If the lines `2x + 3y = 8, 5x-6y + 7 = 0 and px + py = 1` are concurrent, then the line `x + 2y - 1 = 0` passes through

Options

A. `(1,2)`
B. `(rho,1)`
C. `(q, 1)`
D. none of these

Answer

To determine if the line \( x + 2y - 1 = 0 \) passes through the points where the lines \( 2x + 3y = 8 \), \( 5x - 6y + 7 = 0 \), and \( px + qy = 1 \) are concurrent, we will follow these steps:

### Step 1: Rewrite the equations in standard form
The equations of the lines can be rewritten as follows:
1. \( 2x + 3y - 8 = 0 \)
2. \( 5x - 6y + 7 = 0 \)
3. \( px + qy - 1 = 0 \)

### Step 2: Set up the determinant for concurrency
For the lines to be concurrent, the determinant of the coefficients must be zero. The determinant is given by:

\[ \begin{vmatrix} 2 & 3 & -8 \\ 5 & -6 & 7 \\ p & q & -1 \end{vmatrix} = 0 \]

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

\[ D = 2 \begin{vmatrix} -6 & 7 \\ q & -1 \end{vmatrix} - 3 \begin{vmatrix} 5 & 7 \\ p & -1 \end{vmatrix} - 8 \begin{vmatrix} 5 & -6 \\ p & q \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} -6 & 7 \\ q & -1 \end{vmatrix} = (-6)(-1) - (7)(q) = 6 - 7q \)

2. \( \begin{vmatrix} 5 & 7 \\ p & -1 \end{vmatrix} = (5)(-1) - (7)(p) = -5 - 7p \)

3. \( \begin{vmatrix} 5 & -6 \\ p & q \end{vmatrix} = (5)(q) - (-6)(p) = 5q + 6p \)

Now substituting back into the determinant \( D \):

\[ D = 2(6 - 7q) - 3(-5 - 7p) - 8(5q + 6p) \]

Expanding this gives:

\[ D = 12 - 14q + 15 + 21p - 40q - 48p \]

Combining like terms yields:

\[ D = -27p - 54q + 27 \]

### Step 4: Set the determinant to zero
Setting the determinant equal to zero for concurrency:

\[ -27p - 54q + 27 = 0 \]

Dividing the entire equation by -27 gives:

\[ p + 2q - 1 = 0 \]

### Step 5: Check if the line passes through the point
Now we need to check if the line \( x + 2y - 1 = 0 \) passes through the points \( (p, q) \). We can substitute \( p \) and \( q \) into the line equation:

\[ p + 2q - 1 = 0 \]

Since we derived this equation from the concurrency condition, it confirms that the line \( x + 2y - 1 = 0 \) indeed passes through the point \( (p, q) \).

### Conclusion
Thus, the line \( x + 2y - 1 = 0 \) passes through the point where the lines \( 2x + 3y = 8 \), \( 5x - 6y + 7 = 0 \), and \( px + qy = 1 \) are concurrent.

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