The problem asks which of the given lines is perpendicular to the line $x + 2y - 1 = 0$. The options are: (A) $2x - y - 1 = 0$ (B) $x + y + 1 = 0$ (C) $2x + 4y - 1 = 0$ (D) $2x + 4y - 2 = 0$

1. Problem Description

The problem asks which of the given lines is perpendicular to the line $x + 2y - 1 = 0$. The options are:

(A) $2x - y - 1 = 0$

(B) $x + y + 1 = 0$

(C) $2x + 4y - 1 = 0$

(D) $2x + 4y - 2 = 0$

2. Solution Steps

Two lines are perpendicular if the product of their slopes is -

1. First, we need to find the slope of the given line $x + 2y - 1 = 0$.

We rewrite the equation in slope-intercept form, $y = mx + b$, where $m$ is the slope.

$2y = -x + 1$

$y = -\frac{1}{2}x + \frac{1}{2}$

The slope of the given line is $m_1 = -\frac{1}{2}$.

For two lines to be perpendicular, the product of their slopes must be -

1. Let the slope of the perpendicular line be $m_2$. Then,

$m_1 \cdot m_2 = -1$

$-\frac{1}{2} \cdot m_2 = -1$

$m_2 = 2$

Now, we find the slopes of the lines in the options:

(A) $2x - y - 1 = 0$

$y = 2x - 1$

$m_A = 2$

(B) $x + y + 1 = 0$

$y = -x - 1$

$m_B = -1$

(C) $2x + 4y - 1 = 0$

$4y = -2x + 1$

$y = -\frac{1}{2}x + \frac{1}{4}$

$m_C = -\frac{1}{2}$

(D) $2x + 4y - 2 = 0$

$4y = -2x + 2$

$y = -\frac{1}{2}x + \frac{1}{2}$

$m_D = -\frac{1}{2}$

Since we are looking for a line with a slope of 2, option (A) is the correct answer.

3. Final Answer

(A) $2x - y - 1 = 0$