The straight lines `x+2y-9=0,3x+5y-5=0` , and `a x+b y-1=0` are concurrent, if the straight line `35 x-22 y+1=0` passes through the point `(a , b)` (b) `(b ,a)` `(-a ,-b)` (d) none of these

Options

A. (a,b)
B. (b,a)
C. (-a,-b)
D. none of these

Explanation

To solve the problem, we need to determine the conditions under which the three lines are concurrent and then find the point through which the line \(35x - 22y + 1 = 0\) passes.

### Step 1: Write down the equations of the lines
The equations of the three lines are:
1. \(L_1: x + 2y - 9 = 0\)
2. \(L_2: 3x + 5y - 5 = 0\)
3. \(L_3: ax + by - 1 = 0\)

### Step 2: Set up the determinant for concurrency
For the three lines to be concurrent, the determinant of the coefficients must be zero. The determinant can be set up as follows:

\[ \begin{vmatrix} 1 & 2 & -9 \\ 3 & 5 & -5 \\ a & b & -1 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant, we have:

\[ = 1 \begin{vmatrix} 5 & -5 \\ b & -1 \end{vmatrix} - 2 \begin{vmatrix} 3 & -5 \\ a & -1 \end{vmatrix} + (-9) \begin{vmatrix} 3 & 5 \\ a & b \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(= 1(5 \cdot -1 - (-5) \cdot b) = -5 + 5b\)
2. \(= 2(3 \cdot -1 - (-5) \cdot a) = -6 + 10a\)
3. \(= -9(3b - 5a)\)

Putting it all together:

\[ -5 + 5b - (-6 + 10a) - 9(3b - 5a) = 0 \]

### Step 4: Simplify the equation
Now, simplifying the equation:

\[ -5 + 5b + 6 - 10a - 27b + 45a = 0 \]

Combine like terms:

\[ (45a - 10a) + (5b - 27b) + (1) = 0 \]

This simplifies to:

\[ 35a - 22b + 1 = 0 \]

### Step 5: Analyze the line \(35x - 22y + 1 = 0\)
We are given that the line \(35x - 22y + 1 = 0\) passes through the point \((a, b)\).

### Step 6: Compare the equations
By comparing the coefficients of \(x\) and \(y\) from the equation \(35x - 22y + 1 = 0\) with \(ax + by - 1 = 0\), we can conclude:

- \(a = 35\)
- \(b = -22\)

### Conclusion
The point through which the line passes is \((35, -22)\). Thus, the correct option is:

**(a) \((a, b)\)**