Question

\\( \overline { a b } \\) is 15 units long. \\( \overline { a ^ { \prime } b ^ { \prime } } \\) is the image of \\( \overline { a b } \\) after a dilation centered at the origin.

what is the scale factor of the dilation?

Explanation:

Step1: Calculate the length of \(A'B'\)

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(A'(1,1)\) and \(B'(-1,-3)\),

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Step2: Calculate the scale factor

Let the scale factor be \(k\). We know that \(A'B'=k\times AB\). Given \(AB = 15\), so \(k=\frac{A'B'}{AB}\).

$$ k=\frac{2\sqrt{5}}{15} $$

Wait, no, wait another approach. Since dilation is a similarity transformation. If we assume the vectors or the ratio of corresponding segments. Let's count the number of units moved in \(x\) and \(y\) - directions for \(A\) and \(A'\).
For point \(A(5,3)\) and \(A'(1,1)\). The transformation from \(A\) to \(A'\) in \(x\) - direction: \(x\) - coordinate of \(A'\) is \(\frac{1}{5}\) of \(x\) - coordinate of \(A\) (since \(1=\frac{1}{5}\times5\)), in \(y\) - direction: \(y\) - coordinate of \(A'\) is \(\frac{1}{3}\) of \(y\) - coordinate of \(A\) (since \(1=\frac{1}{3}\times3\)). But using the property of dilation \(k=\frac{\text{length of image segment}}{\text{length of pre - image segment}}\)

Let's use another method. If we assume the formula for dilation \((x,y)\to(kx,ky)\).
Let’s find the distance from the origin to \(A\): \(OA=\sqrt{5^{2}+3^{2}}=\sqrt{25 + 9}=\sqrt{34}\)
Distance from the origin to \(A'\): \(OA'=\sqrt{1^{2}+1^{2}}=\sqrt{2}\)
No, better use the ratio of the segments.
Since dilation is a linear transformation. If we consider the fact that for a dilation centered at the origin \(\overrightarrow{OA'}=k\overrightarrow{OA}\)

Count the number of "units" (using the grid). If we assume that the length of \(AB\): from \(A(5,3)\) to \(B(- 5,-9)\) (by counting the grid - based on the line passing through \(A\) and \(B\) and \(A'\) and \(B'\)).
The vector from \(A\) to \(B\): \(\Delta x=-5 - 5=-10\), \(\Delta y=-9 - 3=-12\), \(AB=\sqrt{(-10)^2+(-12)^2}=\sqrt{100 + 144}=\sqrt{244} = 2\sqrt{61}\) (wrong, no, wait the problem says \(AB = 15\))

Wait, the correct way:
Since \(A'B'\) and \(AB\) are related by dilation. Let’s assume the scale factor \(k\). We know that \(A'B'=k\times AB\)
By counting the number of units (using the grid - assume each square is 1 unit).
The length of \(A'B'\): from \(A'(1,1)\) to \(B'(-1,-3)\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), \(A'B'=\sqrt{(1+1)^2+(1 + 3)^2}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5}\)
But wait, no, another approach. If we consider the ratio of the coordinates. For a dilation \((x,y)\to(kx,ky)\)
Take a point \(A(5,3)\) and \(A'(1,1)\), then \(k=\frac{1}{5}\) (for \(x\) - coordinate) and \(k=\frac{1}{3}\) (for \(y\) - coordinate). But since dilation is uniform, we made a mistake in coordinate reading.
Wait, looking at the graph again:
If \(AB = 15\) and assume \(A(6,3)\) (counting 6 units in \(x\) from origin and 3 units in \(y\)) and \(A'(2,1)\) (counting 2 units in \(x\) and 1 unit in \(y\))
The scale factor \(k=\frac{1}{3}\) (because \(2=\frac{1}{3}\times6\) and \(1=\frac{1}{3}\times3\))

Answer:

The scale factor of the dilation is \(\frac{1}{3}\)