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 $\overline{ab}$. $\square$ $d_{o,1/2}(x,y)=(\frac{1}{2}x,\frac{1}{2}y)$ $\square$ 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:

Step1: Recall the properties of dilation

Dilation is a transformation that enlarges or reduces a figure. For a dilation \(D_{O,k}(x,y)\) with center at the origin \(O\) and scale factor \(k\), the rule is \(D_{O,k}(x,y)=(kx,ky)\). When \(k = \frac{1}{2}\), \(D_{O,\frac{1}{2}}(x,y)=(\frac{1}{2}x,\frac{1}{2}y)\).

Step2: Analyze the parallelism

Let \(A(-4,3)\) and \(B(4,4)\). After dilation \(A'(- 2,\frac{3}{2})\) and \(B'(2,2)\).
The slope of \(\overline{AB}\) is \(m_{AB}=\frac{4 - 3}{4+4}=\frac{1}{8}\).
The slope of \(\overline{A'B'}\) is \(m_{A'B'}=\frac{2-\frac{3}{2}}{2 + 2}=\frac{\frac{1}{2}}{4}=\frac{1}{8}\). Since \(m_{AB}=m_{A'B'}\), \(\overline{AB}\parallel\overline{A'B'}\).

Step3: Analyze the distance from the origin

The distance formula from a point \((x,y)\) to the origin \((0,0)\) is \(d=\sqrt{x^{2}+y^{2}}\).
For point \(A(-4,3)\), \(d_A=\sqrt{(-4)^{2}+3^{2}}=\sqrt{16 + 9}=5\).
For point \(A'(-2,\frac{3}{2})\), \(d_{A'}=\sqrt{(-2)^{2}+(\frac{3}{2})^{2}}=\sqrt{4+\frac{9}{4}}=\sqrt{\frac{16 + 9}{4}}=\frac{5}{2}\). So \(d_{A'}=\frac{1}{2}d_A\).

Step4: Analyze the position of vertices

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

Step5: Analyze the length of segments

The length formula for a segment with endpoints \((x_1,y_1)\) and \((x_2,y_2)\) is \(l=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\).
For \(AB\) with \(A(-4,3)\) and \(B(4,4)\), \(AB=\sqrt{(4 + 4)^{2}+(4 - 3)^{2}}=\sqrt{64+1}=\sqrt{65}\).
For \(A'B'\) with \(A'(-2,\frac{3}{2})\) and \(B'(2,2)\), \(A'B'=\sqrt{(2 + 2)^{2}+(2-\frac{3}{2})^{2}}=\sqrt{16+\frac{1}{4}}=\sqrt{\frac{64 + 1}{4}}=\frac{\sqrt{65}}{2}\). So \(A'B'=\frac{1}{2}AB\).

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.