JEE Main 2024 (29 Jan Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Straight Lines: In a $△ A B C$ , suppose $y = x$ is the equation of the bisector of the angle $B$ and the equation of the side $A C$ is $2 x - y = 2$ . If $2 A B = B C$ and the point $A$ and $B$ are respectively $(4, 6)$ and $(α, β)$ , then $α + 2 β$ is equal to

Select Answer:

Visualized Solution

Problem Setup and Visualizing the Given Lines

Given $△ A B C$ with vertex $A (4, 6)$ .
Equation of side $A C$ is $2 x - y = 2$ .
Angle bisector of $∠ B$ is $y = x$ .

Finding the Intersection Point $D$

Let $D$ be the intersection of the side $A C$ and the angle bisector $y = x$ .
Substitute $y = x$ into $2 x - y = 2$ .
$2 x - x = 2 ⟹ x = 2$ .
Since $y = x$ , we get $y = 2$ . Thus, $D (2, 2)$ .

Applying the Internal Angle Bisector Theorem

The internal angle bisector divides the opposite side internally in the ratio of the adjacent sides.
Therefore, $\frac{A D}{D C} = \frac{A B}{B C}$ .
Given $2 A B = B C$ , which means $\frac{A B}{B C} = \frac{1}{2}$ .
So, point $D$ divides $A C$ internally in the ratio $1 : 2$ .

Calculating the Coordinates of Vertex $C$

Let the coordinates of $C$ be $(x, y)$ .
Using the section formula for internal division: $D = \frac{1 \cdot C + 2 \cdot A}{1 + 2}$ .
For the x-coordinate: $2 = \frac{1 \cdot x + 2 \cdot 4}{3} ⟹ 6 = x + 8 ⟹ x = - 2$ .
For the y-coordinate: $2 = \frac{1 \cdot y + 2 \cdot 6}{3} ⟹ 6 = y + 12 ⟹ y = - 6$ .
Thus, vertex $C$ is $(- 2, - 6)$ .

The Reflection Property of Angle Bisectors

Crucial Property: The image of any point on one arm of an angle across its internal bisector lies on the other arm.
Therefore, the reflection of vertex $A (4, 6)$ across the bisector $y = x$ must lie on the line containing side $B C$ .

Calculating the Image $A^{'}$

The reflection of a point $(x_{1}, y_{1})$ across the line $y = x$ is $(y_{1}, x_{1})$ .
So, the image of $A (4, 6)$ across $y = x$ is $A^{'} (6, 4)$ .
This point $A^{'} (6, 4)$ lies on the line $B C$ .

Forming the Equation of Line $B C$

The line $B C$ passes through $C (- 2, - 6)$ and $A^{'} (6, 4)$ .
First, find the slope $m$ : $m = \frac{4 - ( - 6 )}{6 - ( - 2 )} = \frac{10}{8} = \frac{5}{4}$ .
Use point-slope form with $A^{'} (6, 4)$ : $y - 4 = \frac{5}{4} (x - 6)$ .
Cross-multiply: $4 (y - 4) = 5 (x - 6) ⟹ 4 y - 16 = 5 x - 30$ .
Rearranging gives the equation of $B C$ : $5 x - 4 y = 14$ .

Determining the Coordinates of Vertex $B$

Vertex $B (α, β)$ is the intersection of the line $B C$ and the angle bisector $y = x$ .
Since $B$ lies on $y = x$ , its coordinates must be equal: $α = β$ .
Substitute $y = x$ into the equation of $B C$ : $5 x - 4 x = 14 ⟹ x = 14$ .
Therefore, $y = 14$ as well.
The coordinates of $B$ are $(14, 14)$ , meaning $α = 14$ and $β = 14$ .

Calculating the Final Value $α + 2 β$

We have found $α = 14$ and $β = 14$ .
The question asks for the value of $α + 2 β$ .
Substitute the values: $14 + 2 (14)$ .
$14 + 28 = 42$ .
The final answer is $42$ .

00:00 / 00:00

The Sigma Insight: Various Forms of Equations of a Line

Solution Diagram

Analyzing the Setup

We are given a triangle

A B C

with vertex

A (4, 6)

and the side

A C

defined by the line

2 x - y = 2

. The internal angle bisector of

∠ B

is given by the line

y = x

To find the intersection point

D

where the bisector meets

A C

, we solve the system:

2 x - y = 2

y = x

Substituting

y = x

into the first equation, we get

2 x - x = 2

, which yields

x = 2

. Since

y = x

, our anchor point is

D (2, 2)

The Power of the Ratio

We invoke the Internal Angle Bisector Theorem, which states that the bisector divides the opposite side in the ratio of the adjacent sides:

\frac{A D}{D C} = \frac{A B}{B C}

The problem provides the constraint

2 A B = B C

, which implies

\frac{A B}{B C} = \frac{1}{2}

. Therefore, point

D

divides

A C

in the ratio

1 : 2

Using the section formula, where

D = \frac{1 \cdot C + 2 \cdot A}{1 + 2}

, we solve for

C (x, y)

(2, 2) = (\frac{x + 2 ( 4 )}{3}, \frac{y + 2 ( 6 )}{3})

Solving for the coordinates, we find

x + 8 = 6 \Rightarrow x = - 2

and

y + 12 = 6 \Rightarrow y = - 6

. Thus, the vertex is

C (- 2, - 6)

The Reflection Property

A fundamental property of angle bisectors is that the reflection of any point on one arm of an angle across the bisector lies on the other arm. By reflecting vertex

A (4, 6)

across the line

y = x

, we obtain a point

A^{'}

that must lie on the line

B C

Reflecting a point across

y = x

involves swapping the coordinates. Thus, the image is

A^{'} (6, 4)

We now have two points that define the line

B C

C (- 2, - 6)

and

A^{'} (6, 4)

The Final Intersection

The slope

m

of line

B C

is calculated as:

m = \frac{4 - ( - 6 )}{6 - ( - 2 )} = \frac{10}{8} = \frac{5}{4}

Using the point-slope form

y - y_{1} = m (x - x_{1})

with point

A^{'} (6, 4)

y - 4 = \frac{5}{4} (x - 6)

4 y - 16 = 5 x - 30

5 x - 4 y = 14

Vertex

B (α, β)

is the intersection of line

B C

and the bisector

y = x

. Since

α = β

, we substitute into the line equation:

5 x - 4 x = 14 \Rightarrow x = 14

Thus,

B

(14, 14)

, meaning

α = 14

and

β = 14

. The final value requested is

α + 2 β

α + 2 β = 14 + 2 (14) = 14 + 28 = 42

Common Traps & FAQs

Wait, why does the reflection of

A

across the bisector lie on

B C

This is a fundamental property of angle bisectors! Imagine folding the paper along the line

y = x

. The arm

A B

would map onto the arm

B C

. Therefore, any point on

A B

(like

A

) must map onto a point on

B C

. It is a beautiful symmetry that turns a complex coordinate geometry problem into a simple line intersection. It is one of the most elegant tools in your JEE toolkit.

Can I use the distance formula to find

B

You could, but it would lead to a system of quadratic equations that is algebraically exhausting. In JEE Advanced, we look for the 'elegant path.' Using the reflection property allows us to find the equation of line

B C

directly, reducing the problem to a simple linear intersection. Always look for the geometric shortcut before diving into heavy algebra!

Why did we use the section formula for

D

The section formula is the bridge between geometry and algebra. Since

D

lies on

A C

and we know the ratio

A D : D C

from the Angle Bisector Theorem, the section formula is the most direct way to find the coordinates of

C

without needing the slope of

A C

or other complex parameters.

Select Answer:

Visualized Solution

Problem Setup and Visualizing the Given Lines

Finding the Intersection Point $D$

Applying the Internal Angle Bisector Theorem

Calculating the Coordinates of Vertex $C$

The Reflection Property of Angle Bisectors

Calculating the Image $A^{'}$

Forming the Equation of Line $B C$

Determining the Coordinates of Vertex $B$

Calculating the Final Value $α + 2 β$

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Power of the Ratio

The Reflection Property

The Final Intersection

Similar Questions

The vertices of a triangle are $A (- 1, - 7), B (5, 1)$ and $C (1, 4)$ . The equation of the bisector of the angle $∠ A B C$ is .........

The equations of two sides $A B$ and $A C$ of a triangle $A B C$ are $4 x + y = 14$ and $3 x - 2 y = 5$ , respectively. The point $(2, - 4/3)$ divides the third side $B C$ internally in the ratio $2 : 1$ . The equation of the side $B C$ is

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. Then the equation of the bisector of the angle $P QR$ is

Let $A B C$ be an isosceles triangle in which $A$ is at $(- 1, 0)$ , $∠ A = \frac{2 π}{3}$ , $A B = A C$ and $B$ is on the positive $x$ -axis. If $B C = 43$ and the line $B C$ intersects the line $y = x + 3$ at $(α, β)$ , then $\frac{β ^{4}}{α ^{2}}$ is :

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. The equation of the bisector of the angle $P QR$ is

Let B and C be the two points on the line $y + x = 0$ such that B and C are symmetric with respect to the origin. Suppose A is a point on $y - 2 x = 2$ such that $Δ A B C$ is an equilateral triangle. Then, the area of the $Δ A B C$ is

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 $7 x - y = 14$ . If $B (α, β)$ and $D (γ, δ)$ are the other two vertices, then $∣ α + β + γ + δ ∣$ is equal to

Let $α, β, γ, δ \in Z$ and let $A (α, β), B (1, 0), C (γ, δ)$ and $D (1, 2)$ be the vertices of a parallelogram $A B C D$ . If $A B = 10$ and the points $A$ and $C$ lie on the line $3 y = 2 x + 1$ , then $2 (α + β + γ + δ)$ is equal to

A straight line through the point $A (3, 4)$ is such that its intercept between the axes is bisected at $A$ . Its equation is

Let $A$ be a fixed point $(0, 6)$ and $B$ be a moving point $(2 t, 0)$ . Let $M$ be the mid-point of $A B$ and the perpendicular bisector of $A B$ meets the $y$ -axis at $C$ . The locus of the mid-point $P$ of $M C$ is :

Select Answer:

Visualized Solution

Problem Setup and Visualizing the Given Lines

Finding the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Applying the Internal Angle Bisector Theorem

Calculating the Coordinates of Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

The Reflection Property of Angle Bisectors

Calculating the Image .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A′

Forming the Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

Determining the Coordinates of Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Calculating the Final Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α+2β

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Power of the Ratio

The Reflection Property

The Final Intersection

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The vertices of a triangle are A(−1,−7),B(5,1) and C(1,4). The equation of the bisector of the angle ∠ABC is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=(−1,0),Q=(0,0) and R=(3,33​) be three points. Then the equation of the bisector of the angle PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=(−1,0),Q=(0,0) and R=(3,33​) be three points. The equation of the bisector of the angle PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A straight line through the point A(3,4) is such that its intercept between the axes is bisected at A. Its equation is

Finding the Intersection Point $D$

Calculating the Coordinates of Vertex $C$

Calculating the Image $A^{'}$

Forming the Equation of Line $B C$

Determining the Coordinates of Vertex $B$

Calculating the Final Value $α + 2 β$

The vertices of a triangle are $A (- 1, - 7), B (5, 1)$ and $C (1, 4)$ . The equation of the bisector of the angle $∠ A B C$ is .........

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. Then the equation of the bisector of the angle $P QR$ is

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. The equation of the bisector of the angle $P QR$ is

A straight line through the point $A (3, 4)$ is such that its intercept between the axes is bisected at $A$ . Its equation is