Answer :
To determine the pre-image of the line segment \(\overline{E'F'}\) with endpoints at \(E'(1,0)\) and \(F'(1,3)\), dilated by a scale factor of \(\frac{1}{2}\) from the origin, we need to reverse the dilation process.
When a figure is dilated by a scale factor, the coordinates of the points on the figure are multiplied by that scale factor. Here, the scale factor is \(\frac{1}{2}\), meaning each coordinate of the endpoints of the pre-image was multiplied by \(\frac{1}{2}\) to get the corresponding coordinates of the image \(E'\) and \(F'\).
To find the coordinates of the pre-image endpoints \(E\) and \(F\), we need to divide the coordinates of \(E'\) and \(F'\) by the scale factor \(\frac{1}{2}\).
Let's perform these calculations:
1. For point \(E'\) with coordinates \((1, 0)\):
[tex]\[ E_x = \frac{1}{\frac{1}{2}} = 1 \times 2 = 2 \][/tex]
[tex]\[ E_y = \frac{0}{\frac{1}{2}} = 0 \times 2 = 0 \][/tex]
So, \(E(2,0)\).
2. For point \(F'\) with coordinates \((1, 3)\):
[tex]\[ F_x = \frac{1}{\frac{1}{2}} = 1 \times 2 = 2 \][/tex]
[tex]\[ F_y = \frac{3}{\frac{1}{2}} = 3 \times 2 = 6 \][/tex]
So, \(F(2,6)\).
Thus, the pre-image \(\overline{EF}\) of the dilated segment \(\overline{E'F'}\) is located at \(E(2,0)\) and \(F(2,6)\). This pre-image \(\overline{EF}\) is twice the size of the image \(\overline{E'F'}\).
Therefore, the correct statement is:
[tex]\[ \overline{EF} \text{ is located at } E(2,0) \text{ and } F(2,6) \text{ and is twice the size of } \overline{E'F'} \][/tex]
When a figure is dilated by a scale factor, the coordinates of the points on the figure are multiplied by that scale factor. Here, the scale factor is \(\frac{1}{2}\), meaning each coordinate of the endpoints of the pre-image was multiplied by \(\frac{1}{2}\) to get the corresponding coordinates of the image \(E'\) and \(F'\).
To find the coordinates of the pre-image endpoints \(E\) and \(F\), we need to divide the coordinates of \(E'\) and \(F'\) by the scale factor \(\frac{1}{2}\).
Let's perform these calculations:
1. For point \(E'\) with coordinates \((1, 0)\):
[tex]\[ E_x = \frac{1}{\frac{1}{2}} = 1 \times 2 = 2 \][/tex]
[tex]\[ E_y = \frac{0}{\frac{1}{2}} = 0 \times 2 = 0 \][/tex]
So, \(E(2,0)\).
2. For point \(F'\) with coordinates \((1, 3)\):
[tex]\[ F_x = \frac{1}{\frac{1}{2}} = 1 \times 2 = 2 \][/tex]
[tex]\[ F_y = \frac{3}{\frac{1}{2}} = 3 \times 2 = 6 \][/tex]
So, \(F(2,6)\).
Thus, the pre-image \(\overline{EF}\) of the dilated segment \(\overline{E'F'}\) is located at \(E(2,0)\) and \(F(2,6)\). This pre-image \(\overline{EF}\) is twice the size of the image \(\overline{E'F'}\).
Therefore, the correct statement is:
[tex]\[ \overline{EF} \text{ is located at } E(2,0) \text{ and } F(2,6) \text{ and is twice the size of } \overline{E'F'} \][/tex]
