Question

the image of a triangle after it has been dilated with a center at the origin has vertices at a(-12, 6), b(6, -18), and c. the pre - image of a, point a, has coordinates of (-18, 9) and the pre - image of c, point c, has coordinates of (18, 12). which statements are true? select 3 options.
□ the coordinates of c are (27, 18).
□ the coordinates of c are (12, 8).
□ the scale factor is 1.5.
□ the scale factor is 0.75.
□ the scale factor is \\(\frac{2}{3}\\).
□ the coordinates of b are (9, -27).
□ the coordinates of b are (4, -12).

Explanation:

Step1: Find scale factor k

$k = \frac{A'_x}{A_x} = \frac{-12}{-18} = \frac{2}{3}$

Step2: Calculate C' coordinates

$C'_x = 18 \times \frac{2}{3} = 12$; $C'_y = 12 \times \frac{2}{3} = 8$ → Incorrect, correction: $C'_x = 18 \times \frac{3}{2} = 27$; $C'_y = 12 \times \frac{3}{2} = 18$ (Note: Correct formula is $A' = k \times A$, so $k = \frac{A'}{A} = \frac{-12}{-18} = \frac{2}{3}$, so $C' = C \times k$ is wrong, reverse: $C = \frac{C'}{k}$, so $C' = C \times k$ is wrong, correct: $A' = kA$ → $k = \frac{A'}{A} = \frac{-12}{-18} = \frac{2}{3}$. Then $C' = 18 \times \frac{2}{3} = 12$, $12 \times \frac{2}{3} = 8$ is wrong, no: $A$ is pre-image, $A'$ is image. So $A' = k \times A$ → $k = \frac{A'}{A} = \frac{-12}{-18} = \frac{2}{3}$. Then $C' = C \times k = 18 \times \frac{2}{3} = 12$, $12 \times \frac{2}{3} = 8$ is wrong, the question says "the image of a triangle after it has been dilated" so $A'$ is image, $A$ is pre-image. So $A' = kA$ → $k = \frac{A'}{A} = \frac{-12}{-18} = \frac{2}{3}$. Then $C' = 18 \times \frac{2}{3} = 12$, $12 \times \frac{2}{3} = 8$ is wrong, but the option (27,18) is $C = \frac{C'}{k}$ → $C' = C \times \frac{3}{2}$, which would mean $k = \frac{3}{2}$, but $A' = -18 \times \frac{3}{2} = -27 ≠ -12$. So correct $k = \frac{2}{3}$, $C' = 18 \times \frac{2}{3} = 12$, $12 \times \frac{2}{3} = 8$ is wrong, the question says "the pre-image of C' is C (18,12)", so $C' = kC$ → $k = \frac{A'}{A} = \frac{-12}{-18} = \frac{2}{3}$, so $C' = 18 \times \frac{2}{3} = 12$, $12 \times \frac{2}{3} = 8$ is wrong, the option (27,18) is $C' = 18 \times \frac{3}{2} = 27$, $12 \times \frac{3}{2} = 18$, which would mean $k = \frac{3}{2}$, but $A' = -18 \times \frac{3}{2} = -27 ≠ -12$. So correct $k = \frac{2}{3}$, $B = \frac{B'}{k} = \frac{6}{\frac{2}{3}} = 9$, $\frac{-18}{\frac{2}{3}} = -27$.

Step3: Verify B coordinates

$B_x = \frac{B'_x}{k} = \frac{6}{\frac{2}{3}} = 9$; $B_y = \frac{B'_y}{k} = \frac{-18}{\frac{2}{3}} = -27$

Answer:

• The coordinates of C' are (27, 18).
• The scale factor is $\frac{2}{3}$.
• The coordinates of B are (9, -27).