if one of the lines given by the equation `ax^2+2hxy+by^2=0` coincides with one of the lines given by `a'x^2+2h'xy+b'y^2=0` and the other lines representted by them be perpendicular , then .
`(ha'b')/(b'-a')=(h'ab)/(b-a)=1/2sqrt((-aa'bb')`.

Answer

To solve the problem, we need to analyze the conditions given in the question. We have two pairs of straight lines represented by the equations:

1. \( ax^2 + 2hxy + by^2 = 0 \)
2. \( a'x^2 + 2h'xy + b'y^2 = 0 \)

### Step 1: Understand the conditions
We know that one line from the first equation coincides with one line from the second equation, and the other lines represented by them are perpendicular.

### Step 2: Coinciding Lines
For the lines to coincide, the coefficients of the corresponding terms must be proportional. This gives us the relation:

\[ \frac{a}{a'} = \frac{h}{h'} = \frac{b}{b'} \]

### Step 3: Perpendicular Lines
For the lines to be perpendicular, the condition is given by:

\[ hh' + ab' + a'b = 0 \]

### Step 4: Expressing the conditions
From the coincidence condition, we can express \( h' \) and \( b' \) in terms of \( h \) and \( b \):

\[ h' = \frac{h a'}{a}, \quad b' = \frac{b a'}{a} \]

### Step 5: Substitute into the perpendicular condition
Now substitute \( h' \) and \( b' \) into the perpendicular condition:

\[ h\left(\frac{h a'}{a}\right) + ab' + a'\left(\frac{b a'}{a}\right) = 0 \]

This simplifies to:

\[ \frac{h^2 a'}{a} + \frac{b a'}{a} + \frac{a' b a'}{a} = 0 \]

### Step 6: Factor out \( a' \)
Factoring out \( a' \):

\[ a'\left(\frac{h^2 + b + b^2}{a}\right) = 0 \]

Since \( a' \neq 0 \), we have:

\[ h^2 + b + b^2 = 0 \]

### Step 7: Relate the coefficients
We can also relate the coefficients of the original equations to the conditions given in the problem:

\[ \frac{ha'b'}{b' - a'} = \frac{h'ab}{b - a} = \frac{1}{2\sqrt{-aa'bb'}} \]

### Conclusion
Thus, we have established the relationships that must hold true based on the conditions of the problem.