Question

ab has endpoints located at a (0, 0) and b (2, 0). it was dilated at a scale factor of \\(\frac{1}{2}\\) from center (2, 0). which statement describes the pre - image? ab is located at a (-2, 0) and b (2, 0) and is half the length of ab. ab is located at a (-2, 0) and b (2, 0) and is twice the length of ab. ab is located at a (-1, 0) and b (1, 0) and is half the length of ab. ab is located at a (-1, 0) and b (1, 0) and is twice the length of ab.

Explanation:

Step1: Recall dilation formula

The formula for dilation of a point $(x,y)$ from a center $(a,b)$ with scale - factor $k$ is $(x',y')=(a + k(x - a),b + k(y - b))$. Here, $y = y'=0$ and $b = 0$, and the center is $(2,0)$ i.e., $a = 2$ and $k=\frac{1}{2}$.

Step2: Find the pre - image of $A'$

For $A'(0,0)$, let the pre - image be $A(x,0)$. Using the dilation formula $x'=a + k(x - a)$, we substitute $x' = 0$, $a = 2$, and $k=\frac{1}{2}$. So, $0=2+\frac{1}{2}(x - 2)$. First, expand: $0=2+\frac{x}{2}-1$. Then simplify: $0 = 1+\frac{x}{2}$. Solving for $x$ gives $x=-1$.

Step3: Find the pre - image of $B'$

For $B'(2,0)$, let the pre - image be $B(x,0)$. Using the dilation formula $x'=a + k(x - a)$ with $x' = 2$, $a = 2$, and $k=\frac{1}{2}$, we have $2=2+\frac{1}{2}(x - 2)$. Expand: $2=2+\frac{x}{2}-1$. Simplify: $2 = 1+\frac{x}{2}$, and solving for $x$ gives $x = 1$.

Step4: Analyze length relationship

The length of $\overline{A'B'}=2 - 0=2$. The length of $\overline{AB}=1-(-1)=2$. Since the scale factor $k=\frac{1}{2}$, the pre - image $\overline{AB}$ is twice the length of the image $\overline{A'B'}$.

Answer:

D. $\overline{AB}$ is located at $A(-1,0)$ and $B(1,0)$ and is twice the length of $\overline{A'B'}$.