Question

the figure below is dilated by a factor of \\(\frac{1}{2}\\) centered at the origin. plot the resulting image. click twice to plot a segment. click a segment to delete it.

Explanation:

Step1: Identify original coordinates

First, we need to find the coordinates of the original vertices. From the graph:

• \( B(0, 6) \) (since it's on the y - axis, x = 0, y = 6)
• \( F(4, 4) \) (x = 4, y = 4)
• \( E(5, - 4) \) (x = 5, y=-4)
• \( D(-4, - 6) \) (x=-4, y = - 6)
• \( C(-4, 4) \) (x=-4, y = 4)

Step2: Apply dilation formula

The rule for dilation centered at the origin with a scale factor \( k=\frac{1}{2} \) is \( (x,y)\to(kx,ky) \).

• For point \( B(0,6) \):

\( x'=\frac{1}{2}\times0 = 0 \), \( y'=\frac{1}{2}\times6=3 \), so \( B'(0,3) \)

• For point \( F(4,4) \):

\( x'=\frac{1}{2}\times4 = 2 \), \( y'=\frac{1}{2}\times4 = 2 \), so \( F'(2,2) \)

• For point \( E(5,-4) \):

\( x'=\frac{1}{2}\times5=\frac{5}{2}=2.5 \), \( y'=\frac{1}{2}\times(- 4)=-2 \), so \( E'(2.5,-2) \)

• For point \( D(-4,-6) \):

\( x'=\frac{1}{2}\times(-4)=-2 \), \( y'=\frac{1}{2}\times(-6)=-3 \), so \( D'(-2,-3) \)

• For point \( C(-4,4) \):

\( x'=\frac{1}{2}\times(-4)=-2 \), \( y'=\frac{1}{2}\times4 = 2 \), so \( C'(-2,2) \)

Step3: Plot the new points

Plot the points \( B'(0,3) \), \( F'(2,2) \), \( E'(2.5,-2) \), \( D'(-2,-3) \), \( C'(-2,2) \) on the coordinate plane and connect them in the same order as the original figure.

Answer:

The dilated figure has vertices at \( B'(0,3) \), \( F'(2,2) \), \( E'(2.5,-2) \), \( D'(-2,-3) \), \( C'(-2,2) \) (plotted on the coordinate plane as per the dilation rule).