Thread: translation

in a translation, how can you use the relationship between some points and their images to find the images of other points?

Originally Posted by Sally_Math

in a translation, how can you use the relationship between some points and their images to find the images of other points?

Hi Sally,

Let's see if I understand your question. When you translate (slide) a preimage, you want to know the coordinates of the image? A specific example would help, but in general:

$\displaystyle \left(\begin{array}{ccc} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{array}\right)+\left(\begin{array}{ccc} x & x & x \\ y & y & y \end{array}\right)=\left(\begin{array}{ccc} x_1+x & x_2+x & x_3+x \\ y_1+y & y_2+y & y_3+y \end{array}\right)$

In the above generalization, the coordinates in the first matrix (preimage) are shifted (without rotation or dilation) left or right by x units and up or down by y units. The coordinates of the resulting image is indicated by the sum. Of course, if x is negative the shift is to the left, and if y is negative, the shift is down.

Let's look at a specific example.

TRANSLATE TRIANGLE ABC WITH A(-3, 4), B(-1, 2), C(-3,2) TO TRIANGLE A'B'C' BY SLIDING RIGHT 4 UNITS AND DOWN 6 UNITS.

$\displaystyle \left(\begin{array}{ccc} -3 & -1 & -3 \\ 4 & 2 & 2 \end{array}\right)+\left(\begin{array}{ccc} 4 & 4 & 4 \\ -6 & -6 & -6 \end{array}\right)=\left(\begin{array}{ccc} 1 & 3 & 1 \\ -2 & -4 & -4 \end{array}\right)$

TRIANGLE A'B'C' HAS COORDINATES A(1, -2), B(3, -4), C(1, -4)

I hope this helped a little.