Let a point A lie between the parallel lines L1 and L2 such that its distances from L1 and L2 are 6 and 3 units…

Year	2026	2025	2024	Total
Questions	7	13	9	29

More Straight Lines Previous-Year Questions

Q7 jee_main_2026_22_january_evening Orthocenter and Concurrency

Among the statements: (S1): If

A(5,-1)

and

B(-2,3)

are two vertices of a triangle, whose orthocentre is

(0,0)

, then its third vertex is

(-4,-7)

and (S2): If positive numbers

2a, b, c

are three consecutive terms of an A.P., then the lines

ax + by + c = 0

are concurrent at

(2,-2)

A. Only (S1) is correct
B. Only (S2) is correct
C. Both are incorrect
D. Both are correct

Solution

Related Formula

Orthocenter property: Altitude from $A$ $A$ is perpendicular to $BC$ $BC$, Altitude from $B$ $B$ is perpendicular to $AC$ $AC$. Concurrency of lines: $ax+by+c=0$ $ax+by+c=0$ passes through $(x₀, y₀)$ $(x_0, y_0)$ if $ax₀+by₀+c=0$ $ax_0+by_0+c=0$ holds.

Core Logic

Triangle orthocenter diagram for Q7 - JEE Main 2026 Evening

Statement 1: Let third vertex be $C(h,k)$ $C(h,k)$ and orthocenter $O(0,0)$ $O(0,0)$.

Slope $AO \cdot$ $AO \cdot$ Slope $BC = -1 ((-1)/(5)) \cdot ((k-3)/(h+2)) = -1 5h - k + 13 = 0$
Slope $BO \cdot$ $BO \cdot$ Slope $AC = -1 ((3)/(-2)) \cdot ((k+1)/(h-5)) = -1 4k = 7h$

Solving simultaneously gives $h = -4, k = -7 C(-4,-7)$ $h = -4, k = -7 \implies C(-4,-7)$. Statement 1 is correct.

Statement 2: $2a, b, c$ $2a, b, c$ in A.P. $2b = 2a + c 2a - 2b + c = 0$ . Comparing with $ax + by + c = 0$ $ax + by + c = 0$ gives $x = 2, y = -2$ $x = 2, y = -2$. Statement 2 is correct.

Step 1: Final Conclusion

Both (S1) and (S2) are correct statements.

Pattern Recognition

Orthocenter coordinates yield perpendicularity conditions via slopes. Concurrency follows directly from linear relation between line coefficients.

Chapter Mix

Class 11 Maths: Straight Lines

Q4 jee_main_2026_23_january_morning Area and Distance

A rectangle is formed by the lines

x = 0, y = 0, x = 3

and

y = 4

. Let the line

L

be perpendicular to

3x + y + 6 = 0

and divide the area of the rectangle into two equal parts. Then the distance of the point

((1)/(2), -5)

from the line

L

is equal to:

A. $2\sqrt(5)$ $2\sqrt{5}$
B. $3\sqrt(10)$ $3\sqrt{10}$
C. $\sqrt(10)$ $\sqrt{10}$
D. $2\sqrt(10)$ $2\sqrt{10}$

Solution

Related Formula

Distance = |Ax₁ + By₁ + C|\sqrt(A² + B²)

Core Logic

Any line that divides the area of a rectangle into two equal parts must pass through the center of the rectangle.

Area and Distance diagram for Q4 - JEE Main 2026 Morning

The vertices of the rectangle are

(0,0)

(3,0)

(3,4)

, and

(0,4)

. Its center is at

((3)/(2), 2)

Step 1: Equation of Line L

The line $L$ $L$ is perpendicular to $3x + y + 6 = 0$ $3x + y + 6 = 0$. The slope of this reference line is $-3$ $-3$. Thus, the slope of $L$ $L$ is $m = (1)/(3)$ $m = \frac{1}{3}$. The equation of $L$ $L$ passing through $((3)/(2), 2)$ $\left(\frac{3}{2}, 2\right)$ is:

y - 2 = (1)/(3)(x - (3)/(2))

y = (x)/(3) - (1)/(2) + 2 \Rightarrow y = (x)/(3) + (3)/(2)

Multiplying by 6 gives:

6y = 2x + 9 \Rightarrow 2x - 6y + 9 = 0

Step 2: Calculate Distance

We need the perpendicular distance from $((1)/(2), -5)$ $\left(\frac{1}{2}, -5\right)$ to the line $2x - 6y + 9 = 0$ $2x - 6y + 9 = 0$:

D = | 2((1)/(2)) - 6(-5) + 9 |\sqrt(2² + (-6)²)

D = |1 + 30 + 9|\sqrt(4 + 36) = 40\sqrt(40) = \sqrt(40) = 2\sqrt(10)

Pattern Recognition

Area bisectors for symmetric geometric figures (rectangles, circles, ellipses) always pass exactly through their geometric center.

Chapter Mix

Class 11 Maths: Straight Lines

Q4 jee_main_2026_23_january_evening Properties of Rhombus

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(α, β)

and

D(γ, δ)

are the other two vertices, then

|α + β + γ + δ|

is equal to :

A. $9$ $9$
B. $3$ $3$
C. $6$ $6$
D. $1$ $1$

Solution

Related Formula

In a rhombus, diagonals bisect each other. Therefore, the midpoint of diagonal $AC$ $AC$ is the same as the midpoint of diagonal $BD$ $BD$.

Core Logic

Properties of Rhombus diagram for Q4 - JEE Main 2026 Evening

Given the coordinates of vertices

A(1, 2)

and

C(-3, -6)

, we can find the midpoint

O

of the diagonal

AC

xm = (1 + (-3))/(2) = -1

ym = (2 + (-6))/(2) = -2

So, the midpoint is $(-1, -2)$ $(-1, -2)$.

Let the coordinates of $B$ $B$ and $D$ $D$ be $(α, β)$ $(\alpha, \beta)$ and $(γ, δ)$ $(\gamma, \delta)$ respectively. Since the midpoint of $BD$ $BD$ is also $(-1, -2)$ $(-1, -2)$:

(α+γ)/(2) = -1 α+γ = -2

(β+δ)/(2) = -2 β+δ = -4

Step 1: Final Calculation

We need the absolute value of the sum of the coordinates of $B$ $B$ and $D$ $D$:

|α + β + γ + δ| = |(α + γ) + (β + δ)|

= |-2 - 4| = |-6| = 6

Pattern Recognition

The extra information regarding the parallel line $7x - y = 14$ $7x - y = 14$ is completely redundant for finding the sum of the coordinates. Always check if a basic geometric property (like diagonals bisecting) circumvents heavy calculations.

Chapter Mix

Class 11 Maths: Straight Lines

Q13 jee_main_2026_24_january_morning Angle Bisectors of a Triangle

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

α x + β y = 5

, then the value of

α² + β²

A. $8$ $8$
B. $5$ $5$
C. $13$ $13$
D. $10$ $10$

Solution

Related Formula

Angle bisector divides the opposite side in the ratio of adjacent sides: $(BD)/(DC) = (AB)/(BC)$ $\frac{BD}{DC} = \frac{AB}{BC}$ (Internal Bisector Theorem).

Core Logic

Calculate lengths

AB

and

BC

(wait, we need bisector of angle

ABC

, so it intersects

AC

D

, so

(AD)/(DC) = (AB)/(BC)

Step 1: Side Lengths

AB = \sqrt((2-1)² + (-1-0)²) = \sqrt(1 + 1) = \sqrt(2)

BC = \sqrt(((7)/(3)-2)² + ((4)/(3)-(-1))²) = \sqrt((1)/(9) + (49)/(9)) = \sqrt((50)/(9)) = 5\sqrt(2)3

Ratio $(AB)/(BC) = \sqrt(2) 5\sqrt(2)3 = (3)/(5)$ .

Step 2: Coordinates of D

Point $D$ $D$ divides $AC$ $AC$ internally in ratio $3:5$ $3:5$.

x = (3(7/3) + 5(1))/(3+5) = (7+5)/(8) = (12)/(8) = (3)/(2)

y = (3(4/3) + 5(0))/(3+5) = (4)/(8) = (1)/(2)

$D = ((3)/(2), (1)/(2))$

Step 3: Equation of Bisector

The bisector passes through $B(2,-1)$ $B(2,-1)$ and $D(3/2, 1/2)$ $D(3/2, 1/2)$. Slope $m = (1/2 - (-1))/(3/2 - 2) = (3/2)/(-1/2) = -3$ Equation: $y - (-1) = -3(x - 2)$ $y - (-1) = -3(x - 2)$ $y + 1 = -3x + 6 \Rightarrow 3x + y = 5$

Step 4: Final Value Calculation

Comparing $3x + y = 5$ $3x + y = 5$ with $α x + β y = 5$ $\alpha x + \beta y = 5$, we get $α = 3$ $\alpha = 3$, $β = 1$ $\beta = 1$.

α² + β² = 9 + 1 = 10

Pattern Recognition

When asked for a specific angle bisector in a coordinate triangle, using the Angle Bisector Theorem (ratio trick) to find a second point is much faster than computing formulas of bisectors between lines.

Chapter Mix

Class 11 Maths: Straight Lines

Solution & Explanation

Related Formula

Core Logic

Step 1: Set up geometric projections

Step 2: Solve the trigonometric system

Step 3: Calculate Area

Pattern Recognition

Chapter Mix

Reference Study Guides

More Straight Lines Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Equation of Line L

Step 2: Calculate Distance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Side Lengths

Step 2: Coordinates of D

Step 3: Equation of Bisector

Step 4: Final Value Calculation

Pattern Recognition

Chapter Mix

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