Question

triangle klm was dilated according to the rule ( d_{o,0.75}(x,y) ). what is true about the image ( \triangle klm )? select three options. ( square d_{o,0.75}(x,y)=(0.75x,0.75y) ) ( square overline{lm} ) is parallel to ( overline{lm} ). ( square overline{km} ) is shorter than ( overline{km} ). ( square ) the vertices of the image are closer to the origin than those of the pre - image. ( square ) the distance from ( m ) to the origin is exactly half the distance from ( m ) to the origin.

Explanation:

Step1: Analyze the dilation rule

The dilation rule \(D_{O,0.75}(x,y)\) means that each coordinate of a point \((x,y)\) is multiplied by \(0.75\). So \(D_{O,0.75}(x,y)=(0.75x,0.75y)\).

Step2: Check parallelism

When a figure is dilated, corresponding line - segments are parallel. For line - segments \(\overline{LM}\) and \(\overline{L'M'}\), since dilation is a similarity transformation (a non - rigid transformation that preserves the shape of the figure), \(\overline{LM}\parallel\overline{L'M'}\).

Step3: Analyze the length of line - segments

The length of a line - segment \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). If we have a point \(P(x,y)\) and its image \(P'(0.75x,0.75y)\) after dilation with scale factor \(k = 0.75\). The distance from the origin \(O(0,0)\) to \(P(x,y)\) is \(d=\sqrt{x^{2}+y^{2}}\), and the distance from the origin \(O(0,0)\) to \(P'(0.75x,0.75y)\) is \(d'=\sqrt{(0.75x)^{2}+(0.75y)^{2}}=0.75\sqrt{x^{2}+y^{2}}\). So the vertices of the image are closer to the origin than those of the pre - image.

Step4: Analyze the length of \(\overline{KM}\) and \(\overline{K'M'}\)

Since the scale factor \(k = 0.75\lt1\), the length of \(\overline{K'M'}=0.75\times\) length of \(\overline{KM}\). So \(\overline{KM}\) is longer than \(\overline{K'M'}\).

Step5: Analyze the distance from \(M\) and \(M'\) to the origin

The distance from \(M(-2,2)\) to the origin \(d_M=\sqrt{(- 2)^{2}+2^{2}}=\sqrt{4 + 4}=\sqrt{8}\). The distance from \(M'(-2\times0.75,2\times0.75)=M'(-1.5,1.5)\) to the origin \(d_{M'}=\sqrt{(-1.5)^{2}+1.5^{2}}=\sqrt{2.25 + 2.25}=\sqrt{4.5}\), and \(0.75\times\sqrt{8}=\sqrt{0.75^{2}\times8}=\sqrt{4.5}\), not half.

Answer:

\(\text{D}_{O,0.75}(x,y)=(0.75x,0.75y)\), \(\overline{LM}\) is parallel to \(\overline{L'M'}\), The vertices of the image are closer to the origin than those of the pre - image.