Question

1. dilate efg by a scale factor of $\frac{2}{3}$. pre - image image algebraic rule: ____________

Explanation:

Step1: Identify the coordinates of pre - image

Let's assume \(E(3,3)\), \(F(4,9)\), \(G(9,3)\). The algebraic rule for dilation with a scale factor \(k\) centered at the origin is \((x,y)\to(kx,ky)\). Here \(k = \frac{2}{3}\).

Step2: Calculate the coordinates of the image

For point \(E\):
\(x_E'=\frac{2}{3}\times3 = 2\), \(y_E'=\frac{2}{3}\times3=2\), so \(E'(2,2)\)
For point \(F\):
\(x_F'=\frac{2}{3}\times4=\frac{8}{3}\), \(y_F'=\frac{2}{3}\times9 = 6\), so \(F'(\frac{8}{3},6)\)
For point \(G\):
\(x_G'=\frac{2}{3}\times9 = 6\), \(y_G'=\frac{2}{3}\times3=2\), so \(G'(6,2)\)
The algebraic rule is \((x,y)\to(\frac{2}{3}x,\frac{2}{3}y)\)

Answer:

Pre - image: \(E(3,3)\), \(F(4,9)\), \(G(9,3)\)
Image: \(E'(2,2)\), \(F'(\frac{8}{3},6)\), \(G'(6,2)\)
Algebraic rule: \((x,y)\to(\frac{2}{3}x,\frac{2}{3}y)\)