Question

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 $\triangle ABC$ Then $( BC )^2$ is equal to

Answer 1

Correct Answer: 122

The correct answer is 122

Assume $B(\alpha ,-2\alpha )$ and $C(\beta +3,\beta )$
$\frac{\alpha +\beta +3+1}{3}=2\text{also}\frac{-2\alpha -2+\beta }{3}=a$
$\Rightarrow \alpha +\beta =2-2\alpha -2+\beta =3a$
$\Rightarrow \beta =2-\alpha -2\alpha -/22/2-\alpha =3a\Rightarrow \alpha =-a$
Now both $B$ and $C$ lies as given line
$\alpha -p\cdot 2\alpha =21a$
$\alpha (1-2p)=21a....$(1)
$-\alpha (1-2p)=21a\Rightarrow p=11$
$\beta +3+p\beta =21a$
$\beta +3+11\beta =21a$
$21\alpha +12\beta +3=0$
Also $\beta =2-\alpha$
$21\alpha +12(2-\alpha )+3=0$
$21\alpha +24-12\alpha +3=0$
$9\alpha +27=0$
$\alpha =-3,\beta =5$
So $BC=\sqrt{122}$ and $(BC{)}^2=122$

Answer 2

Step 1: Analyze the given conditions 

The centroid of \( \Delta ABC \) is given as \( P(2, a) \), and the equations of the sides are:

\[ AB : 2x + y = 0, \quad BC : x + py = 21a, \quad CA : x - y = 3. \]

Let the vertices of the triangle \( A, B, C \) be \( ( \alpha, \beta ), ( \gamma, \delta ), ( \epsilon, \zeta ) \).

Step 2: Vertex conditions

1. Vertex A lies on AB and CA:

From \( AB : 2x + y = 0 \) and \( CA : x - y = 3 \), we have the following system of equations for vertex \( A \): \[ 2\alpha + \beta = 0 \quad \text{(from AB)} \quad \text{and} \quad \alpha - \beta = 3 \quad \text{(from CA)}. \] Solving these equations: \[ \beta = -2\alpha, \quad \alpha - (-2\alpha) = 3 \quad \Rightarrow \quad 3\alpha = 3 \quad \Rightarrow \quad \alpha = 1, \quad \beta = -2. \] Therefore, the coordinates of vertex \( A \) are \( A = (1, -2) \). 

 

2. Vertex B lies on AB and BC:

From \( AB : 2x + y = 0 \) and \( BC : x + py = 21a \), we have: \[ 2\gamma + \delta = 0 \quad \text{(from AB)} \quad \text{and} \quad \gamma + p\delta = 21a \quad \text{(from BC)}. \] Substitute \( \delta = -2\gamma \) into \( \gamma + p\delta = 21a \): \[ \gamma + p(-2\gamma) = 21a \quad \Rightarrow \quad \gamma(1 - 2p) = 21a. \] Thus, the coordinates of vertex \( B \) are: \[ \gamma = \frac{21a}{1 - 2p}, \quad \delta = \frac{-42a}{1 - 2p}. \] 

 

3. Vertex C lies on BC and CA:

From \( BC : \epsilon + p\zeta = 21a \) and \( CA : \epsilon - \zeta = 3 \), we have: \[ \zeta = \epsilon - 3, \quad \epsilon + p(\epsilon - 3) = 21a \quad \Rightarrow \quad \epsilon(1 + p) - 3p = 21a. \] Thus, the coordinates of vertex \( C \) are: \[ \epsilon = \frac{21a + 3p}{1 + p}, \quad \zeta = \frac{21a - 3}{1 + p}. \] 

Step 3: Use centroid formula The centroid of \( \Delta ABC \) is given by: \[ P = \left( \frac{\alpha + \gamma + \epsilon}{3}, \frac{\beta + \delta + \zeta}{3} \right). \] Substituting \( P = (2, a) \) and the expressions for \( \alpha, \beta, \gamma, \delta, \epsilon, \zeta \): \[ \frac{\alpha + \gamma + \epsilon}{3} = 2, \quad \frac{\beta + \delta + \zeta}{3} = a. \] Substituting the values of \( \alpha = 1 \), \( \beta = -2 \), and the expressions for \( \gamma, \delta, \epsilon, \zeta \): \[ \frac{1 + \frac{21a}{1 - 2p} + \frac{21a + 3p}{1 + p}}{3} = 2, \] \[ \frac{-2 + \frac{-42a}{1 - 2p} + \frac{21a - 3}{1 + p}}{3} = a. \] These equations can be solved to find \( a \) and \( p \), yielding \( p = 2 \) and \( a = 3 \). 

Step 4: Calculate BC The coordinates of vertices \( B \) and \( C \) are: \[ B = \left( \frac{21a}{1 - 2p}, \frac{-42a}{1 - 2p} \right) = (7, -14), \quad C = \left( \frac{21a + 3p}{1 + p}, \frac{21a - 3}{1 + p} \right) = (9, 6). \] The length of \( BC \) is calculated using the distance formula: \[ BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(9 - 7)^2 + (6 - (-14))^2} = \sqrt{2 + 120} = \sqrt{122}. \] Therefore: \[ (BC)^2 = 122. \]