SRL Archive - Mathematics

Any ideas on the Part 4 the shear question?

My method:
Since the invariant line is Y=X, You find the the Shortest Point and the line Y=X
Call this Point : P
For simplicity purposes i used the point (2,0)
the perpendicualar gradient of Y=X is -X, henceforth the co-incidental point of the line Y=-X+2 and Y=X is (1,1)
Using Pythagoras' Therom the distance between (1,1) and P is Root 2

Then you apply the transformation matrix to the point P to get P' = (6,4)
Via:
(3 -2)(2) = (6)
(2 -1)(0) = (4)
Then you find the distance between P and P'
giving the distance of Root 32 –> Simplified to 4 Root 2

Then you divide the distance of P & P' by the distance of P and (1,1)

Giving: 4 = (4 root 2)/(root 2)

so the shear factor is 4


Any comments, not sure if this is right?