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ARST is dilated with the rule D1,1/3 (x, y), where the
center of dilation is T(3,-2).
The distance between the x-coordinates of R and T is
The distance between the y-coordinates of R and T is
V
from T, so the
R' is
coordinates of R' are

ARST is dilated with the rule D113 x y where the center of dilation is T32 The distance between the xcoordinates of R and T is The distance between the ycoordin class=

Respuesta :

Answer: first one is 3, second is 6 third is 1 unit left, 2units up

Step-by-step explanation:

The transformation of ΔRST by a third, is a contraction, such that the

coordinates of point R from the center of dilation is reduced.

The correct responses are;

  • The distance between the x-coordinate of R and T  is -3
  • The distance between the y-coordinate of R and is 6
  • R' is 1 unit to the left and 2 units up from T, so the coordinates of R' are (2, 0)

Reasons:

The dilation of ΔRST with the rule [tex]D_{T, \ 1/3} \, (x, \, y)[/tex], requires that we find a third

of the rise and run of the vertices from the center of dilation as follows;

  • The coordinates of the vertices of ΔRST are; R(0, 4), S(0, -2), and T(3, -2).

The run;

  • The distance between the x-coordinate of R and T  = 0 - 3 = -3

The rise;

  • The distance between the y-coordinate of R and T  = 4 - (-2) = 6
  • R' is (-3, 6) × [tex]\left(\frac{1}{3} \right)[/tex] = (-1, 2) from T

The coordinates of R' are

  • [tex]D_{T, \ 1/3} \, R(0, \, 4)[/tex] = (3 + (-3) × [tex]\left(\frac{1}{3} \right)[/tex], -2 + 6 × [tex]\left(\frac{1}{3} \right)[/tex]) = (2, 0)

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