Question

triangle efg is dilated by a scale factor of 3 centered at (0, 1) to create triangle efg. which statement is true about the dilation?
eh and eh both pass through the center of dilation.
the slope of ef is the same as the slope of eh.
eg will overlap eg.
eh = eh

Explanation:

Step1: Recall properties of dilation

Dilation is a transformation that enlarges or reduces a figure. When a figure is dilated from a center of dilation, the lines connecting corresponding points (pre - image and image) pass through the center of dilation.

Step2: Analyze each option

• Option 1: \( \overline{EH}\cong\overline{E'H'} \)

Since the scale factor is \(3\), \(E'H' = 3EH\) (by the definition of dilation \(d(A,A')=k\cdot d(A,O)\) where \(k\) is the scale factor and \(O\) is the center of dilation). So \( \overline{EH}\) and \( \overline{E'H'}\) are not congruent.

• Option 2: \(\overline{E'G'}\) will overlap \(\overline{EG}\)

Because of the scale factor \(k = 3\) and the center of dilation \((0,1)\), the lines \(EG\) and \(E'G'\) are parallel (not overlapping) since \(E'G'=3EG\) and they are in the same direction but different lengths.

• Option 3: The slope of \(\overline{EF}\) is the same as the slope of \(\overline{E'H'}\)

Dilation is a similarity transformation. Similarity transformations (including dilation) preserve the slope of lines. Corresponding line segments in a pre - image and its dilated image (with the same center of dilation) are parallel. Parallel lines have the same slope.

• Option 4: \(\overline{EH}\) and \(\overline{E'H'}\) both pass through the center of dilation

The center of dilation is \((0,1)\). Let's assume \(E\) has coordinates \((x_1,y_1)\) and \(H\) has coordinates \((x_2,y_2)\). The line \(EH\) does not pass through \((0,1)\) (from the graph), and \(E'H'\) is a dilated line segment. The lines connecting non - center points of the pre - image and image pass through the center of dilation. But \(EH\) is a segment of the pre - image that does not pass through the center \((0,1)\)

Answer:

The slope of \(\overline{EF}\) is the same as the slope of \(\overline{E'H'}\)