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Chapter 3: Problem 7

Find a formula for reflection about the vertical line \(x=k\).

Short Answer

Expert verified
The formula for reflection about the vertical line \(x=k\) is \((2k - x, y)\).

Step by step solution

01

- Understand the problem

We need to find a formula that reflects a point across a vertical line given by the equation \(x = k\). The point \((x, y)\) will be transformed into a new point \((x', y')\) such that the distance from \(x\) to \(k\) is the same as the distance from \(x'\) to \(k\).
02

- Formula for horizontal reflection

The reflection principle states that for any point \((x, y)\) and its reflection \((x', y')\) across \(x = k\), the horizontal distance should be preserved but mirrored. Thus, the formula for the new x-coordinate should satisfy: \[ |x - k| = |x' - k| \]
03

- Determine the new x-coordinate

To find the new x-coordinate, solve the equation \[ x' = 2k - x \] This ensures that the point \((x, y)\) is reflected over the line \(x = k\).
04

- New coordinates

Since the reflection is about a vertical line, the y-coordinate does not change. Therefore, the new coordinates after reflection are: \[ (x', y') = (2k - x, y) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertical Line Reflection
Reflecting a point across a vertical line, such as the line given by the equation \( x = k \), involves transforming the point \( (x, y) \) to a new point \( (x', y') \). The key idea is that only the horizontal position (the x-coordinate) of the point changes, while the vertical position (the y-coordinate) remains the same. This is because the vertical line acts as a mirror, flipping the point horizontally but not vertically. This concept is very useful in solving geometry problems involving reflections in a coordinate plane.
Coordinate Transformation
In geometry, reflecting a point across the vertical line involves a coordinate transformation. To transform the point \( (x, y) \), we move it to a new position \( (x', y') \) such that it maintains a specific relationship with the line \( x = k \). We'll set the new x-coordinate to be compared to the original x-coordinate, adjusting it to across the vertical line \( x = k \). So, the new point is calculated using the formula: \[ x' = 2k - x \]. Note that the y-coordinate does not change when reflected, so \[ (x', y') = (2k - x, y) \]. This straightforward transformation ensures that the point's distance to the line \( x = k \) remains preserved but mirrored.
Distance Preservation
One fundamental aspect of reflecting a point in geometry is preserving the distance in the reflection process. When a point \( (x, y) \) is reflected across the vertical line \( x = k \), the critical idea is to maintain equal distance from both sides of this line. This means the original and reflected points must be equally distant from \( k \). To achieve this, we use the absolute value concept: \[ |x - k| = |x' - k| \]. By solving for \ x' \ in this equation, we retain the same distance but in the mirrored position horizontally across the line, ensuring the geometry's distance-preserving properties are kept intact.
Reflection Formula
The reflection formula for a point \ (x, y) \ over a vertical line \ x = k \ is vital in geometry. This formula helps us calculate the exact position of the reflected point by flipping the x-coordinate while keeping the y-coordinate constant. The formula used is: \[ (x', y') = (2k - x, y) \]. Here:
  • \ x' \ is calculated as \ 2k - x \ to ensure the correct reflection across the vertical line.
  • The value of \ y \ remains unchanged since vertical positions are not affected by reflection along a vertical axis.
Using this formula, we can determine the reflected coordinates precisely, which is pivotal in solving complex problems involving reflections in the coordinate system.

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Most popular questions from this chapter

Determine the inverse stereographic projection function \(\phi^{-1}: \mathbb{C}^{+} \rightarrow \mathbb{S}^{2} .\) In particular, show that for \(z=x+y i \neq \infty\), $$ \phi^{-1}(x, y)=\left(\frac{2 x}{x^{2}+y^{2}+1}, \frac{2 y}{x^{2}+y^{2}+1}, \frac{x^{2}+y^{2}-1}{x^{2}+y^{2}+1}\right). $$

Prove that the cross ratio of four distinct complex numbers is a real number if and only if the four points lie on the same cline.

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