Answer:
The image of the point (1, -2) under a dilation of 3 is (3, -6).
Step-by-step explanation:
Correct statement is:
What are the coordinates of the image of the point (1, -2) under a dilation of 3 with the origin.
From Linear Algebra we get that dilation of a point with respect to another point is represented by:
[tex]\vec P' = \vec R + r\cdot (\vec P-\vec R)[/tex] (Eq. 1)
Where:
[tex]\vec R[/tex] - Reference point with respect to origin, dimensionless.
[tex]\vec P[/tex] - Original point with respect to origin, dimensionless.
[tex]r[/tex] - Dilation factor, dimensionless.
If we know that [tex]\vec R = (0,0)[/tex], [tex]\vec P = (1, -2)[/tex] and [tex]r = 3[/tex], then the coordinates of the image of the original point is:
[tex]\vec P' = (0,0) +3\cdot [(1,-2)-(0,0)][/tex]
[tex]\vec P' = (0,0) + 3\cdot (1,-2)[/tex]
[tex]\vec P' = (3,-6)[/tex]
The image of the point (1, -2) under a dilation of 3 is (3, -6).
