Question

If the points A(2,3), B(3,2) form a triangle with a variable point \( p(t, t^2) \), where t is a parameter, then the equation of the locus of the centroid of triangle ABC is

• A. \( 9x^2 - 30x - 3y + 20 = 0 \)
• B. \( 3x^2 - 10x - y + 10 = 0 \)
• C. \( 9y^2 - 30y - 3x + 20 = 0 \)
• D. \( 3y^2 - 10y - x + 10 = 0 \)

Hint:

Eliminating the parameter is the standard technique for finding loci. Ensure you simplify the linear variable first (here, \( t \) from the x-coordinate) to substitute into the higher-degree equation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

The locus of the centroid is found by expressing the centroid coordinates \( (x, y) \) in terms of the parameter \( t \) and then eliminating \( t \) to find the relationship between \( x \) and \( y \).
Step 2: Key Formula or Approach:

Centroid \( G(x, y) \) of triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \): \[ x = \frac{x_1 + x_2 + x_3}{3}, \quad y = \frac{y_1 + y_2 + y_3}{3} \]
Step 3: Detailed Explanation:

Given vertices: \( A(2,3), B(3,2), P(t, t^2) \). Let the centroid be \( (x, y) \). 1. \( x = \frac{2 + 3 + t}{3} = \frac{5 + t}{3} \implies 3x = 5 + t \implies t = 3x - 5 \) 2. \( y = \frac{3 + 2 + t^2}{3} = \frac{5 + t^2}{3} \implies 3y = 5 + t^2 \) Substitute \( t = 3x - 5 \) into the equation for \( y \): \[ 3y = 5 + (3x - 5)^2 \] \[ 3y = 5 + (9x^2 - 30x + 25) \] \[ 3y = 9x^2 - 30x + 30 \] Divide by 3: \[ y = 3x^2 - 10x + 10 \] Rearrange to standard form: \[ 3x^2 - 10x - y + 10 = 0 \]
Step 4: Final Answer:

The equation of the locus is \( 3x^2 - 10x - y + 10 = 0 \).