Today's lesson is from Ch 9, Ex 15, Q14 David Rayner (http://www.amazon.com/Extended-Maths-Camb-Igcse-3e/dp/0199138745/ref=dp_ob_title_bk)

a) Find and
draw the image of the square O (0, 0), P (1, 1), Q (0, 2), R (-1, 1) under the

Square transformed to Parallelogram (Geogebra 4.0,
www. geogebra.org)

b) Show that the transformation is a shear and find the equation of the invariant line.

Mary says, from
the drawing, it is evident that O’ = O and R’ = R. Hence the invariant line is
y = -x. If you need convincing, the
gradient of the line is (-1 - 0)/(1 - 0)
=
-1 and its y intercept is 0. Next, look
at the determinant: det M = (4)(-2) – (3)(-3) = -8 + 9 = 1. For IGCSE purposes, we can understand the determinant
as an area scaling factor. Since the scale factor is unity, the image parallelogram
must have the same area as the original square. M is not a congruency, since a square is
transformed into to a parallelogram.
There is just one line of invariant points, y = -x. Finally, M preserves areas. M must
be a shear.

Johnny says, but perhaps there is another invariant line? Actually, a little bit more algebra will prove this transformation has only one line of invariant points. For an invariant point, P’ = P by definition. That is, after the transformation,

__the point has not moved__. That is exactly what we mean by invariant!! Note that for any matrix transformation, the point O (0, 0) must always be an invariant point since

This is exactly
the same as the equation that we found previously. Thus we can say safely there is

__only one__line of invariant points.
I summarise the
lesson by reminding the students that, of the transformations we know, both
reflections and shears have one invariant line. However, the matrix
representing the reflection will have det R

_{e}= -1 whereas, as we have already seen, the shear will have det H = 1.
It works. Supposing that something exists, and then
seeing where it leads us is quite characteristic of algebraic geometry. Our motto is “Only a contradiction implies a
contradiction.” That is, if our supposition does not lead us into a
contradiction, it is valid. Equating elements: p =4, q = 3, k = -¾, s = ¼ . As
the first matrix has the form of a shear, the transformation must be a shear.

A parting question: Do you think the author, David Rayner, just played around with different values for a, b, c and d until he found a shear matrix?

A parting question: Do you think the author, David Rayner, just played around with different values for a, b, c and d until he found a shear matrix?