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graph the image of δbcd after a dilation with a scale factor of \\(\\fr…

Question

graph the image of δbcd after a dilation with a scale factor of \\(\frac{1}{3}\\), centered at the origin.

Explanation:

Step1: Identify coordinates

First, find the coordinates of points \( B \), \( C \), and \( D \) from the graph.

  • \( B \): \( (6, 9) \) (since it's 6 units right on x - axis and 9 units up on y - axis)
  • \( C \): \( (9, 9) \) (9 units right on x - axis and 9 units up on y - axis)
  • \( D \): \( (3, - 6) \) (3 units right on x - axis and 6 units down on y - axis)

Step2: Apply dilation formula

The dilation formula with scale factor \( k=\frac{1}{3} \) centered at the origin is \( (x,y)\to(kx,ky) \).

  • For point \( B(6,9) \):
  • New \( x \) - coordinate: \( \frac{1}{3}\times6 = 2 \)
  • New \( y \) - coordinate: \( \frac{1}{3}\times9 = 3 \)
  • So, \( B'=(2,3) \)
  • For point \( C(9,9) \):
  • New \( x \) - coordinate: \( \frac{1}{3}\times9 = 3 \)
  • New \( y \) - coordinate: \( \frac{1}{3}\times9 = 3 \)
  • So, \( C'=(3,3) \)
  • For point \( D(3, - 6) \):
  • New \( x \) - coordinate: \( \frac{1}{3}\times3 = 1 \)
  • New \( y \) - coordinate: \( \frac{1}{3}\times(-6)=-2 \)
  • So, \( D'=(1, - 2) \)

Step3: Graph the new triangle

Plot the points \( B'(2,3) \), \( C'(3,3) \), and \( D'(1, - 2) \) on the coordinate plane and connect them to form the dilated triangle \( \triangle B'C'D' \).

Answer:

The dilated triangle \( \triangle B'C'D' \) has vertices at \( B'(2,3) \), \( C'(3,3) \), and \( D'(1, - 2) \). To graph it, plot these points and connect them.