Question

ixl.com/math/grade - 8/dilations - in - the - coordinate - plane
t(-1/2, 1)
u(2, 1)
v( , )
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Explanation:

Step1: Determine the scale - factor

We know that if a point $(x,y)$ is dilated about the origin by a scale - factor $k$, the new point $(x',y')$ is given by $(x',y')=(kx,ky)$. Let's find the scale - factor using the transformation of point $T$. The original point $T$ has coordinates $(- 4,4)$ and the new point $T'$ has coordinates $(-\frac{1}{2},1)$. If we assume the scale - factor is $k$, then for the $x$ - coordinate: $kx=x'$, so $k\times(-4)=-\frac{1}{2}$, and $k=\frac{1}{8}$; for the $y$ - coordinate: $ky = y'$, so $k\times4 = 1$, and $k=\frac{1}{4}$. Let's use another approach. We can find the relationship between the original and new points of two known points to determine the scale - factor. The vector from the origin to $T(-4,4)$ and to $T'(-\frac{1}{2},1)$. The scale - factor $k$ for dilation about the origin can be found by looking at the ratio of the coordinates of the corresponding points. For $T$ and $T'$, if we consider the $x$ - coordinates: $\frac{x_{T'}}{x_T}=\frac{-\frac{1}{2}}{-4}=\frac{1}{8}$, and for the $y$ - coordinates: $\frac{y_{T'}}{y_T}=\frac{1}{4}$. Let's assume the dilation is about the origin. The original point $U$ has coordinates $(8,4)$ and the new point $U'$ has coordinates $(2,1)$. The scale - factor $k$ for the $x$ - coordinate is $\frac{x_{U'}}{x_U}=\frac{2}{8}=\frac{1}{4}$, and for the $y$ - coordinate is $\frac{y_{U'}}{y_U}=\frac{1}{4}$. So the scale - factor $k = \frac{1}{4}$.

Step2: Apply the scale - factor to point $V$

The original point $V$ has coordinates $(-8,-8)$. To find the new point $V'$ after dilation about the origin by a scale - factor $k=\frac{1}{4}$, we use the formula $(x',y')=(kx,ky)$. For the $x$ - coordinate of $V'$: $x_{V'}=k\times x_V=\frac{1}{4}\times(-8)=-2$. For the $y$ - coordinate of $V'$: $y_{V'}=k\times y_V=\frac{1}{4}\times(-8)=-2$.

Answer:

$V'(-2,-2)$