Let `sqrta+sqrtd=sqrtc+sqrtb and ad=bc`, where `a, b, c, in R^(+)`. If the family of lines `(a^(2)x+b^(2)y+c^(2))+d^(2)x=0` passes through a fixed point `(x_(0), y_(0))`, then the value of `(x_(0)+y_(0))` is

Answer

To solve the problem, we need to analyze the given equations and derive the fixed point through which the family of lines passes. Let's break down the solution step by step.

### Step 1: Understand the Given Equations
We are given two equations:
1. \( \sqrt{a} + \sqrt{d} = \sqrt{c} + \sqrt{b} \)
2. \( ad = bc \)

Where \( a, b, c, d \) are positive real numbers.

### Step 2: Analyze the Second Equation
From the second equation \( ad = bc \), we can conclude that both sides are products of positive real numbers, hence \( d \) must also be a positive real number.

### Step 3: Square Both Sides of the First Equation
We square both sides of the first equation:
\[ (\sqrt{a} + \sqrt{d})^2 = (\sqrt{c} + \sqrt{b})^2 \]
Expanding both sides:
\[ a + d + 2\sqrt{ad} = c + b + 2\sqrt{bc} \]

### Step 4: Substitute \( \sqrt{ad} \) and \( \sqrt{bc} \)
Since \( ad = bc \), we have \( \sqrt{ad} = \sqrt{bc} \). Thus, we can substitute:
\[ a + d + 2\sqrt{ad} = c + b + 2\sqrt{ad} \]
This simplifies to:
\[ a + d = c + b \]

### Step 5: Square the Resulting Equation
Now we square the equation \( a + d = c + b \):
\[ (a + d)^2 = (c + b)^2 \]
Expanding both sides:
\[ a^2 + d^2 + 2ad = c^2 + b^2 + 2bc \]

### Step 6: Substitute \( ad = bc \)
Using \( ad = bc \) again, we can substitute:
\[ a^2 + d^2 + 2bc = c^2 + b^2 + 2bc \]
This simplifies to:
\[ a^2 + d^2 = c^2 + b^2 \]

### Step 7: Analyze the Family of Lines
The family of lines is given by:
\[ a^2 x + b^2 y + c^2 + d^2 x = 0 \]
This can be rearranged as:
\[ (a^2 + d^2)x + b^2 y + c^2 = 0 \]

### Step 8: Find the Fixed Point
To find the fixed point \((x_0, y_0)\), we can substitute \( x = -1 \) and \( y = 1 \):
\[ (a^2 + d^2)(-1) + b^2(1) + c^2 = 0 \]
This leads to:
\[ -(a^2 + d^2) + b^2 + c^2 = 0 \]
Thus, we have:
\[ b^2 + c^2 = a^2 + d^2 \]

### Step 9: Conclusion
Since the equation holds true for \( x = -1 \) and \( y = 1 \), we have found that the fixed point is:
\[ (x_0, y_0) = (-1, 1) \]
Now, we need to find \( x_0 + y_0 \):
\[ x_0 + y_0 = -1 + 1 = 0 \]

### Final Answer
The value of \( x_0 + y_0 \) is \( \boxed{0} \).