If the pairs of lines `ax^2+2hxy+by^2=0` and `a'x^2+2h'xy+b'y^2=0` have one line in common, then `(ab'-a'b)^2` is equal to

Options

A. `(h'b-hb')(ha'-h'a)`
B. `4(h'b-hb')(ha'-h'a)`
C. `2(h'b-hb')(ha'-h'a)`
D. `4(h'b+hb')(ha'+h'a)`

Explanation

To solve the problem, we need to find the value of \((ab' - a'b)^2\) given that the pairs of lines \(ax^2 + 2hxy + by^2 = 0\) and \(a'x^2 + 2h'xy + b'y^2 = 0\) have one line in common.

### Step-by-Step Solution:

1. **Assume the Common Line**:
Let's assume the common line can be represented as \(y = mx\).

2. **Substitute into the First Equation**:
Substitute \(y = mx\) into the first equation:
\[ ax^2 + 2h(mx)x + b(mx)^2 = 0 \]
This simplifies to:
\[ ax^2 + 2hmx^2 + bm^2x^2 = 0 \]
Factoring out \(x^2\):
\[ (a + 2hm + bm^2)x^2 = 0 \]
Thus, we have:
\[ a + 2hm + bm^2 = 0 \quad \text{(Equation 1)} \]

3. **Substitute into the Second Equation**:
Now substitute \(y = mx\) into the second equation:
\[ a'x^2 + 2h'(mx)x + b'(mx)^2 = 0 \]
This simplifies to:
\[ a'x^2 + 2h'mx^2 + b'm^2x^2 = 0 \]
Factoring out \(x^2\):
\[ (a' + 2h'm + b'm^2)x^2 = 0 \]
Thus, we have:
\[ a' + 2h'm + b'm^2 = 0 \quad \text{(Equation 2)} \]

4. **Set Up the System of Equations**:
We now have two equations:
- \(a + 2hm + bm^2 = 0\) (Equation 1)
- \(a' + 2h'm + b'm^2 = 0\) (Equation 2)

5. **Eliminate \(m^2\)**:
From both equations, we can express \(m^2\):
From Equation 1:
\[ m^2 = -\frac{a + 2hm}{b} \]
From Equation 2:
\[ m^2 = -\frac{a' + 2h'm}{b'} \]

6. **Equate the Two Expressions for \(m^2\)**:
Setting the two expressions for \(m^2\) equal gives:
\[ -\frac{a + 2hm}{b} = -\frac{a' + 2h'm}{b'} \]
Cross-multiplying leads to:
\[ (a + 2hm)b' = (a' + 2h'm)b \]

7. **Rearranging the Equation**:
Rearranging this gives:
\[ ab' + 2hbm - a'b - 2h'a'm = 0 \]
This can be rearranged to:
\[ ab' - a'b + 2hbm - 2h'a'm = 0 \]

8. **Finding the Result**:
From the above equation, we can isolate the term \(ab' - a'b\):
\[ (ab' - a'b) = 2h'a'm - 2hbm \]
Squaring both sides gives:
\[ (ab' - a'b)^2 = (2h'a'm - 2hbm)^2 \]

9. **Final Result**:
After simplification, we find that:
\[ (ab' - a'b)^2 = 4(h'a'm - hbm)^2 \]

### Conclusion:
Thus, the value of \((ab' - a'b)^2\) is equal to \(4(h'a'm - hbm)^2\).