1. Let F(x, y) = (3x – 2, 3y – 10). Draw a triangle ABC on graph paper and then draw the image of the triangle A'B'C' under F. Draw the lines (not segments) AA', BB', CC'. Are they concurrent? Where do they intersect? Is F a dilation?

ANSWER: In the figure, the sides of ABC should be parallel to the corresponding sides of A'B'C' and the lines AA', BB', CC' should be concurrent. The figures will be different, depending on the choice of ABC, but the point of concurrency will be the same for each figure. It is the fixed point in the answer to (b).

2. Investigate F algebraically. Find the fixed point of F by solving (u, v) = F(u, v). Use the fixed point to regroup the formula for F so that it is clearly a dilation.

ANSWER: Solve u = 3u –2 and v = 3v – 10 to get u = 1 and v = 5. So (1, 5) is the only fixed point of F.

If we change variables so that (1, 5) is the origin, we rewrite the formula for F as an expression in (x – 1) and (y – 5). So if F(x, y) = (x', y') then (x' – 1) + 1 = 3((x – 1) + 1) – 2 and (y' – 5) + 5 = 3((y – 5) + 5) – 10. This can be rewritten as (x' – 1) = 3(x – 1) and (y' – 5) = 3(y – 5). In vector notation, if (x, y ) = X and P = (1, 5) , then F(X) – P = 3(X – P). This is the vector language formulation of the definition of a dilation, since it says that P, X and F(X) are collinear and PF(X)/PX = 3.

On graph paper, draw a line through (0,0) with nonzero slope. Pick two points on the line and draw the feet of the perpendiculars to the x-axis and y-axis. Show how the proportional relations in the figure lead to the equation of the line in terms of x and y.

ANSWER. Suppose the line is y = mx. (On your graph paper, you will have chosen a specific slope to draw.). Let P = (a, ma) and Q = (b, mb) be two points on the line. Also, let P1 = (a, 0) and P2 = (0, ma) and Q1 = (b, 0) and Q2 = (0, mb). Then if O = (0, 0), the dilation with center O and ratio b/a maps figure OP1PP2 to OQ1QQ2.

Now where does the equation y = mx come from? Let us set aside the equation and start with points O and P = (a, ma). What is the equation of the line OP? Let Z = (x, y) be a point on this line, with Z1 = (x, 0) and Z2 = (0, y). Then since O, P and Z are collinear, there is a dilation with center O that takes P to Z. This takes the line PP1 to the parallel line ZZ1. It also takes the x-axis to itself, since O is on this line. Thus P1, which the intersection of line PP1 with the x-axis must be taken to Z1, which is the intersection of the images of these two lines. The ratio of the dilation that takes P1 to Z1 is x/a. The same argument applied to P2 and Z2 says that the same ratio is y/ma, so y/ma = x/a, or y = mx.

In each case, draw the figures and find the centers of any dilations that take one of the two figures to the other.

ANSWER: See Eye for Similarity handout

1. Two parallel segments of different length.
2. Two circles of different size.
3. A triangle ABC with midpoint triangle A'B'C.
4. The same triangle ABC and triangle AC'B'.
5. A parallelogram and itself.