Thread: Equivalent transformations by matrices

I have done the first part. I can't think of a way to do the second part other than by making guesses. The answers are y=2x and y=-0.5x.

I have done it now. The second part is done by starting similarly to the first part.

Yes. To do the first part, look at .

So x'= 5x and y'= -4x. x= x'/5 so y'= -4(x'/5)= (-4/5)x'.

Any "line through the origin" is of the form y= ax with slope a. The line at right angles to that is y= (-1/a)x so you want to find a such that


That is, we must have x'= (2+a)x and y'= (-1/a)x'= (2- 2a)x. Dividing one equation by the other so that we eliminate x' on both sides of the equation, . Multiply on both sides by (2+ 2a)a to get rid of the fractions.

which reduces to .

Ouch, I divided the two equations incorrectly. Dividing x'= (2+ a)x by (-1/a)x'= (2- 2a)x gives or . Multiplying both sides by 2- 2a gives -(2- 2a)a= -2a+ 2a^2= 2+ a. which has roots a= 2 and a= -1/2. the lines, y= ax, are y= 2x and y= (-1/2)x as you said.

I used a slightly different method, still observing that multiplied gradients of perpendicular lines equals -1. I think you might have a mistake. You haven't finished with the equations of lines required.

Could you also fix the missing equals sign that is missing from the second row of matrix multiplications? Thanks.