If `x_(1), x_(2), x_(3) and y_(1), y_(2), y_(3)` are both in G.P. with the same common ratio, then the points `A(x_(1), y_(1)), B(x_(2), y_(2)), C(x_(3), y_(3))` :

Options

A. lie on a st. line
B. lie on a circle
C. lie on an ellipse
D. vertices of a `Delta`

Answer

To solve the problem, we need to show that the points \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) are collinear, given that \( x_1, x_2, x_3 \) and \( y_1, y_2, y_3 \) are in geometric progression (G.P.) with the same common ratio.

### Step-by-Step Solution:

1. **Understanding the Geometric Progression**:
Since \( x_1, x_2, x_3 \) are in G.P., we can express them as:
\[ x_2 = x_1 \cdot r \quad \text{and} \quad x_3 = x_1 \cdot r^2 \]
where \( r \) is the common ratio.

Similarly, for \( y_1, y_2, y_3 \):
\[ y_2 = y_1 \cdot r \quad \text{and} \quad y_3 = y_1 \cdot r^2 \]

2. **Finding the Slopes**:
We will calculate the slopes of the lines formed by the points \( A, B, C \).

- **Slope of line AB**:
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{y_1 \cdot r - y_1}{x_1 \cdot r - x_1} = \frac{y_1(r - 1)}{x_1(r - 1)} = \frac{y_1}{x_1} \]

- **Slope of line BC**:
\[ m_{BC} = \frac{y_3 - y_2}{x_3 - x_2} = \frac{y_1 \cdot r^2 - y_1 \cdot r}{x_1 \cdot r^2 - x_1 \cdot r} = \frac{y_1(r^2 - r)}{x_1(r^2 - r)} = \frac{y_1}{x_1} \]

- **Slope of line AC**:
\[ m_{AC} = \frac{y_3 - y_1}{x_3 - x_1} = \frac{y_1 \cdot r^2 - y_1}{x_1 \cdot r^2 - x_1} = \frac{y_1(r^2 - 1)}{x_1(r^2 - 1)} = \frac{y_1}{x_1} \]

3. **Comparing the Slopes**:
Now we can see that:
\[ m_{AB} = m_{BC} = m_{AC} = \frac{y_1}{x_1} \]
Since all the slopes are equal, it implies that the points \( A, B, C \) are collinear.

4. **Conclusion**:
Therefore, we conclude that the points \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) lie on a straight line.

### Final Answer:
The points \( A, B, C \) are collinear.

---