Four distinct points (a, 0), (0, b), (c , 0) and (0, d) are lie on a plane in such a way that ac = bd, they will

Options

A. Form a trapezium
B. Form a triangle
C. Lie on a circle
D. Form a quadrilateral, whose areas is zero

Answer

To solve the problem, we need to analyze the given points and the condition \( ac = bd \). The points are \( (a, 0) \), \( (0, b) \), \( (c, 0) \), and \( (0, d) \). We will determine whether these points lie on a circle, form a trapezium, form a triangle, or form a quadrilateral whose area is zero.

### Step-by-Step Solution:

1. **Identify the Points**:
The four points given are:
- \( A(a, 0) \)
- \( B(0, b) \)
- \( C(c, 0) \)
- \( D(0, d) \)

2. **Understand the Condition**:
We are given the condition \( ac = bd \). This implies a relationship between the x-coordinates and y-coordinates of the points.

3. **Check for Collinearity**:
To check if the points can lie on a circle, we need to ensure they are not collinear. If they are collinear, they cannot lie on a circle.

4. **Use the Circle Equation**:
The general equation of a circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
For the points to lie on the circle, they must satisfy this equation.

5. **Substituting Points into the Circle Equation**:
- For point \( A(a, 0) \):
\[ a^2 + 0^2 + 2ga + 2f(0) + c = 0 \implies a^2 + 2ga + c = 0 \quad \text{(1)} \]
- For point \( B(0, b) \):
\[ 0^2 + b^2 + 2g(0) + 2fb + c = 0 \implies b^2 + 2fb + c = 0 \quad \text{(2)} \]
- For point \( C(c, 0) \):
\[ c^2 + 0^2 + 2gc + 2f(0) + c = 0 \implies c^2 + 2gc + c = 0 \quad \text{(3)} \]
- For point \( D(0, d) \):
\[ 0^2 + d^2 + 2g(0) + 2fd + c = 0 \implies d^2 + 2fd + c = 0 \quad \text{(4)} \]

6. **Subtracting Equations**:
By subtracting equations (1) and (3):
\[ (a^2 - c^2) + 2g(a - c) = 0 \implies (a - c)(a + c + 2g) = 0 \]
Since \( a \neq c \) (the points are distinct), we have:
\[ a + c + 2g = 0 \implies g = -\frac{a+c}{2} \]

Similarly, we can subtract equations (2) and (4) to find \( f \):
\[ (b^2 - d^2) + 2f(b - d) = 0 \implies (b - d)(b + d + 2f) = 0 \]
Since \( b \neq d \), we have:
\[ b + d + 2f = 0 \implies f = -\frac{b+d}{2} \]

7. **Final Condition**:
We have established that the points can satisfy the circle equation if \( ac = bd \) holds true. Thus, the condition \( ac = bd \) ensures that the points \( A, B, C, D \) lie on the same circle.

### Conclusion:
The four distinct points \( (a, 0) \), \( (0, b) \), \( (c, 0) \), and \( (0, d) \) will lie on a circle if \( ac = bd \).