JEE Main 2023 (24 January Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Straight Lines: The equations of the sides AB, BC and CA of a triangle ABC are: $2 x + y = 0$ , $x + p y = 21 a$ , $(a \neq = 0)$ and $x - y = 3$ respectively. Let $P (2, a)$ be the centroid of $Δ A B C$ . Then $(B C)^{2}$ is equal to

Q: Why do we parameterize the vertices $B$ and $C$ instead of just solving for the intersection points directly?

That is a great question! If we tried to solve for $B$ and $C$ directly, we would have a system of equations with too many unknowns ($p$ and $a$ are also unknown). By parameterizing $B$ and $C$ using the equations of the lines they lie on, we reduce the number of variables and create a system of equations that we can solve step-by-step using the centroid information.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): To solve this problem, visualize the triangle as a system of three intersecting lines. The vertices

A

B

, and

C

are the points where these lines meet. Since the centroid

P (2, a)

is given, you should express the coordinates of vertices

B

and

C

in terms of parameters (like

α

and

β

) based on the lines they lie on.

Hint 2

Hint 2 (The Arsenal): Recall that the centroid

P (x_{G}, y_{G})

of a triangle with vertices

(x_{1}, y_{1})

(x_{2}, y_{2})

, and

(x_{3}, y_{3})

is given by

x_{G} = \frac{x _{1} + x _{2} + x _{3}}{3}

and

y_{G} = \frac{y _{1} + y _{2} + y _{3}}{3}

. Use this to relate the parameters

α

and

β

to the centroid coordinates.

Hint 3

Hint 3 (The Execution): Start by finding vertex

A

by solving the system of equations for lines

A B

and

C A

. Then, parameterize

B

(α, - 2 α)

and

C

(β, β - 3)

. Use the centroid formula to find a relationship between

α, β,

and

a

. Finally, substitute these into the equation for line

B C

to solve for the unknowns.

Visualized Solution

Visualizing the Triangle $A B C$

Triangle $A B C$ with sides:
$A B : 2 x + y = 0$
$B C : x + p y = 21 a$
$C A : x - y = 3$
Centroid: $P (2, a)$

Finding Vertex $A$

Solve $2 x + y = 0$ and $x - y = 3$ :
From $A B$ , $y = - 2 x$
Substitute in $C A$ : $x - (- 2 x) = 3 \Rightarrow 3 x = 3 \Rightarrow x = 1$
Then $y = - 2 (1) = - 2$
Vertex $A (1, - 2)$

Parametrizing Vertex $B$

$B$ lies on $2 x + y = 0$ .
Let $x$ -coordinate be $α$ .
Then $y = - 2 α$ .
Vertex $B (α, - 2 α)$

Parametrizing Vertex $C$

$C$ lies on $x - y = 3$ .
Let $x$ -coordinate be $β$ .
Then $y = β - 3$ .
Vertex $C (β, β - 3)$

Using the Centroid Formula for $x$

Centroid $P (2, a)$ $x$ -coordinate:
$\frac{1 + α + β}{3} = 2$
$1 + α + β = 6$
$α + β = 5 —(1)$

Using the Centroid Formula for $y$

Centroid $P (2, a)$ $y$ -coordinate:
$\frac{- 2 - 2 α + β - 3}{3} = a$
$\frac{β - 2 α - 5}{3} = a$
$β - 2 α - 5 = 3 a —(2)$

Points $B$ and $C$ on line $B C$

$B (α, - 2 α)$ lies on $x + p y = 21 a$ .
$α + p (- 2 α) = 21 a$
$α (1 - 2 p) = 21 a —(3)$

Substituting Point $C$ into $B C$

$C (β, β - 3)$ lies on $x + p y = 21 a$ .
$β + p (β - 3) = 21 a$
$β (1 + p) - 3 p = 21 a —(4)$

Solving for $α$ and $a$

From (1), $β = 5 - α$ . Substitute in (2):
$(5 - α) - 2 α - 5 = 3 a$
$- 3 α = 3 a$
$a = - α$

Finding the value of $p$

Substitute $a = - α$ into (3):
$α (1 - 2 p) = 21 (- α)$
Since $a \neq = 0$ , $α \neq = 0$ . Divide by $α$ :
$1 - 2 p = - 21$
$2 p = 22 \Rightarrow p = 11$

Finding coordinates of $B$ and $C$

Substitute $p = 11$ , $β = 5 - α$ , $a = - α$ into (4):
$(5 - α) (1 + 11) - 3 (11) = 21 (- α)$
$12 (5 - α) - 33 = - 21 α$
$60 - 12 α - 33 = - 21 α$
$27 = - 9 α \Rightarrow α = - 3$
$β = 5 - (- 3) = 8$
Vertices: $B (- 3, 6)$ and $C (8, 5)$

Final Calculation of $(B C)^{2}$

Using distance formula for $(B C)^{2}$ :
$(B C)^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
$(B C)^{2} = (8 - (- 3))^{2} + (5 - 6)^{2}$
$(B C)^{2} = 1 1^{2} + (- 1)^{2}$
$(B C)^{2} = 121 + 1 = 122$

00:00 / 00:00

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Solution Diagram

Analyzing the Setup

We are given the equations of the sides of triangle

A B C

A B : 2 x + y = 0

B C : x + p y = 21 a

C A : x - y = 3

The centroid of the triangle is given as

P (2, a)

. Our objective is to determine the value of

(B C)^{2}

Locating the Anchor Point

Vertex

A

is the intersection of lines

A B

and

C A

. We solve the system:

2 x + y = 0

x - y = 3

From the first equation, we have

y = - 2 x

. Substituting this into the second equation:

x - (- 2 x) = 3

3 x = 3 \Rightarrow x = 1

Substituting

x = 1

back into

y = - 2 x

, we find

y = - 2

. Thus, the vertex

A

is located at

(1, - 2)

The Art of Parameterization

We define the coordinates of vertices

B

and

C

based on the lines they lie upon. Since

B

lies on

A B

(

2 x + y = 0

), we parameterize it as

B (α, - 2 α)

Since

C

lies on

C A

(

x - y = 3

), we parameterize it as

C (β, β - 3)

. This transformation reduces the complexity of the vertices to two variables,

α

and

β

The Centroid Connection

The centroid

P (2, a)

is the arithmetic mean of the vertices

A (1, - 2)

B (α, - 2 α)

, and

C (β, β - 3)

. Using the centroid formula for the

x

-coordinate:

\frac{1 + α + β}{3} = 2 \Rightarrow α + β = 5

Using the centroid formula for the

y

-coordinate:

\frac{- 2 - 2 α + β - 3}{3} = a \Rightarrow β - 2 α - 5 = 3 a

The Final Convergence

Both

B

and

C

must satisfy the equation of line

B C

x + p y = 21 a

. Substituting

B (α, - 2 α)

α - 2 p α = 21 a \Rightarrow α (1 - 2 p) = 21 a

Substituting

C (β, β - 3)

β + p (β - 3) = 21 a

By solving this system of equations, we determine the relationship

a = - α

. Solving the system yields the coordinates

B (- 3, 6)

and

C (8, 5)

Final Calculation

To find

(B C)^{2}

, we apply the distance formula between

B (- 3, 6)

and

C (8, 5)

(B C)^{2} = (8 - (- 3))^{2} + (5 - 6)^{2}

(B C)^{2} = (11)^{2} + (- 1)^{2}

(B C)^{2} = 121 + 1

The final result is:

(B C)^{2} = 122

Common Traps & FAQs

Why do we parameterize the vertices

B

and

C

instead of just solving for the intersection points directly?

That is a great question! If we tried to solve for

B

and

C

directly, we would have a system of equations with too many unknowns (

p

and

a

are also unknown). By parameterizing

B

and

C

using the equations of the lines they lie on, we reduce the number of variables and create a system of equations that we can solve step-by-step using the centroid information.

Can we use the property that the centroid divides the median in a 2:1 ratio?

Absolutely! You could use that property, but it would involve finding the midpoint of

B C

first, which would require knowing the coordinates of

B

and

C

. In this case, the direct application of the centroid coordinate formula is more efficient and leads directly to the system of equations we need.

The problem states

a \neq = 0

. Why is this condition important?

This is a crucial observation! In our derivation, we eventually reach a point where we need to divide by

α

(or

a

). If

a

were zero, then

α

would also be zero, and we would not be able to perform that division. The condition

a \neq = 0

ensures that our algebraic steps are valid and that we don't accidentally divide by zero.

Animated Solution for Mathematics - Straight Lines: The equations of the sides AB, BC and CA of a triangle ABC are: $2 x + y = 0$ , $x + p y = 21 a$ , $(a \neq = 0)$ and $x - y = 3$ respectively. Let $P (2, a)$ be the centroid of $Δ A B C$ . Then $(B C)^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Triangle $A B C$

Finding Vertex $A$

Parametrizing Vertex $B$

Parametrizing Vertex $C$

Using the Centroid Formula for $x$

Using the Centroid Formula for $y$

Points $B$ and $C$ on line $B C$

Substituting Point $C$ into $B C$

Solving for $α$ and $a$

Finding the value of $p$

Finding coordinates of $B$ and $C$

Final Calculation of $(B C)^{2}$

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Setup

Locating the Anchor Point

The Art of Parameterization

The Centroid Connection

The Final Convergence

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

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

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

Using the Centroid Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Using the Centroid Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Substituting Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Finding the value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Finding coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (BC)2

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Setup

Locating the Anchor Point

The Art of Parameterization

The Centroid Connection

The Final Convergence

Final Calculation

Similar Questions

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Visualizing the Triangle $A B C$

Finding Vertex $A$

Parametrizing Vertex $B$

Parametrizing Vertex $C$

Using the Centroid Formula for $x$

Using the Centroid Formula for $y$

Substituting Point $C$ into $B C$

Solving for $α$ and $a$

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Finding coordinates of $B$ and $C$

Final Calculation of $(B C)^{2}$

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