Question

in the diagram, the circle will be dilated by a scale factor of 3 about the origin. the points c, a, and b map to c, a, and b after the dilation. what is the length of overline{cb}? use the distance formula to help you decide: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. a = (8, 15), b = (12, 13), c = (8, 10) \
a. 5 units \
b. 45 units \
c. 21 units \
d. 24 units \
e. 15 units

Explanation:

Step1: Find coordinates of \( C' \) and \( B' \)

Dilation about origin with scale factor 3: \((x,y)\to(3x,3y)\).
For \( C=(8,10) \), \( C'=(3\times8, 3\times10)=(24,30) \).
For \( B=(12,13) \), \( B'=(3\times12, 3\times13)=(36,39) \).

Step2: Apply distance formula

Distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
Substitute \( (x_1,y_1)=(24,30) \), \( (x_2,y_2)=(36,39) \):
\( d = \sqrt{(36 - 24)^2 + (39 - 30)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 \)? Wait, no—wait, first find \( CB \) length, then scale.

Wait, alternative: Dilation preserves shape, so \( C'B' = 3 \times CB \).
Find \( CB \): \( C=(8,10) \), \( B=(12,13) \).
\( CB = \sqrt{(12 - 8)^2 + (13 - 10)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \).
Thus, \( C'B' = 3 \times 5 = 15 \)? Wait, but option E is 15. Wait, did I miscalculate earlier?

Wait, recheck \( CB \): \( x \)-difference: \( 12 - 8 = 4 \), \( y \)-difference: \( 13 - 10 = 3 \). \( \sqrt{4^2 + 3^2} = 5 \). Dilation scale 3: \( 5 \times 3 = 15 \). So length of \( \overline{C'B'} \) is 15.

Answer:

E. 15 units