Question

If the plane \[ 2x + 3y + 5z = 1 \] intersects the coordinate axes at the points A, B, C, then the centroid of \( \triangle ABC \) is

• A. \( \left( \frac{3}{2}, 1, \frac{3}{5} \right) \)
• B. \( \left( \frac{1}{2}, \frac{1}{3}, \frac{1}{5} \right) \)
• C. \( \left( \frac{1}{6}, \frac{1}{9}, \frac{1}{15} \right) \)
• D. \( (2, 3, 5) \)

Hint:

The centroid of a triangle formed by intercepts on the coordinate axes can be found by averaging the intercepts.

Answer 1

The Correct Option is C

Step 1: Find the intercepts.
The intercepts of the plane are found by setting two variables to 0 and solving for the third. For \( 2x + 3y + 5z = 1 \): - For the x-intercept, set \( y = 0 \) and \( z = 0 \), solving for \( x \) gives \( x = \frac{1}{2} \). - For the y-intercept, set \( x = 0 \) and \( z = 0 \), solving for \( y \) gives \( y = \frac{1}{3} \). - For the z-intercept, set \( x = 0 \) and \( y = 0 \), solving for \( z \) gives \( z = \frac{1}{5} \).
Step 2: Calculate the centroid.
The centroid of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is given by \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) \). Substituting the intercepts, we get the centroid as \( \left( \frac{1}{6}, \frac{1}{9}, \frac{1}{15} \right) \).

Step 3: Conclusion.
Thus, the centroid of \( \triangle ABC \) is \( \left( \frac{1}{6}, \frac{1}{9}, \frac{1}{15} \right) \), corresponding to option (C).