What are the coordinates of the image of the point #(–3, 6)# after a dilation with a center of #(0, 0)# and scale factor of #1/3#?

1 Answer
Dec 25, 2017

Multiply the scale factor, #1/3#, into the coordinates #(-3, 6)#, to get the coordinates of the image point, #(-1, 2)#.


The idea of dilation, scaling, or "resizing", is to make something either bigger or smaller, but when doing this to a shape, you would have to somehow "scale" each coordinate.

Another thing is that we're not sure how the object would "move"; when scaling to make something bigger, the area/volume becomes larger, but that would mean the distances between points should become longer, so, which point goes where? A similar question arises when scaling to make things smaller.

An answer to that would be to set a "center of dilation", where all lengths are transformed in a way that makes their new distances from this center proportional to their old distances from this center.

Luckily, the dilation being centered at the origin #(0, 0)# makes this simpler: we simply multiply the scale factor to the #x# and #y#-coordinates to obtain the image point coordinates.

#1/3 * (-3, 6) = (1/3 * -3, 1/3 * 6) = ((-3)/(3), (6)/(3)) = (-1, 2)#

That way, if it gets bigger, it should move away from the origin, and if it gets smaller (as is the case here), it should move closer to the origin.

Fun fact: one way to dilate something if the center is not at the origin, is to somehow subtract the coordinates to make the center at the origin, then add them back later on once the dilation is done. The same can be done for rotation. Clever, right?