Recently I had a very interesting discussion with one of my grade 9 students around the idea of why co-ordinates switch positions when reflecting over the line y = x . This section is covered in GR9 Maths under Transformations. However a lot of the time students tend to memorise the formula with very little curiosity as to why it actually works. What follows in the rest of this blog post is bits from our discussion, together with artifacts of the learning process.

My drawing to understand why the rule actually works (Geogebra)

Consider object A, when it gets reflected over the line y = x , it follows a path perpendicular to that line and reappears on the opposite side as A’ the image. It is a reflection therefore AE = A’E.

Now consider triangles AEB & A’EB, they must be congruent due to (SAS)

1) BE is common

2) Angle BEA = Angle BEA’ = 90 degrees

3) AE = A’E

Now consider Triangle ABC & A’BD, they must also be congruent due to (AACS)

1) Length g = Length h – congruency above

2) Angle ABC = Angle A’BD – sum of angles

3) Angle ACB = Angle A’BD = 90 degrees – construction

So why do co-ordinates switch when going from getting the reflection over the line y = x

eg A(5;1) to A’ (1;5)

Take ABC and imagine trying to place it on top of congruent triangle A’BD. Can you see BC now lines up with BD & AC with A’D. When that happens, BC which used to be on the x axis goes to the y axis and AC which used to be parallel to the y axis is now parallel to the x axis.

The effect of this is that x becomes the y and the y becomes the x, hence the rule is:

**(x;y) -> (y;x)**

I will have a crack at a formula for the more general concept of a formula for reflecting over any arbitary line in a later post , but I have a feeling it may have some fairly interesting mathematics in it … (:

If you enjoyed the post or just want to clarify anything you read, please feel to leave a comment.

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