Question

nc.8.g.3

1. triangle abc has vertices a(4, 2), b (2,4), c (6, 6)

a. draw and label the image of jklm with a
center of dilation at (0,0) and a scale factor of
0.5.
a(4,2) a’( , )
b (2,4) b’( , )
c (6,6) c’( , )
explain how you could determine the
coordinates of the dilated image a’b’c
without graphing.

...
tion of triangle abc with center (0,0) and a scale factor of 2. label the new tr
(1 poi

Explanation:

Step1: Recall dilation rule

For a dilation with center at \((0,0)\) and scale factor \(k\), the coordinates \((x,y)\) of a point transform to \((kx,ky)\). Here, \(k = 0.5\).

Step2: Find \(A'\)

For point \(A(4,2)\), multiply \(x\) and \(y\) by \(0.5\):
\(x' = 4\times0.5 = 2\), \(y' = 2\times0.5 = 1\). So \(A'(2,1)\).

Step3: Find \(B'\)

For point \(B(2,4)\), multiply \(x\) and \(y\) by \(0.5\):
\(x' = 2\times0.5 = 1\), \(y' = 4\times0.5 = 2\). So \(B'(1,2)\).

Step4: Find \(C'\)

For point \(C(6,6)\), multiply \(x\) and \(y\) by \(0.5\):
\(x' = 6\times0.5 = 3\), \(y' = 6\times0.5 = 3\). So \(C'(3,3)\).

To determine coordinates without graphing: Use the dilation rule \((x,y)\to(kx,ky)\) with \(k = 0.5\). Multiply each original coordinate’s \(x\) and \(y\) by \(0.5\) to get the dilated coordinates.

Answer:

\(A'(2,1)\), \(B'(1,2)\), \(C'(3,3)\)
(Explanation: Multiply each original \(x,y\) by \(0.5\) for dilation about \((0,0)\).)