Question

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polygon mnopq is dilated by a scale factor of 5.4 with point m as the center of dilation to form the image mnopq. compare \\( \overline{no} \\) and \\( \overline{no} \\).
\\( \overline{no} \\) through point m. the slopes of \\( \overline{no} \\) and \\( \overline{no} \\) are

Explanation:

Step1: Properties of Dilation

When a figure is dilated with a center of dilation, the corresponding line segments (in this case, \(\overline{NO}\) and \(\overline{N'O'}\)) are collinear with the center of dilation. So, \(\overline{N'O'}\) passes through point \(M\).

Step2: Slope of Dilated Line Segments

Dilation is a similarity transformation. Similarity transformations (including dilation) preserve the slope of line segments. If we have two points \((x_1,y_1)\) and \((x_2,y_2)\) on \(\overline{NO}\) and the corresponding points \((x_1',y_1')\) and \((x_2',y_2')\) on \(\overline{N'O'}\) after dilation with center \(M=(x_0,y_0)\), the slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\) and \(m'=\frac{y_2' - y_1'}{x_2' - x_1'}\). Since \(x_1'=x_0+(x_1 - x_0)\times k\), \(y_1'=y_0+(y_1 - y_0)\times k\), \(x_2'=x_0+(x_2 - x_0)\times k\), \(y_2'=y_0+(y_2 - y_0)\times k\) (where \(k = 5.4\) is the scale factor), then \(m'=\frac{(y_0+(y_2 - y_0)\times k)-(y_0+(y_1 - y_0)\times k)}{(x_0+(x_2 - x_0)\times k)-(x_0+(x_1 - x_0)\times k)}=\frac{(y_2 - y_1)\times k}{(x_2 - x_1)\times k}=\frac{y_2 - y_1}{x_2 - x_1}=m\).

Answer:

\(\overline{N'O'}\) passes through point \(M\). The slopes of \(\overline{N'O'}\) and \(\overline{NO}\) are equal.