UpStudy Free Solution:

To solve the problem, we need to use the properties of enlargement with a given center and scale factor.

Part (a): Finding the coordinates of vertex \(C' \)

The center of enlargement is \(( 2, 9) \) and the scale factor is \(\frac { 1} { 2} \).

The original coordinates of \(C\) are \(( 6, 1) \).

To find \(C' \), we use the formula for enlarging a point:

\[C' = \left ( x_ c' , y_ c' \right ) \]

where:

\[x_ c' = x_ c \cdot k + ( 1 - k) \cdot x_ e\]

\[y_ c' = y_ c \cdot k + ( 1 - k) \cdot y_ e\]

Here, \(k\) is the scale factor, \(( x_ e, y_ e) \) is the center of enlargement, and \(( x_ c, y_ c) \) are the coordinates of the original vertex.

Substituting the values:

\[x_ c' = 6 \cdot \frac { 1} { 2} + ( 1 - \frac { 1} { 2} ) \cdot 2\]

\[y_ c' = 1 \cdot \frac { 1} { 2} + ( 1 - \frac { 1} { 2} ) \cdot 9\]

Calculating:

\[x_ c' = 3 + 1 = 4\]

\[y_ c' = 0.5 + 4.5 = 5\]

So, the coordinates of \(C' \) are \(( 4, 5) \).

Part (b): Finding the length of the side \(A' B' \)

The original coordinates of \(A\) are \(( 4, 8) \) and \(B\) are \(( 8, 9) \).

First, calculate the length of \(AB\):

\[AB = \sqrt { ( x_ B - x_ A) ^ 2 + ( y_ B - y_ A) ^ 2} \]

\[AB = \sqrt { ( 8 - 4) ^ 2 + ( 9 - 8) ^ 2} \]

\[AB = \sqrt { 4^ 2 + 1^ 2} \]

\[AB = \sqrt { 16 + 1} \]

\[AB = \sqrt { 17} \]

Since the scale factor is \(\frac { 1} { 2} \), the length of \(A' B' \) will be half the length of \(AB\):

\[A' B' = \frac { 1} { 2} \times \sqrt { 17} \]

\[A' B' = \frac { \sqrt { 17} } { 2} \]

So, the length of the side \(A' B' \) is \(\frac { \sqrt { 17} } { 2} \).

Supplemental Knowledge

In geometry, transformations such as enlargements or reductions involve changing the size of a shape while maintaining its proportions. When a shape is enlarged or reduced by a scale factor from a specific center, each point on the shape moves along the line connecting it to the center by a distance proportional to the scale factor.

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