If equations `a^2x+b^2y+c^2=0,a^4x+b^4y+c^4=0 and x+y+1=0` are consistent then

Options

A. a=-b
B. `|b|=|c|`
C. c=a
D. none of these

Answer

To determine the relationship between \( a \), \( b \), and \( c \) for the given equations to be consistent, we will analyze the conditions under which the three equations intersect or overlap.

### Step-by-Step Solution:

1. **Identify the Given Equations:**
The equations given are:
\[ a^2x + b^2y + c^2 = 0 \quad \text{(1)} \]
\[ a^4x + b^4y + c^4 = 0 \quad \text{(2)} \]
\[ x + y + 1 = 0 \quad \text{(3)} \]

2. **Rewrite the Equations in Standard Form:**
We can rewrite the equations in the standard linear form \( Ax + By + C = 0 \):
- For equation (1): \( a^2x + b^2y + c^2 = 0 \) implies \( A_1 = a^2, B_1 = b^2, C_1 = c^2 \)
- For equation (2): \( a^4x + b^4y + c^4 = 0 \) implies \( A_2 = a^4, B_2 = b^4, C_2 = c^4 \)
- For equation (3): \( x + y + 1 = 0 \) implies \( A_3 = 1, B_3 = 1, C_3 = 1 \)

3. **Set Up the Determinant for Consistency:**
The equations are consistent if the determinant of the coefficients is zero:
\[ \Delta = \begin{vmatrix} a^2 & b^2 & c^2 \\ a^4 & b^4 & c^4 \\ 1 & 1 & 1 \end{vmatrix} \]

4. **Calculate the Determinant:**
We can compute the determinant \( \Delta \):
\[ \Delta = a^2 \begin{vmatrix} b^4 & c^4 \\ 1 & 1 \end{vmatrix} - b^2 \begin{vmatrix} a^4 & c^4 \\ 1 & 1 \end{vmatrix} + c^2 \begin{vmatrix} a^4 & b^4 \\ 1 & 1 \end{vmatrix} \]

This simplifies to:
\[ \Delta = a^2(b^4 - c^4) - b^2(a^4 - c^4) + c^2(a^4 - b^4) \]

5. **Set the Determinant to Zero:**
For the equations to be consistent:
\[ a^2(b^4 - c^4) - b^2(a^4 - c^4) + c^2(a^4 - b^4) = 0 \]

6. **Analyze the Result:**
The above equation implies that:
\[ (b^2 - a^2)(b^2 - c^2)(c^2 - a^2) = 0 \]
This means that:
\[ a^2 = b^2, \quad b^2 = c^2, \quad \text{or} \quad c^2 = a^2 \]
Hence, we conclude:
\[ |a| = |b| = |c| \]

### Final Result:
The relationship between \( a \), \( b \), and \( c \) is:
\[ |a| = |b| = |c| \]