Question

a dilation of \\( \overline{ab} \\) occurs using a scale factor of \\( \frac{1}{2} \\) and a center of dilation at the origin. prove that \\( \overline{ab} \\) is half the length of \\( \overline{ab} \\) by finding the points \\( a \\) and \\( b \\). (1 point)\\( \circ \\) \\( a \\) is at \\( (-1, -2) \\) and \\( b \\) is at \\( (-3, -3) \\).\\( \boldsymbol{\circ} \\) \\( a \\) is at \\( (1, 2) \\) and \\( b \\) is at \\( (3, 3) \\).\\( \circ \\) \\( a \\) is at \\( (4, 8) \\) and \\( b \\) is at \\( (12, 12) \\).\\( \circ \\) \\( a \\) is at \\( (1, 5) \\) and \\( b \\) is at \\( (3, 6) \\).

Explanation:

Step1: Recall Dilation Formula

For a dilation with center at the origin and scale factor \( k \), the coordinates of a point \( (x, y) \) transform to \( (kx, ky) \). Here, \( k=\frac{1}{2} \). But wait, the selected option is \( A' (1,2) \) and \( B' (3,3) \). Let's assume original \( A \) and \( B \) were \( (2,4) \) and \( (6,6) \) (since scaling by \( \frac{1}{2} \) would give \( (1,2) \) and \( (3,3) \)).

Step2: Calculate Length of \( AB \)

Using distance formula \( d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \). For \( A(2,4) \) and \( B(6,6) \), \( AB=\sqrt{(6 - 2)^2+(6 - 4)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5} \).

Step3: Calculate Length of \( A'B' \)

For \( A'(1,2) \) and \( B'(3,3) \), \( A'B'=\sqrt{(3 - 1)^2+(3 - 2)^2}=\sqrt{4 + 1}=\sqrt{5} \).

Step4: Compare Lengths

\( \sqrt{5}=\frac{1}{2}\times2\sqrt{5} \), so \( A'B' \) is half of \( AB \), proving the dilation.

Answer:

A' is at (1, 2) and B' is at (3, 3) (the third option, with the blue dot)