If lines `a^(2)x^(2)+bcy^(2)=a(b+c)xy` are mutually perpendicular then

Options

A. `c^(2)+ab=0`
B. `b^(2)+ca=0`
C. `a^(2)+bc=0`
D. `a^(2)+b^(2)+c^(2)=0`

Explanation

To solve the problem of determining the condition under which the lines represented by the equation \( a^2 x^2 + b c y^2 = a(b+c)xy \) are mutually perpendicular, we can follow these steps:

### Step-by-Step Solution:

1. **Rearranging the Equation**:
Start with the given equation:
\[ a^2 x^2 + b c y^2 - a(b+c)xy = 0 \]
This is the standard form of a pair of straight lines.

2. **Identifying Coefficients**:
Compare the equation with the general form of a pair of straight lines:
\[ A x^2 + 2H xy + B y^2 = 0 \]
Here, we identify:
- \( A = a^2 \)
- \( 2H = -a(b+c) \) (thus, \( H = -\frac{a(b+c)}{2} \))
- \( B = bc \)

3. **Condition for Perpendicular Lines**:
The condition for two lines to be mutually perpendicular is given by:
\[ H^2 = AB \]
Substituting the values of \( A \), \( H \), and \( B \):
\[ \left(-\frac{a(b+c)}{2}\right)^2 = a^2 \cdot bc \]

4. **Expanding the Equation**:
Squaring the left side:
\[ \frac{a^2(b+c)^2}{4} = a^2bc \]

5. **Simplifying**:
Multiply both sides by 4 to eliminate the fraction:
\[ a^2(b+c)^2 = 4a^2bc \]
Now, assuming \( a^2 \neq 0 \), we can divide both sides by \( a^2 \):
\[ (b+c)^2 = 4bc \]

6. **Rearranging the Equation**:
This can be rearranged to:
\[ (b+c)^2 - 4bc = 0 \]
Which simplifies to:
\[ b^2 - 2bc + c^2 = 0 \]
This can be factored as:
\[ (b - c)^2 = 0 \]

7. **Conclusion**:
Therefore, we conclude that:
\[ b - c = 0 \quad \Rightarrow \quad b = c \]

### Final Result:
The lines represented by the equation \( a^2 x^2 + b c y^2 = a(b+c)xy \) are mutually perpendicular if \( b = c \).

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