Find the equation of the line passing through the point $(-1, 2)$ and perpendicular to the line $y = \frac{1}{2}x + 3$.

1. Problem Description

Find the equation of the line passing through the point $(-1, 2)$ and perpendicular to the line $y = \frac{1}{2}x + 3$.

2. Solution Steps

The given line is $y = \frac{1}{2}x + 3$. The slope of this line is $m_1 = \frac{1}{2}$.

If two lines are perpendicular, then the product of their slopes is $-1$. Let $m_2$ be the slope of the line we want to find. Then,

$m_1 \cdot m_2 = -1$

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

$m_2 = -2$

The slope of the line we want to find is $-2$. We are also given that the line passes through the point $(-1, 2)$. Using the point-slope form of a line, we have

$y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the given point and $m$ is the slope.

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

$y - 2 = -2(x + 1)$

$y - 2 = -2x - 2$

$y = -2x - 2 + 2$

$y = -2x$

3. Final Answer

The equation of the line is $y = -2x$. The correct answer is (a).