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Linear transformation

  1. Feb 25, 2004 #1
    Linear transformation in Maths
    (a) If a triangle ABC with coordiantes A(2,7), B(2,9) and C(6,7) has a rotation and maps to triangle PQR with coordinates P(6,5), Q(8,5) and R(6,1), what is the centre of rotation? I want to ask in general, what's the way to find the answer?

    (b) An enlargement maps the triangle ABC with coordiantes A(2,7), B(2,9) and C(6,7) onto triangle XYZ with coordiantes X(12,12), Y(12,13) and Z(14,12). How to find the centre of enlargement? And this question, the enlargment scale factor is 1/2, right?

    (c) A shear maps triangle ABC with coordiantes A(2,7), B(2,9) and C(6,7) onto triangle LMN with coordinates L(2,10), M(2,12) and N(6,16). How to find the shear factor? And is this transformation first with a reflection and then a shear?

    (d) A triangle ABC wiht coordinates A(1,1), B(0,2) and C(3,1) is reflected in the line y=-x. How to find the matrix which represents the reflection?

    Please help me for these. I need to have examples of this to solve other problems.
  2. jcsd
  3. Feb 26, 2004 #2


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    There may be an easier or more elegant way to do it but a striaght forward method would be (after having initially verified that tri-ABC and tri-PQR are indeed isomorphic with correspondance between A and P, B and Q etc ) to look for a point (x,y) that is equi-distant from A and P and also equi-distant from B and Q and also equi-distant from C and R.

    There is no garantee that such a point (x,y) will exist however as the movement may involve both rotation and translation. In other words, the three "equi-distant" equations above will give you three equations in two unknowns and therefore may be inconsistant. In fact the particular example you give does correspond to an inconsistant set of equations (no solution)
  4. Feb 26, 2004 #3
    Can the solution be found by graphical method, uart? And uart, do you know how to solve other questions?
  5. Feb 26, 2004 #4


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    I imagine that there would be graphical techniques possible, though I dont know of any off hand. This stuff is not really my field. :)

    Anyway the algrebraic approach is simple enough. For example in the first problem the equations of interest are as follows :

    (x-2)^2 + (y-7)^2 = (x-6)^2 + (y-5)^2

    (x-2)^2 + (y-9)^2 = (x-8)^2 + (y-5)^2

    (x-6)^2 + (y-7)^2 = (x-6)^2 + (y-1)^2

    You can verify that these equation have a common solution at x=3, y=4, which is the point of the rotation.

    BTW. I made a transciption error before when I said that the three equations were inconsistant, in this case they actually do have a consistant solution as given above. Note that the point that I previously made about not generally having a guaranted consistant solution is still valid.

    Hey, I cant do all your homework for you. Here's a hint for the second one though : Look for the point of intersection of the three lines joining the corresponding nodes of the original and enlarged triangles. This point is the center of enlargement you seek.
    Last edited: Feb 26, 2004
  6. Feb 27, 2004 #5
    For the second question, I have also tried this way. But those lines joined do not have any intersection.
    And for the first question, how can you get those equations? Aren't those equations linear?
  7. Feb 28, 2004 #6


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    Hey, I said it might not be the most elegant solution, it worked though :). The equations came from simply equating the (squared) distance from the point of rotation (x,y) to each vertice both before and after the rotation.

    I just thought of an alternate method that is both linear and also easily amenable to graphical methods. What if you take the locus of all points that are equi-distant from the two points (vertices before and after rotation). This is just a straight line, in particular the perpendicular bisector of the two points. So there you have just the intersection of three straight lines.

    That's ok, there isn't a guarented solution to this type of question. I haven't done the calculations but if you recheck them and are sure that the lines dont intersect then the enlargement was not done via a projection from a point and that should be your answer.

    BTW, I was assuming that the "center of enlargement" refered to the point of projection, do you have a definition for the terms about which you are enquiring ?

    Also in the third question what is the definition of a "shear". I am assuming that it is a "translation", can you clarify all the definitions.
    Last edited: Feb 28, 2004
  8. Feb 28, 2004 #7
    Thank you, uart! The first two questions and the fourth question, I have solved.
    About the definition of "shear":
    A transformation in which all points along a given line L remain fixed while other points are shifted parallel to L by a distance proportional to their perpendicular distance from L. Shearing a plane figure does not change its area. The shear can also be generalized to three dimensions, in which planes are translated instead of lines.
  9. Feb 28, 2004 #8
    For the shear, you can immediately see that the shear direction is parallel to the y-axis, because the x coordinates do not move. So:
    x' = x
    y' = y + xf

    where f is the shear factor, so plug in the points and solve f, and you get 3/2 for f.
  10. Feb 29, 2004 #9
    What should be the value of x, it should be the distance from the moving point to the invariant line, right?
    If so, I can solve that the shear factor is 3/2.
  11. Feb 29, 2004 #10
    That is correct; (x-0)=x is the distance from the line in this case where the line for the shear is x=0. I'm not sure if the line is always x=0 or y=0 (my linear algebra book seems to think so). In this case, I set the shear equations as y'=y+(x-n)f. (x-n) would be the distance from the invariant line, since we know that the line is parallel to the x axis. Then plugging in two of the points (with different x values) and solving, I found n = 0. I'm not sure if this is the correct way.
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