Let ABC be the triangle such that the equations of lines AB and AC be 3y - x = 2 and x + y = 2 , respectively, and the…

Year	2026	2025	2024	Total
Questions	7	13	9	29

Q jee_main_2025_08_april_evening Rotation of Coordinate Axes / Distance from Origin

A line passing through the point

P(a, 0)

makes an acute angle

α

with the positive x-axis. Let this line be rotated about the point

P

through an angle

(α)/(2)

in the clock-wise direction. If in the new position, the slope of the line is

2 - \sqrt(3)

and its distance from the origin is

1\sqrt(2)

, then the value of

3a² ²α - 2\sqrt(3)

is

A. $4$ $4$
B. $6$ $6$
C. $5$ $5$
D. $8$ $8$

Solution

Related Formula

Distance from origin d = |C|\sqrt(A²+B²)

Core Logic

Decode slope angle fields to define vector shifts. Match variables to isolate focal root segments accurately as provided in the reference layout steps.

Step 1: Identify Phase Shift Angles

Given final tracking position slope component is:

m = 2 - \sqrt(3) = 15^°

Because clockwise translation turns angular coordinates down by half increments:

α - (α)/(2) = 15^° α = 30^°

Step 2: Construct Rotated Vector Path Line

The line equation through point $(a,0)$ $(a,0)$ with final angle scaling is:

y = (2-\sqrt(3))(x-a) (2-\sqrt(3))x - y - a(2-\sqrt(3)) = 0

Step 3: Solve Origin Geometric Profiles

Impose perpendicular bounds tracking distance rule:

| \sqrt(3)a - 2a 4+3-4\sqrt(3)+1 | = 1\sqrt(2) a² = 2(2+\sqrt(3))

Target formulation output text string solution:

3a² ² 30^° - 2\sqrt(3) = 3 \times 2(2+\sqrt(3)) \times (1)/(3) - 2\sqrt(3) = 4

Pattern Recognition

Recognizing standard trigonometric slopes ( $15^° = 2-\sqrt(3)$ $\tan 15^\circ = 2-\sqrt{3}$) bypasses computational bottlenecks instantly when working with line updates.

Chapter Mix

Class 11 Mathematics: Straight Lines

Q57 jee_main_2025_08_april_evening Properties of Square and Diagonals

Let

a

be the length of a side of a square OABC with

O

being the origin. Its side

OA

makes an acute angle

α

with the positive x-axis and the equations of its diagonals are

(\sqrt(3) + 1)x + (\sqrt(3) - 1)y = 0 and

(\sqrt(3) - 1)x - (\sqrt(3) + 1)y + 8\sqrt(3) = 0.

Then

a²

is equal to

A. $48$ $48$
B. $32$ $32$
C. $16$ $16$
D. $24$ $24$

Solution

Related Formula

Slope m = -(A)/(B)

Core Logic

Identify which diagonal passes through the origin. Use its directional configuration slope to determine the exact angle orientation layout map of the geometric system elements.

Step 1: Determine Diagonal Orientation

The line passing through the origin is diagonal $OB$ $OB$:

mOB = - \sqrt(3)+1\sqrt(3)-1 = 105^°

Since diagonals of a square split the vertex paths evenly at $45^°$ $45^\circ$, the side $OA$ $OA$ makes an angle:

α = 105^° - 45^° = 60^°

Step 2: Coordinate Vector Definition

Coordinates of vertex $A$ $A$ matching side value $a$ $a$ are:

A(a 60^°, a 60^°) = ( (a)/(2), \sqrt(3)a2 )

Step 3: Diagonal Intersection Point Solver

Vertex $A$ $A$ lies upon the alternative structural path line:

(\sqrt(3)-1)(a)/(2) - (\sqrt(3)+1) \sqrt(3)a2 + 8\sqrt(3) = 0

a [ \sqrt(3) - 1 - 3 - \sqrt(3)2 ] = -8\sqrt(3) -2a = -8\sqrt(3) a = 4\sqrt(3)

$a² = 48$ $a^2 = 48$

Pattern Recognition

Slopes involving $\sqrt(3) \pm 1$ $\sqrt{3} \pm 1$ explicitly alert you to geometric orientations built on multiples of $15^circ$ $15^circ$ or $75^circ$ $75^circ$ configurations.

Chapter Mix

Class 11 Mathematics: Straight Lines

Q58 jee_main_2025_29_jan_evening Area of Triangles and Inscribed Shapes

Let the line

x + y = 1

meet the axes of

x

and

y

at A and B, respectively. A right angled triangle AMN is inscribed in the triangle OAB, where O is the origin and the points M and N lie on the lines OB and AB, respectively. If the area of the triangle AMN is

(4)/(9)

of the area of the triangle OAB and AN: NB = \lambda : 1, then the sum of all possible value(s) of

λ

is :

A. $(1)/(2)$ $\frac{1}{2}$
B. $(13)/(6)$ $\frac{13}{6}$
C. $(5)/(2)$ $\frac{5}{2}$
D. $2$ $2$

Solution

Related Formula

Area of a right-angled triangle:

Area = (1)/(2) \times base \times height

Core Logic

The line equation is $x + y = 1$ $x + y = 1$, giving intercept coordinates $A(1, 0)$ $A(1, 0)$ and $B(0, 1)$ $B(0, 1)$.

Area of Triangles and Inscribed Shapes diagram for Q58 - JEE Main 2025 Evening

Area of $Δ OAB = (1)/(2) \times 1 \times 1 = (1)/(2)$ . Given area condition:

Area of Δ AMN = (4)/(9) \times (1)/(2) = (2)/(9)

Step 1: Set up Trigonometric Tracing

Let $∠ MAO = 45^° - θ$ $\angle MAO = 45^\circ - \theta$. Since $Δ OAB$ $\Delta OAB$ is isosceles right-angled at $O$ $O$, $∠ OAB = 45^°$ $\angle OAB = 45^\circ$. This establishes:

OA = 1, AM = (45^° - θ)

AN = (45^° - θ) θ

MN = (45^° - θ) θ

Step 2: Solve for Angles and Ratios

Area(Δ AMN) = (1)/(2) \times ²(45^° - θ) θ θ = (2)/(9)

Solving the trigonometric ratio yields:

θ = 2 or (1)/(2)

Rejecting $θ = 2$ $\tan\theta = 2$ based on physical boundaries within the triangle limits:

(AN)/(NB) = (λ)/(1) = θ = 2

Thus, the valid evaluation matches the option sequence value of 2.

Pattern Recognition

When dealing with inscribed right triangles inside symmetric linear bounds, parameterizing coordinates with angles matching the axis slope simplifies configuration variables dramatically.

Chapter Mix

Class 11 Mathematics: Straight Lines

Q72 jee_main_2025_28_jan_morning Distance of a Point from a Line

If

α = 1 + Σr=1⁶ (-3)r-1 ¹² C₂ᵣ₋₁

, then the distance of the point

(12, \sqrt(3))

from the line

α x - \sqrt(3) y + 1 = 0

is

Numerical Answer. Answer: 5 to 5

Solution

Related Formula

Perpendicular distance from a point $(x₁, y₁)$ $(x_1, y_1)$ to a line $Ax + By + C = 0$ $Ax + By + C = 0$:

d = |Ax₁ + By₁ + C|\sqrt(A² + B²)

Core Logic

Rewrite the summation for $α$ $\alpha$ by embedding complex roots to isolate alternating terms:

α = 1 + 1\sqrt(3)i [ (1 + \sqrt(3)i)¹² - (1 - \sqrt(3)i)¹²2 ]

Using complex cube roots of unity conversions ( $1 + \sqrt(3)i = -2ω²$ $1 + \sqrt{3}i = -2\omega^2$), this simplifies directly to $α = 1$ $\alpha = 1$.

Step 1: Calculating Perpendicular Distance

Substitute $α = 1$ $\alpha = 1$ into the line equation, yielding $x - \sqrt(3)y + 1 = 0$ $x - \sqrt{3}y + 1 = 0$.

Evaluating the distance from point $(12, \sqrt(3))$ $(12, \sqrt{3})$:

d = |12 - \sqrt(3)(\sqrt(3)) + 1| 1² + (-\sqrt(3))² = (|12 - 3 + 1|)/(2) = (10)/(2) = 5

Pattern Recognition

Binomial series containing terms like $\sqrt(3)2r-1$ $\sqrt{3}^{2r-1}$ collapse neatly into simple geometric forms using standard complex rotations ( $eiπ/3$ $e^{i\pi/3}$).

Chapter Mix

Class 11 Maths: Binomial Theorem Class 11 Maths: Straight Lines

Q63 jee_main_2025_03_april_morning Distance between Parallel Lines

A line passes through the origin and makes equal angles with the positive coordinate axes[cite: 602]. It intersects the lines

L₁:2x+y+6=0

and

L₂:4x+2y-p=0

,

p>0

, at the points

A

and

B

, respectively[cite: 603]. If

AB = 9\sqrt(2)

[cite: 605] and the foot of the perpendicular from the point

A

on the line

L₂

is

M

[cite: 606, 607], then

(AM)/(BM)

is equal to[cite: 619, 620]:

A. 5
B. 4
C. 2
D. 3

Solution

Related Formula

Slope angle interaction tracking: In right triangle $AMB$ $\triangle AMB$, the tangent of inclination satisfies:

θ = (AM)/(BM)

Distance between Parallel Lines diagram for Q63 - JEE Main 2025 Morning

Core Logic

A line tracking equal angles through positive coordinate frames has equation $y=x$ $y=x$ [cite: 1353]. Its slope parameter is $m₁ = 1$ $m_1 = 1$ [cite: 1353].

The parallel boundary lines $L₁$ $L_1$ and $L₂$ $L_2$ have a uniform slope value of $m₂ = -2$ $m_2 = -2$ [cite: 1353].

Let $θ$ $\theta$ represent the precise geometric crossing angle matching the line $y=x$ $y=x$ intersecting line $L₂$ $L_2$[cite: 1353].

Step 1: Finding the requested ratio

Using the slope interaction formula to calculate the tangent angle metric[cite: 1353]: $θ = | (m₁ - m₂)/(1 + m₁ m₂) |$ [cite: 1353] $θ = | (1 - (-2))/(1 + (1)(-2)) | = | (3)/(-1) | = 3$ [cite: 1353]

By observing right-angled geometry configuration $AMB$ $\triangle AMB$ [cite: 1353]: $(AM)/(BM) = θ = 3$ $$\frac{AM}{BM} = \tan \theta = 3$$ [cite: 1353]

Pattern Recognition

Ratios of side projections of lines intersecting parallel setups depend completely on direction orientations, bypassing raw variable solving steps entirely.

Chapter Mix

Class 11 Mathematics: Straight Lines

Let ABC be the triangle such that the equations of lines AB and AC be 3y - x = 2$3y - x = 2$ and x + y = 2$x + y = 2$ , respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Find Equations of Altitudes

Step 2: Solve for Orthocentre coordinates P

Step 3: Compute Area of Triangle PBC

Pattern Recognition

Chapter Mix

Reference Study Guides

More Straight Lines Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Identify Phase Shift Angles

Step 2: Construct Rotated Vector Path Line

Step 3: Solve Origin Geometric Profiles

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine Diagonal Orientation

Step 2: Coordinate Vector Definition

Step 3: Diagonal Intersection Point Solver

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set up Trigonometric Tracing

Step 2: Solve for Angles and Ratios

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Perpendicular Distance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding the requested ratio

Pattern Recognition

Chapter Mix

More Straight Lines Questions — jee_main_2025_07_april_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

Let ABC be the triangle such that the equations of lines AB and AC be $3y - x = 2$ $3y - x = 2$ and $x + y = 2$ $x + y = 2$ , respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to