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The lines 2x + y -1=0, ax+ 3y-3=0 and 3x...

The lines `2x + y -1=0`, `ax+ 3y-3=0` and `3x + 2y -2=0` are concurrent for -

A

all a

B

a = 4 only

C

`- 1 le a le 3`

D

`a gt 0` only

Text Solution

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The correct Answer is:
To determine the value of \( a \) for which the lines \( 2x + y - 1 = 0 \), \( ax + 3y - 3 = 0 \), and \( 3x + 2y - 2 = 0 \) are concurrent, we need to set up the condition for concurrency. The lines are concurrent if the determinant formed by their coefficients is equal to zero. ### Step-by-Step Solution: 1. **Write the equations in standard form**: - The first line is \( 2x + y - 1 = 0 \). - The second line is \( ax + 3y - 3 = 0 \). - The third line is \( 3x + 2y - 2 = 0 \). 2. **Identify coefficients**: - From the first line: \( A_1 = 2, B_1 = 1, C_1 = -1 \) - From the second line: \( A_2 = a, B_2 = 3, C_2 = -3 \) - From the third line: \( A_3 = 3, B_3 = 2, C_3 = -2 \) 3. **Set up the determinant**: The determinant for the coefficients is given by: \[ \begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} = 0 \] Substituting the values: \[ \begin{vmatrix} 2 & 1 & -1 \\ a & 3 & -3 \\ 3 & 2 & -2 \end{vmatrix} = 0 \] 4. **Calculate the determinant**: Expanding the determinant: \[ = 2 \begin{vmatrix} 3 & -3 \\ 2 & -2 \end{vmatrix} - 1 \begin{vmatrix} a & -3 \\ 3 & -2 \end{vmatrix} - 1 \begin{vmatrix} a & 3 \\ 3 & 2 \end{vmatrix} \] Calculate each of the 2x2 determinants: - First determinant: \[ \begin{vmatrix} 3 & -3 \\ 2 & -2 \end{vmatrix} = (3)(-2) - (-3)(2) = -6 + 6 = 0 \] - Second determinant: \[ \begin{vmatrix} a & -3 \\ 3 & -2 \end{vmatrix} = (a)(-2) - (-3)(3) = -2a + 9 \] - Third determinant: \[ \begin{vmatrix} a & 3 \\ 3 & 2 \end{vmatrix} = (a)(2) - (3)(3) = 2a - 9 \] 5. **Substituting back into the determinant**: Now substituting back: \[ 2(0) - 1(-2a + 9) - 1(2a - 9) = 0 \] Simplifying: \[ 0 + 2a - 9 - 2a + 9 = 0 \] Which simplifies to: \[ 0 = 0 \] 6. **Conclusion**: Since the determinant equals zero for all values of \( a \), the lines are concurrent for all values of \( a \). ### Final Answer: The lines are concurrent for all values of \( a \).
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