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Help with Isometry

  • 15-03-2009 10:26AM
    #1
    Registered Users, Registered Users 2 Posts: 1,137 ✭✭✭


    Hi folks, I know I must be missing something obvious here, but help appreciated...

    A triangle has at A(7,5), B(5,1) and C(-4,2). It has to be moved so that C is at the origin and CB lies on the positive X axis.

    First you have to do the translation that moves C to the origin, write down the formal definition and find the images of A and B under the translation. I've done that bit.

    Next bit I'm stumped on - Let r theta be the rotation that completes the required transformation, where theta lies in the interval (-pi, pi]. Find the exact values of tan theta, sin theta and cos theta and hence write down a formal definition of r theta. (There is no need to work out the value of the angle theta)

    Apologies if I am being really stupid....but I have no idea how to do that and have been filling sheets of paper for hours.


Comments

  • Closed Accounts Posts: 6,151 ✭✭✭Thomas_S_Hunterson


    I think this is about forming and solving a couple of linear equations, keeping in mind that in 2 dimensions, a rotation matrix will look like this:

    fb2fd35e006acf8996cd07e45de27310.png

    /edit: more info here http://en.wikipedia.org/wiki/Rotation_matrix


  • Registered Users, Registered Users 2 Posts: 1,137 ✭✭✭Monkey61


    Ah but we haven't encountered matrices yet....yikes.


  • Closed Accounts Posts: 6,151 ✭✭✭Thomas_S_Hunterson


    Well if you take it back to simple coordinate geometry and trig, you can find the tan of the angle between the translated bc (b'c' say) and the x axis simply as the slope. The tan of the angle of the rotation will be the negation of this I think.

    Now if you look at the right angled triangle formed by b'c' and the x-axis you calculate the length of the 3 sides and can get sin(-theta) and cos(-theta) easily enough from that (it's minus theta as you want your rotation to go the opposite way)


  • Registered Users, Registered Users 2 Posts: 1,137 ✭✭✭Monkey61


    Lovely, thanks a million


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