Q. what does (y = -x) mean in reflections

Understanding what the line \(y = -x\) means for reflections

We explain step by step what it means to reflect a point across the line \(y = -x\), how to compute the reflected point, why the formula is correct, and give the matrix form. Each step is explained separately and in detail.

1. Describe the mirror line:

   The line \(y = -x\) passes through the origin and has slope \(-1\). Its direction vector is \((1, -1)\). Geometrically it is a \(45^\circ\) line slanting down from left to right. When we reflect a point across this line, the line acts like a mirror: the original point and its reflected image are symmetric with respect to this line.

2. Set up the unknown reflected point:

   Let the original point be \(P = (x, y)\). Let its reflection across the line \(y = -x\) be \(P’ = (x’, y’)\). Reflection means two things:

   • The midpoint of the segment joining \(P\) and \(P’\) lies on the mirror line.
   • The segment joining \(P\) and \(P’\) is perpendicular to the mirror line.
3. Write the midpoint condition:

   The midpoint \(M\) of \(P\) and \(P’\) is

   \[ M = \left(\dfrac{x + x’}{2}, \dfrac{y + y’}{2}\right). \]

   Because \(M\) lies on the line \(y = -x\), its coordinates satisfy

   \[ \dfrac{y + y’}{2} = -\dfrac{x + x’}{2}. \]

   Multiply by 2 to obtain the linear relation

   \[ y + y’ = -x – x’. \]

   Rearrange to a convenient form:

   \[ x’ + y’ = -x – y. \]

4. Write the perpendicularity condition:

   The vector from \(P\) to \(P’\) is

   \[ \overrightarrow{PP’} = (x’ – x, \, y’ – y). \]

   The mirror line has direction vector \((1, -1)\). For \(\overrightarrow{PP’}\) to be perpendicular to the mirror line, their dot product must be zero:

   \[ (x’ – x)\cdot 1 + (y’ – y)\cdot(-1) = 0. \]

   Simplify this to:

   \[ x’ – x – y’ + y = 0. \]

   Rearrange to get

   \[ x’ – y’ = x – y. \]

5. Solve the linear system for \(x’\) and \(y’\):

   We now have the two linear equations:

   \[ x’ + y’ = -x – y \]
   \[ x’ – y’ = x – y \]

   Add the two equations to eliminate \(y’\):

   \[ 2x’ = (-x – y) + (x – y) = -2y. \]

   Therefore

   \[ x’ = -y. \]

   Subtract the second equation from the first to eliminate \(x’\):

   \[ 2y’ = (-x – y) – (x – y) = -2x. \]

   Therefore

   \[ y’ = -x. \]

6. State the reflection formula and interpret it:

   The reflection across the line \(y = -x\) sends a point \((x, y)\) to the point

   \[ (x, y) \text{ reflected across } y=-x \text{ equals } (-y,-x). \]

   In words: swap the coordinates and negate each one. Geometrically, the reflected point lies on the line perpendicular to \(y = -x\) through the original point, and the mirror line is the midpoint location between the point and its image.

7. Matrix form of the reflection:

   The linear transformation that performs reflection across the line \(y = -x\) is given by the matrix

   \[
   \begin{pmatrix}
   0 & -1 \\
   -1 & 0
   \end{pmatrix}.
   \]

   Applying this matrix to the column vector \(\begin{pmatrix} x \\ y \end{pmatrix}\) yields \(\begin{pmatrix} -y \\ -x \end{pmatrix}\), consistent with the formula above.

8. Examples:

   Reflect \((3, 1)\) across \(y = -x\): the image is \((-1, -3)\).

   Reflect \((0, 2)\) across \(y = -x\): the image is \((-2, 0)\).

Summary: Reflecting across the line \(y = -x\) takes any point \((x, y)\) to \((-y, -x)\). This can be derived by enforcing that the midpoint of the point and its image lies on \(y = -x\) and that the segment joining them is perpendicular to \(y = -x\). The matrix of reflection is \(\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\).