For what value of k do the pair of linear equations 2x+3y-1=0 and 4x+ky+2=0 have no solution?

Correct option is A

Given:

2x + 3y - 1 = 0 and 4x + ky + 2 = 0 

Concept Used: ​

To determine the value of k for which the given pair of linear equations has no solution, we use the condition for parallel lines. If the lines are parallel, they have no solution, and their coefficients must satisfy the following condition:

​$$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$

Solution:

For given equation:

2x + 3y - 1 = 0

Coefficients: $$a_1 = 2, \, b_1 = 3, \, c_1 = -1$$​​

4x + ky + 2 = 0

Coefficients: $$a_2 = 4, \, b_2 = k, \, c_2 = 2$$​​

Substitute into the condition:

$$\frac{2}{4} = \frac{3}{k} \neq \frac{-1}{2}$$ 

$$\frac{a_1}{a_2} = \frac{b_1}{b_2}$$ = $$\frac{2}{4} = \frac{3}{k}$$ 

$$\frac{1}{2} = \frac{3}{k}$$ (Cross-multiply): $$k = 6$$

Verify the inequality $$\frac{a_1}{a_2} \neq \frac{c_1}{c_2}$$

$$\frac{2}{4} \neq \frac{-1}{2}$$

Simplify:

$$\frac{1}{2} \neq \frac{-1}{2}$$

This condition is satisfied.

The value of $$k$$ is 6.

Thus the correct answer is (A)