Question

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
( square overline{ab} ) is parallel to ( overline{ab} )
( square d_{o,1/2}(x,y)=left( \frac{1}{2}x,\frac{1}{2}y
ight) )
( square ) the distance from ( a ) to the origin is half the distance from a to the origin
( square ) the vertices of the image are farther from the origin than those of the pre - image
( square ab ) is greater than ab

Explanation:

Step 1: Understanding dilation properties

A dilation with center at the origin and scale factor \(k = \frac{1}{2}\) has the rule \(D_{O,k}(x,y)=(kx,ky)\). So for \(k=\frac{1}{2}\), \(D_{O,\frac{1}{2}}(x,y)=(\frac{1}{2}x,\frac{1}{2}y)\).

Step 2: Parallelism in dilation

Under a dilation, corresponding line segments of the pre - image and the image are parallel. So \(\overline{AB}\parallel\overline{A'B'}\) because dilation is a similarity transformation that preserves parallelism.

Step 3: 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 \(P'(x',y')\) which is the image of \(P(x,y)\) under \(D_{O,\frac{1}{2}}\), \(x'=\frac{1}{2}x\) and \(y'=\frac{1}{2}y\). The distance from \(P'\) 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.

Step 4: Length of line segments

Since the scale factor \(k = \frac{1}{2}<1\), the length of the image line segment \(A'B'\) is \(k\times AB\). So \(A'B'=\frac{1}{2}AB\), which means \(A'B'

Answer:

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