Answer:
The coordinate of C' is (0,7).
Step-by-step explanation:
Relative coordinates of each point of the ABC-Triangle are obtained first:
[tex]A_{rel} = A - O_{dil}[/tex]
[tex]B_{rel} = B - O_{dil}[/tex]
[tex]C_{rel} = C - O_{dil}[/tex]
Where:
[tex]A, B, C[/tex] - Absolute coordinates of the vertices of the ABC-Triangle.
[tex]O_{dil}[/tex] - Coordinates of the center of dilation.
[tex]A_{rel}, B_{rel}, C_{rel}[/tex] - Relative coordinates of the vertices of the ABC-Triangle.
If [tex]O_{dil} = (-4, -9)[/tex], [tex]A = (-4, -6)[/tex], [tex]B = (3, -6)[/tex] and [tex]C = (-2, -1)[/tex], the relative coordinates are now computed:
[tex]A_{rel} = (-4,-6) - (-4,-9)[/tex]
[tex]A_{rel} = (-4 + 4, -6 + 9)[/tex]
[tex]A_{rel} = (0, 3)[/tex]
[tex]B_{rel} = (3, -6) - (-4,-9)[/tex]
[tex]B_{rel} = (3+4, -6 +9)[/tex]
[tex]B_{rel} = (7,3)[/tex]
[tex]C_{rel} = (-2, -1) - (-4,-9)[/tex]
[tex]C_{rel} = (-2+4, -1 +9)[/tex]
[tex]C_{rel} = (2, 8)[/tex]
Each outcome is consequently dilated:
[tex]A'_{rel} = 2\cdot (0,3)[/tex]
[tex]A'_{rel} = (0,6)[/tex]
[tex]B'_{rel} = 2 \cdot (7,3)[/tex]
[tex]B'_{rel} = (14, 6)[/tex]
[tex]C'_{rel} = 2 \cdot (2,8)[/tex]
[tex]C'_{rel} = (4,16)[/tex]
The absolute coordinates of A', B' and C' are, respectively:
[tex]A' = O_{dil} + A'_{rel}[/tex]
[tex]A' = (-4,-9) + (0,6)[/tex]
[tex]A' = (-4+0, -9 + 6)[/tex]
[tex]A' = (-4, 3)[/tex]
[tex]B' = O_{dil} + B'_{rel}[/tex]
[tex]B' = (-4,-9) + (14,6)[/tex]
[tex]B' = (-4+14, -9+6)[/tex]
[tex]B' = (10, -3)[/tex]
[tex]C' = O_{dil} + C'_{rel}[/tex]
[tex]C' = (-4,-9) + (4,16)[/tex]
[tex]C' = (-4 + 4, -9 + 16)[/tex]
[tex]C' = (0, 7)[/tex]
The coordinate of C' is (0,7).
