The family of straight lines
`x(a+b)+y(a-b)=2a` where a and b are parameters, are

Options

A. concurrent at (1,-1)
B. concurrent at (1,1)
C. not concurrent
D. None of these

Answer

To analyze the family of straight lines given by the equation \( x(a+b) + y(a-b) = 2a \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ x(a+b) + y(a-b) = 2a \]
This can be rearranged to express \( y \) in terms of \( x \):
\[ y(a-b) = 2a - x(a+b) \]
\[ y = \frac{2a - x(a+b)}{a-b} \]

### Step 2: Identify the parameters
In this equation, \( a \) and \( b \) are parameters. We need to analyze how the lines behave as we change these parameters.

### Step 3: Find the intersection point
To find the intersection point of these lines, we can set \( a \) and \( b \) to specific values. For instance, if we set \( a = 1 \) and \( b = -1 \):
\[ x(1 + (-1)) + y(1 - (-1)) = 2(1) \]
This simplifies to:
\[ 0 + 2y = 2 \]
\[ y = 1 \]
Thus, when \( a = 1 \) and \( b = -1 \), we find that the intersection point is \( (x, y) = (1, 1) \).

### Step 4: Check for concurrency
To check if all lines are concurrent, we can analyze the coefficients of \( x \) and \( y \) in the original equation. If we can express the lines in a form where they all intersect at a single point, they are concurrent.

### Step 5: Conclusion
Since we found that the lines intersect at the point \( (1, 1) \) for specific values of \( a \) and \( b \), we conclude that the family of straight lines given by the equation \( x(a+b) + y(a-b) = 2a \) are concurrent at the point \( (1, 1) \).

### Summary
The family of straight lines \( x(a+b) + y(a-b) = 2a \) is concurrent at the point \( (1, 1) \).

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