Question

practice with dilations and similar figures. a has the coordinates (-4, 3) and b has the coordinates (4, 4). if ( d_{o,1/2}(x,y) ) is a dilation of ( \triangle abc ), what is true about the image ( \triangle abc )? check all that apply. ( overline{ab} ) is parallel to ( overline{ab} ). ( d_{o,1/2}(x,y)=(\frac{1}{2}x,\frac{1}{2}y) ) the distance from ( a ) to the origin is half the distance from a to the origin. the vertices of the image are farther from the origin than those of the pre - image. ( ab ) is greater than ab.

Explanation:

Step1: Recall the properties of dilation

A dilation \(D_{O,k}(x,y)=(kx,ky)\) where \(k = \frac{1}{2}\). For a dilation centered at the origin with scale factor \(k\), if \(0

Step2: Analyze the parallelism

For a dilation centered at the origin, the lines in the pre - image and the corresponding lines in the image are parallel. So, if we have a dilation \(D_{O,\frac{1}{2}}\), \(\overline{AB}\parallel\overline{A'B'}\) because dilation is a similarity transformation that preserves parallelism.

Step3: Analyze the dilation formula

By the definition of dilation \(D_{O,k}(x,y)=(kx,ky)\), when \(k=\frac{1}{2}\), \(D_{O,\frac{1}{2}}(x,y)=(\frac{1}{2}x,\frac{1}{2}y)\)

Step4: Analyze the distance from a point to the origin

The distance from a point \(P(x,y)\) to the origin \(O(0,0)\) is \(d=\sqrt{x^{2}+y^{2}}\). For a point \(A(x,y)\) and its image \(A'(\frac{1}{2}x,\frac{1}{2}y)\) after dilation \(D_{O,\frac{1}{2}}\), the distance from \(A'\) to the origin \(d'=\sqrt{(\frac{1}{2}x)^{2}+(\frac{1}{2}y)^{2}}=\frac{1}{2}\sqrt{x^{2}+y^{2}}\). So the distance from \(A'\) to the origin is half the distance from \(A\) to the origin.

Step5: Analyze the position of vertices relative to the origin

Since \(k=\frac{1}{2}<1\), the vertices of the image \(\triangle A'B'C'\) are closer to the origin than those of the pre - image \(\triangle ABC\)

Step6: Analyze the length of segments

The length of a segment \(AB\) with \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is \(l=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\). For \(A'( \frac{1}{2}x_1,\frac{1}{2}y_1)\) and \(B'(\frac{1}{2}x_2,\frac{1}{2}y_2)\), the length of \(A'B'\) is \(l'=\sqrt{(\frac{1}{2}x_2-\frac{1}{2}x_1)^{2}+(\frac{1}{2}y_2 - \frac{1}{2}y_1)^{2}}=\frac{1}{2}\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\). So \(A'B'=\frac{1}{2}AB\), and \(A'B'

Answer:

• \(\overline{AB}\) is parallel to \(\overline{A'B'}\)
• \(D_{O,1/2}(x,y)=(\frac{1}{2}x,\frac{1}{2}y)\)
• The distance from \(A'\) to the origin is half the distance from \(A\) to the origin.