Consider a triange ABC, where B and C are `(-a,0) and (a,0)` respectively . A be any point (h,k) P,Q,R divides the sides of this triangle in same ratio. Then centroid of `Delta PQR` is

Options

A. `(0,0)`
B. `((h)/(3), 0)`
C. `(0,(k)/(3))`
D. `((h)/(3), (k)/(3))`

Answer

To find the centroid of triangle PQR, where points P, Q, and R divide the sides of triangle ABC in the same ratio, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the vertices of triangle ABC:**
- Given points B and C are at coordinates (-a, 0) and (a, 0) respectively, and point A is at (h, k).
- Thus, the vertices of triangle ABC are:
- A(h, k)
- B(-a, 0)
- C(a, 0)

2. **Determine the coordinates of points P, Q, and R:**
- Let’s assume points P, Q, and R divide the sides AB, AC, and BC in the same ratio \( m:1 \).
- For point P on side AB:
\[ P = \left( \frac{-am + h}{m + 1}, \frac{0 + k}{m + 1} \right) = \left( \frac{-am + h}{m + 1}, \frac{k}{m + 1} \right) \]
- For point Q on side AC:
\[ Q = \left( \frac{am + h}{m + 1}, \frac{0 + k}{m + 1} \right) = \left( \frac{am + h}{m + 1}, \frac{k}{m + 1} \right) \]
- For point R on side BC:
\[ R = \left( \frac{0 + a}{m + 1}, \frac{0 + 0}{m + 1} \right) = \left( \frac{0}{m + 1}, 0 \right) \]

3. **Calculate the coordinates of the centroid of triangle PQR:**
- The centroid \( G \) of triangle PQR can be calculated using the formula:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
- Substituting the coordinates of P, Q, and R:
\[ G_x = \frac{\frac{-am + h}{m + 1} + \frac{am + h}{m + 1} + 0}{3} \]
\[ G_y = \frac{\frac{k}{m + 1} + \frac{k}{m + 1} + 0}{3} \]
- Simplifying \( G_x \):
\[ G_x = \frac{\frac{-am + h + am + h}{m + 1}}{3} = \frac{\frac{2h}{m + 1}}{3} = \frac{2h}{3(m + 1)} \]
- Simplifying \( G_y \):
\[ G_y = \frac{2k}{3(m + 1)} \]

4. **Final coordinates of the centroid of triangle PQR:**
- Thus, the coordinates of the centroid \( G \) of triangle PQR are:
\[ G = \left( \frac{2h}{3(m + 1)}, \frac{2k}{3(m + 1)} \right) \]