Understanding the Problem: Finding the Line Mirror Equation

The question asks us to find the equation of a line that acts as a mirror, reflecting a given point to another point. We are given the original point, let's call it \(A\), and its image after reflection, let's call it \(A'\). The key properties of a line mirror in relation to a point and its image are:

1. The line mirror is the perpendicular bisector of the line segment connecting the point and its image.
2. This means the midpoint of the segment \(AA'\) lies on the line mirror.
3. The line mirror is perpendicular to the segment \(AA'\).

We are given the point \(A(-4, 2)\) and its image \(A'(4, -2)\). We will use the properties mentioned above to find the equation of the line mirror.

Step 1: Finding the Midpoint of the Segment AA'

The line mirror passes through the midpoint of the segment connecting the point \(A(-4, 2)\) and its image \(A'(4, -2)\). Let \(M\) be the midpoint. We use the midpoint formula:

\( M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \)

Here, \((x_1, y_1) = (-4, 2)\) and \((x_2, y_2) = (4, -2)\).

Substituting these values into the formula:

\( M = \left(\frac{-4 + 4}{2}, \frac{2 + (-2)}{2}\right) \) \( M = \left(\frac{0}{2}, \frac{0}{2}\right) \) \( M = (0, 0) \)

So, the midpoint of the segment \(AA'\) is \((0, 0)\). This tells us that the line mirror passes through the origin.

Step 2: Finding the Slope of the Segment AA'

The line mirror is perpendicular to the segment \(AA'\). First, let's find the slope of the segment \(AA'\). We use the slope formula:

\( m_{AA'} = \frac{y_2 - y_1}{x_2 - x_1} \)

Using the points \(A(-4, 2)\) and \(A'(4, -2)\):

\( m_{AA'} = \frac{-2 - 2}{4 - (-4)} \) \( m_{AA'} = \frac{-4}{4 + 4} \) \( m_{AA'} = \frac{-4}{8} \) \( m_{AA'} = -\frac{1}{2} \)

The slope of the segment \(AA'\) is \(-\frac{1}{2}\).

Step 3: Finding the Slope of the Line Mirror

Since the line mirror is perpendicular to the segment \(AA'\), the product of their slopes is -1. If the slope of the line mirror is \(m_{mirror}\), then:

\( m_{mirror} \times m_{AA'} = -1 \) \( m_{mirror} \times \left(-\frac{1}{2}\right) = -1 \)

To find \(m_{mirror}\), we can multiply both sides by -2:

\( m_{mirror} = (-1) \times (-2) \) \( m_{mirror} = 2 \)

The slope of the line mirror is 2.

Step 4: Finding the Equation of the Line Mirror

We know the line mirror passes through the point \((0, 0)\) (the midpoint) and has a slope of 2. We can use the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\):

Here, \((x_1, y_1) = (0, 0)\) and \(m = 2\).

\( y - 0 = 2(x - 0) \) \( y = 2x \)

This is the equation of the line mirror.

Comparing with Options

Let's compare our derived equation with the given options:

• Option 1: \(y = x\)
• Option 2: \(y = 2x\)
• Option 3: \(4y = x\) (which is \(y = \frac{1}{4}x\))
• Option 4: \(y = 4x\)

Our derived equation, \(y = 2x\), matches Option 2.

Conclusion: Equation of the Line Mirror

Based on our calculations using the properties of reflection, the equation of the line mirror that transforms the point \((-4, 2)\) into the point \((4, -2)\) is \(y = 2x\).

Revision Table: Key Concepts for Line Mirror Equation

Additional Information: Reflection in Coordinate Geometry

Reflection is a fundamental transformation in geometry. When a point is reflected across a line, the line acts like a mirror. Every point on the original object (or point, in this case) is mapped to a corresponding point on the image. The distance from the original point to the line mirror is equal to the distance from the image point to the line mirror.

In coordinate geometry, finding the equation of the line mirror between a point \(A\) and its image \(A'\) always involves these two steps:

• Finding the midpoint \(M\) of \(AA'\). Since \(M\) is equidistant from \(A\) and \(A'\), and lies on the segment \(AA'\), it must lie on the line mirror.
• Finding the slope of \(AA'\). The line mirror is perpendicular to \(AA'\), so its slope is the negative reciprocal of the slope of \(AA'\).

Once you have the midpoint (a point on the line) and the slope, you can easily write the equation of the line using standard forms like point-slope form or slope-intercept form.

If the line mirror were a vertical line (e.g., \(x=c\)), the image of \((x_1, y_1)\) would be \((2c-x_1, y_1)\). If the line mirror were a horizontal line (e.g., \(y=c\)), the image would be \((x_1, 2c-y_1)\). For a general line \(y=mx+c\) (where \(m \neq 0\)), the calculation is more complex but still relies on the perpendicular bisector property. In this specific problem, the line mirror passes through the origin, simplifying the final equation form.