Dilation Examples

Given the preimage ∆ABC, identify the coordinates of and draw the image of ∆A'B'C' under dilation about the origin of ratio 3.

Our scale factor is 3, meaning each vertex in ∆A'B'C' will be three times the distance from the origin as its preimage vertex. Conveniently, we can identify the coordinates by multiplying the preimage coordinates by 3. So, A' will be at (4 × 3, 5 × 3) or (12, 15). B' will be at (3 × 3, 2 × 3) or (9, 6). And C' will be at (6 × 3, 3 × 3) or (18, 9). (This neat little trick only works when the origin is the center of dilation.)

As for drawing the image, start by drawing a ray from the origin through each vertex and extending the rays on for quite a ways. Locate each new vertex along those rays. Since A is over 4, up 5, we know that A' will be over 12, up 15. B' will be over 9, up 6. And C' will be over 18, up 9. Connect the points to make the image.

Given the preimage ∆DEF, draw the image ∆D'E'F' under dilation about F of ratio .

This time our center of dilation is one of the vertices, F, so we know that F' will be in the same place. D' and E' will be half the distance from F as D and E. Mark the midpoints of EF and DF, and they'll be the vertices of the image.

Given the preimage ∆GHI, draw the image ∆G'H'I' under dilation about P of ratio , and classify the image as an enlargement or reduction.

Okay, we've got a sticky little problem here. Our center of dilation is no longer the origin, so we can't just use the trick of multiplying the scale factor by the coordinates. Instead, we'll just identify how far over and up or down we travel to get from the center to each vertex. Then, since our scale factor is , we'll travel half the distances to find the new vertices.

From the center to G, we move 4 to the left and down 2. So to find G', we'll go 2 to the left and down 1. That'll put us at (4, 2). For H, we travel right 2 and down 1, so now just go right 1 and down  to H'. For I, we would go right 3 and up 5, so to get to I' we'll just go right 1.5 and up 2.5.

Oh, oops, we're not done yet. This image is a reduction. (Okay, now we're done.)

What happens when you dilate a line?

Depends on where the line goes. Let's look first at what happens with just any old line.

Quite simply, it's going to be another line that's parallel to k but a different distance from P. To be sure you've got it in the right place, construct a ray that's perpendicular to k and goes through P. Then you can measure the distance from P to k and extend it to find where k' should be. And since k and k' are parallel, you know that k' needs to be perpendicular to ray P as well.

But what happens when the line goes through the center of dilation?

Frankly, we're done. Think about it. We'd basically be extending the line twice its length in both directions from P. But lines go on forever. So l is l'.

So to answer the question, if we dilate a line that passes through the center, it is unchanged, but if we dilate a line that does not pass through the center, it gets transformed to another line that's parallel to the first.

Use the concept of dilation to explain why ∆JKL ~ ∆MNO.

We're given that the two triangles share three pairs of congruent angles. That' enough to prove that they're similar, but we're supposed to talk about dilation in our answer. One of the properties of dilation is that it leaves angles unchanged, so the angles in the image are congruent to the ones in the preimage.

We also can see that the three pairs of corresponding sides are parallel. Another property of dilation is that it "maps" a line to a parallel line, so long as the line doesn't go through the center of dilation. In this case, that happened three times, to each of the sides of the preimage.

Let's go ahead and draw in the center of dilation, just to confirm our thoughts.

So pretty.