) 
so that the distance from ?? to the line ?? = 0 is 1 unit (and its distance to the line ?? = 5 is 4 
units). 
Let point 
?? = ( ?? , 0) 
be on the line ?? = 0, and let point 
?? = ( ?? , 5) 
be on the line ?? = 5. Since triangle ?????? is equilateral with side length ?? , we require: 
???? = ???? = ???? = ?? . 
A convenient method is to "rotate" ?? about ?? by an angle of 60
°
 to obtain ?? . In complex-
number (or vector) terms, if we translate so that ?? is at the origin, then the rotation is given by 
?? ?? 60
°
= cos 60
°
+ ?? sin 60
°
=
1
2
+ ?? v 3
2
. 
Thus, writing ?? in vector form relative to ?? , we have 
?? - ?? = ( ?? , -1) . 
Rotating this by 60
°
 gives 
?? - ?? = ( ?? cos 60
°
- ( -1) sin 60
°
, ?? sin 60
°
+ ( -1) cos 60
°
) . 
Substituting the values cos 60
°
=
1
2
 and sin 60
°
=
v 3
2
, we obtain 
?? - ?? = (
?? 2
+
v 3
2
,
?? v 3
2
-
1
2
) , 
so  ?? = (
?? +v 3
2
, 1 +
?? v 3
2
-
1
2
)= (
?? +v 3
2
,
?? v 3+1
2
) . 
Since ?? lies on ?? = 5, its ?? -coordinate must equal 5 : 
?? v 3+1
2
= 5. 
Solve for ?? : 
?? v 3 + 1 = 10
?? v 3 = 9
?? =
9
v 3
= 3v 3
 
Now, the side length ?? (which is the distance ???? ) is given by 
?? 2
 = ?? ?? 2
= ( 3v 3 - 0)
2
+ ( 0 - 1)
2
 = ( 3v 3)
2
+ 1
2
 = 27 + 1
 = 28
 
Thus, the square of side ???? is 
( ???? )
2
= ?? 2
= 28. 
Q2: Let the triangle PQR be the image of the triangle with vertices ( ?? , ?? ) , ( ?? , ?? ) and 
( ?? , ?? ) in the line ?? + ?? ?? = ?? . If the centroid of ? ?????? is the point ( ?? , ?? ) , then 
???? ( ?? - ?? ) is equal to : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 21 
B. 19 
C. 22 
D. 24 
Ans: C 
Solution: 
Let ' ?? ' be the centroid of ? formed by ( 1,3) ( 3,1)\ &( 2,4) 
G ? ( 2,
8
3
) 
Image of G w.r.t. ?? + 2?? - 2 = 0 
?? - 2
1
=
?? -
8
3
2
= -2
( 2 +
16
3
- 2)
1 + 4
 
=
-2
5
(
16
3
) 
? ?? =
-32
15
+ 2 =
-2
15
, ?? =
-32 × 2
15
+
8
3
=
-24
15
 
15( ?? - ?? )= -2 + 24 = 22 
Q3: A rod of length eight units moves such that its ends ?? and ?? always lie on the lines 
?? - ?? + ?? = ?? and ?? + ?? = ?? , respectively. If the locus of the point ?? , that divides the 
rod ???? internally in the ratio ?? : ?? is ?? ( ?? ?? + ?? ?? ?? + ?????? + ???? + ???? ?? )- ???? = ?? , then 
?? - ?? - ?? is equal to 
Page 3


JEE Main Previous Year Questions 
(2025): Straight Lines 
 
Q1: Let the distance between two parallel lines be 5 units and a point ?? lie between 
the lines at a unit distance from one of them. An equilateral triangle ?????? is formed 
such that ?? lies on one of the parallel lines, while ?? lies on the other. Then ( ???? )
?? is 
equal to ____ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 28 
Solution: 
We set up a coordinate system so that the two parallel lines are given by 
?? = 0  and  ?? = 5, 
since their distance is 5 units. Choose point 
?? = ( 0,1) 
so that the distance from ?? to the line ?? = 0 is 1 unit (and its distance to the line ?? = 5 is 4 
units). 
Let point 
?? = ( ?? , 0) 
be on the line ?? = 0, and let point 
?? = ( ?? , 5) 
be on the line ?? = 5. Since triangle ?????? is equilateral with side length ?? , we require: 
???? = ???? = ???? = ?? . 
A convenient method is to "rotate" ?? about ?? by an angle of 60
°
 to obtain ?? . In complex-
number (or vector) terms, if we translate so that ?? is at the origin, then the rotation is given by 
?? ?? 60
°
= cos 60
°
+ ?? sin 60
°
=
1
2
+ ?? v 3
2
. 
Thus, writing ?? in vector form relative to ?? , we have 
?? - ?? = ( ?? , -1) . 
Rotating this by 60
°
 gives 
?? - ?? = ( ?? cos 60
°
- ( -1) sin 60
°
, ?? sin 60
°
+ ( -1) cos 60
°
) . 
Substituting the values cos 60
°
=
1
2
 and sin 60
°
=
v 3
2
, we obtain 
?? - ?? = (
?? 2
+
v 3
2
,
?? v 3
2
-
1
2
) , 
so  ?? = (
?? +v 3
2
, 1 +
?? v 3
2
-
1
2
)= (
?? +v 3
2
,
?? v 3+1
2
) . 
Since ?? lies on ?? = 5, its ?? -coordinate must equal 5 : 
?? v 3+1
2
= 5. 
Solve for ?? : 
?? v 3 + 1 = 10
?? v 3 = 9
?? =
9
v 3
= 3v 3
 
Now, the side length ?? (which is the distance ???? ) is given by 
?? 2
 = ?? ?? 2
= ( 3v 3 - 0)
2
+ ( 0 - 1)
2
 = ( 3v 3)
2
+ 1
2
 = 27 + 1
 = 28
 
Thus, the square of side ???? is 
( ???? )
2
= ?? 2
= 28. 
Q2: Let the triangle PQR be the image of the triangle with vertices ( ?? , ?? ) , ( ?? , ?? ) and 
( ?? , ?? ) in the line ?? + ?? ?? = ?? . If the centroid of ? ?????? is the point ( ?? , ?? ) , then 
???? ( ?? - ?? ) is equal to : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 21 
B. 19 
C. 22 
D. 24 
Ans: C 
Solution: 
Let ' ?? ' be the centroid of ? formed by ( 1,3) ( 3,1)\ &( 2,4) 
G ? ( 2,
8
3
) 
Image of G w.r.t. ?? + 2?? - 2 = 0 
?? - 2
1
=
?? -
8
3
2
= -2
( 2 +
16
3
- 2)
1 + 4
 
=
-2
5
(
16
3
) 
? ?? =
-32
15
+ 2 =
-2
15
, ?? =
-32 × 2
15
+
8
3
=
-24
15
 
15( ?? - ?? )= -2 + 24 = 22 
Q3: A rod of length eight units moves such that its ends ?? and ?? always lie on the lines 
?? - ?? + ?? = ?? and ?? + ?? = ?? , respectively. If the locus of the point ?? , that divides the 
rod ???? internally in the ratio ?? : ?? is ?? ( ?? ?? + ?? ?? ?? + ?????? + ???? + ???? ?? )- ???? = ?? , then 
?? - ?? - ?? is equal to 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 24 
B. 22 
C. 21 
D. 23 
Ans: D 
Solution: 
 
 
h =
3?? + ?? 3
 
k =
-4 + ?? + 2
3
 
?? = 3k + 2 
2?? = 3 h - a = 3 h - 3k - 2 
so AB = 8 
( ?? - ?? )
2
+ ( ?? + 4)
2
= 64 
(3k + 2 - (
3 h - 3k - 2
2
) )
2
+ ( 3k + 2 + 4)
2
= 64 
( 9k - 3 h + 6)
2
4
+ ( 3k + 6)
2
= 64 
9[( 3k - h + 2)
2
+ 4( k + 2)
2
] = 64 × 4 
9( x
2
+ 13y
2
- 6xy - 4x+ 28y)= 76 
?? - ?? - ?? = 13 + 6 + 4 = 23 
Q4: Let the lines ?? ?? - ?? ?? - ?? = ?? , ?? ?? - ???? ?? - ???? = ?? , and ?? ?? - ?? ?? + ?? = ?? be 
concurrent. If the image of the point( 1,2 ) in the line ?? ?? - ?? ?? + ?? = ?? is (
????
????
,
-????
????
) , 
then |???? | is equal to 
Page 4


JEE Main Previous Year Questions 
(2025): Straight Lines 
 
Q1: Let the distance between two parallel lines be 5 units and a point ?? lie between 
the lines at a unit distance from one of them. An equilateral triangle ?????? is formed 
such that ?? lies on one of the parallel lines, while ?? lies on the other. Then ( ???? )
?? is 
equal to ____ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 28 
Solution: 
We set up a coordinate system so that the two parallel lines are given by 
?? = 0  and  ?? = 5, 
since their distance is 5 units. Choose point 
?? = ( 0,1) 
so that the distance from ?? to the line ?? = 0 is 1 unit (and its distance to the line ?? = 5 is 4 
units). 
Let point 
?? = ( ?? , 0) 
be on the line ?? = 0, and let point 
?? = ( ?? , 5) 
be on the line ?? = 5. Since triangle ?????? is equilateral with side length ?? , we require: 
???? = ???? = ???? = ?? . 
A convenient method is to "rotate" ?? about ?? by an angle of 60
°
 to obtain ?? . In complex-
number (or vector) terms, if we translate so that ?? is at the origin, then the rotation is given by 
?? ?? 60
°
= cos 60
°
+ ?? sin 60
°
=
1
2
+ ?? v 3
2
. 
Thus, writing ?? in vector form relative to ?? , we have 
?? - ?? = ( ?? , -1) . 
Rotating this by 60
°
 gives 
?? - ?? = ( ?? cos 60
°
- ( -1) sin 60
°
, ?? sin 60
°
+ ( -1) cos 60
°
) . 
Substituting the values cos 60
°
=
1
2
 and sin 60
°
=
v 3
2
, we obtain 
?? - ?? = (
?? 2
+
v 3
2
,
?? v 3
2
-
1
2
) , 
so  ?? = (
?? +v 3
2
, 1 +
?? v 3
2
-
1
2
)= (
?? +v 3
2
,
?? v 3+1
2
) . 
Since ?? lies on ?? = 5, its ?? -coordinate must equal 5 : 
?? v 3+1
2
= 5. 
Solve for ?? : 
?? v 3 + 1 = 10
?? v 3 = 9
?? =
9
v 3
= 3v 3
 
Now, the side length ?? (which is the distance ???? ) is given by 
?? 2
 = ?? ?? 2
= ( 3v 3 - 0)
2
+ ( 0 - 1)
2
 = ( 3v 3)
2
+ 1
2
 = 27 + 1
 = 28
 
Thus, the square of side ???? is 
( ???? )
2
= ?? 2
= 28. 
Q2: Let the triangle PQR be the image of the triangle with vertices ( ?? , ?? ) , ( ?? , ?? ) and 
( ?? , ?? ) in the line ?? + ?? ?? = ?? . If the centroid of ? ?????? is the point ( ?? , ?? ) , then 
???? ( ?? - ?? ) is equal to : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 21 
B. 19 
C. 22 
D. 24 
Ans: C 
Solution: 
Let ' ?? ' be the centroid of ? formed by ( 1,3) ( 3,1)\ &( 2,4) 
G ? ( 2,
8
3
) 
Image of G w.r.t. ?? + 2?? - 2 = 0 
?? - 2
1
=
?? -
8
3
2
= -2
( 2 +
16
3
- 2)
1 + 4
 
=
-2
5
(
16
3
) 
? ?? =
-32
15
+ 2 =
-2
15
, ?? =
-32 × 2
15
+
8
3
=
-24
15
 
15( ?? - ?? )= -2 + 24 = 22 
Q3: A rod of length eight units moves such that its ends ?? and ?? always lie on the lines 
?? - ?? + ?? = ?? and ?? + ?? = ?? , respectively. If the locus of the point ?? , that divides the 
rod ???? internally in the ratio ?? : ?? is ?? ( ?? ?? + ?? ?? ?? + ?????? + ???? + ???? ?? )- ???? = ?? , then 
?? - ?? - ?? is equal to 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 24 
B. 22 
C. 21 
D. 23 
Ans: D 
Solution: 
 
 
h =
3?? + ?? 3
 
k =
-4 + ?? + 2
3
 
?? = 3k + 2 
2?? = 3 h - a = 3 h - 3k - 2 
so AB = 8 
( ?? - ?? )
2
+ ( ?? + 4)
2
= 64 
(3k + 2 - (
3 h - 3k - 2
2
) )
2
+ ( 3k + 2 + 4)
2
= 64 
( 9k - 3 h + 6)
2
4
+ ( 3k + 6)
2
= 64 
9[( 3k - h + 2)
2
+ 4( k + 2)
2
] = 64 × 4 
9( x
2
+ 13y
2
- 6xy - 4x+ 28y)= 76 
?? - ?? - ?? = 13 + 6 + 4 = 23 
Q4: Let the lines ?? ?? - ?? ?? - ?? = ?? , ?? ?? - ???? ?? - ???? = ?? , and ?? ?? - ?? ?? + ?? = ?? be 
concurrent. If the image of the point( 1,2 ) in the line ?? ?? - ?? ?? + ?? = ?? is (
????
????
,
-????
????
) , 
then |???? | is equal to 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 91 
B. 113 
C. 101 
D. 84 
Ans: A 
Solution: 
 
 
? PM = QM 
So, ?? (
57
13
+1
2
,
-40
13
+2
2
) 
= (
35
13
,
-7
13
) 
? M lies on the time 
2?? - 3?? + ?? = 0 
2 (
35
13
)- 3 (
-7
13
)+ ?? = 0 
?? = -
70
13
+
21
13
 
=
-91
13
= -7 
|
3 -4 -?? 8 -11 -33
2 3 ?? | = 0 
? 3( -11?? - 99)+ 4( 8?? + 66)- ?? ( -24 + 22)= 0 
? 33?? - 297+ 32?? + 264+ 24?? - 22?? = 0 
? -?? + 2?? - 33 = 0 ( 1) 
Page 5


JEE Main Previous Year Questions 
(2025): Straight Lines 
 
Q1: Let the distance between two parallel lines be 5 units and a point ?? lie between 
the lines at a unit distance from one of them. An equilateral triangle ?????? is formed 
such that ?? lies on one of the parallel lines, while ?? lies on the other. Then ( ???? )
?? is 
equal to ____ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 28 
Solution: 
We set up a coordinate system so that the two parallel lines are given by 
?? = 0  and  ?? = 5, 
since their distance is 5 units. Choose point 
?? = ( 0,1) 
so that the distance from ?? to the line ?? = 0 is 1 unit (and its distance to the line ?? = 5 is 4 
units). 
Let point 
?? = ( ?? , 0) 
be on the line ?? = 0, and let point 
?? = ( ?? , 5) 
be on the line ?? = 5. Since triangle ?????? is equilateral with side length ?? , we require: 
???? = ???? = ???? = ?? . 
A convenient method is to "rotate" ?? about ?? by an angle of 60
°
 to obtain ?? . In complex-
number (or vector) terms, if we translate so that ?? is at the origin, then the rotation is given by 
?? ?? 60
°
= cos 60
°
+ ?? sin 60
°
=
1
2
+ ?? v 3
2
. 
Thus, writing ?? in vector form relative to ?? , we have 
?? - ?? = ( ?? , -1) . 
Rotating this by 60
°
 gives 
?? - ?? = ( ?? cos 60
°
- ( -1) sin 60
°
, ?? sin 60
°
+ ( -1) cos 60
°
) . 
Substituting the values cos 60
°
=
1
2
 and sin 60
°
=
v 3
2
, we obtain 
?? - ?? = (
?? 2
+
v 3
2
,
?? v 3
2
-
1
2
) , 
so  ?? = (
?? +v 3
2
, 1 +
?? v 3
2
-
1
2
)= (
?? +v 3
2
,
?? v 3+1
2
) . 
Since ?? lies on ?? = 5, its ?? -coordinate must equal 5 : 
?? v 3+1
2
= 5. 
Solve for ?? : 
?? v 3 + 1 = 10
?? v 3 = 9
?? =
9
v 3
= 3v 3
 
Now, the side length ?? (which is the distance ???? ) is given by 
?? 2
 = ?? ?? 2
= ( 3v 3 - 0)
2
+ ( 0 - 1)
2
 = ( 3v 3)
2
+ 1
2
 = 27 + 1
 = 28
 
Thus, the square of side ???? is 
( ???? )
2
= ?? 2
= 28. 
Q2: Let the triangle PQR be the image of the triangle with vertices ( ?? , ?? ) , ( ?? , ?? ) and 
( ?? , ?? ) in the line ?? + ?? ?? = ?? . If the centroid of ? ?????? is the point ( ?? , ?? ) , then 
???? ( ?? - ?? ) is equal to : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 21 
B. 19 
C. 22 
D. 24 
Ans: C 
Solution: 
Let ' ?? ' be the centroid of ? formed by ( 1,3) ( 3,1)\ &( 2,4) 
G ? ( 2,
8
3
) 
Image of G w.r.t. ?? + 2?? - 2 = 0 
?? - 2
1
=
?? -
8
3
2
= -2
( 2 +
16
3
- 2)
1 + 4
 
=
-2
5
(
16
3
) 
? ?? =
-32
15
+ 2 =
-2
15
, ?? =
-32 × 2
15
+
8
3
=
-24
15
 
15( ?? - ?? )= -2 + 24 = 22 
Q3: A rod of length eight units moves such that its ends ?? and ?? always lie on the lines 
?? - ?? + ?? = ?? and ?? + ?? = ?? , respectively. If the locus of the point ?? , that divides the 
rod ???? internally in the ratio ?? : ?? is ?? ( ?? ?? + ?? ?? ?? + ?????? + ???? + ???? ?? )- ???? = ?? , then 
?? - ?? - ?? is equal to 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 24 
B. 22 
C. 21 
D. 23 
Ans: D 
Solution: 
 
 
h =
3?? + ?? 3
 
k =
-4 + ?? + 2
3
 
?? = 3k + 2 
2?? = 3 h - a = 3 h - 3k - 2 
so AB = 8 
( ?? - ?? )
2
+ ( ?? + 4)
2
= 64 
(3k + 2 - (
3 h - 3k - 2
2
) )
2
+ ( 3k + 2 + 4)
2
= 64 
( 9k - 3 h + 6)
2
4
+ ( 3k + 6)
2
= 64 
9[( 3k - h + 2)
2
+ 4( k + 2)
2
] = 64 × 4 
9( x
2
+ 13y
2
- 6xy - 4x+ 28y)= 76 
?? - ?? - ?? = 13 + 6 + 4 = 23 
Q4: Let the lines ?? ?? - ?? ?? - ?? = ?? , ?? ?? - ???? ?? - ???? = ?? , and ?? ?? - ?? ?? + ?? = ?? be 
concurrent. If the image of the point( 1,2 ) in the line ?? ?? - ?? ?? + ?? = ?? is (
????
????
,
-????
????
) , 
then |???? | is equal to 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 91 
B. 113 
C. 101 
D. 84 
Ans: A 
Solution: 
 
 
? PM = QM 
So, ?? (
57
13
+1
2
,
-40
13
+2
2
) 
= (
35
13
,
-7
13
) 
? M lies on the time 
2?? - 3?? + ?? = 0 
2 (
35
13
)- 3 (
-7
13
)+ ?? = 0 
?? = -
70
13
+
21
13
 
=
-91
13
= -7 
|
3 -4 -?? 8 -11 -33
2 3 ?? | = 0 
? 3( -11?? - 99)+ 4( 8?? + 66)- ?? ( -24 + 22)= 0 
? 33?? - 297+ 32?? + 264+ 24?? - 22?? = 0 
? -?? + 2?? - 33 = 0 ( 1) 
? ?? = -7 
-( -7)+ 2?? - 33 = 0 
2?? = 26 
?? = 13 
? |???? | = |13 × ( -7) | 
= 91 
Q5: Let the points (
????
?? , ?? ) lie on or inside the triangle with sides ?? + ?? = ???? , ?? + ?? ?? =
???? and ?? ?? + ?? ?? = ???? . Then the product of the smallest and the largest values of ?? is 
equal to : 
JEE Main 2025 (Online) 24th January Evening Shift 
Options: 
A. 22 
B. 33 
C. 55 
D. 44 
Ans: B 
Solution: 
 
Point of intersection of ?? =
11
2
 with ?? 1
&?? 3
 gives, 
?? min 
=
11
2
 
and ?? max 
= 6 
? ?? min
· ?? max
=
11
2
× 6 = 33