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

By the definition of dilation \(D_{O,k}(x,y)=(kx,ky)\), when \(k = 0.75\), \(D_{O,0.75}(x,y)=(0.75x,0.75y)\).

Step2: Check parallelism

For a dilation with center \(O\), if a line segment is not passing through the center of dilation, the pre - image and image of a line segment are parallel. Since \(\overline{LM}\) does not pass through the origin (center of dilation \(O\)), \(\overline{LM}\parallel\overline{L'M'}\) (dilation preserves parallelism for non - collinear - with - center line segments).

Step3: Analyze the distance from vertices to the origin

Let the coordinates of a point \(P(x,y)\) in the pre - image. After dilation \(P'(0.75x,0.75y)\). The distance from \(P(x,y)\) to the origin \(d=\sqrt{x^{2}+y^{2}}\), and the distance from \(P'(0.75x,0.75y)\) to the origin \(d'=\sqrt{(0.75x)^{2}+(0.75y)^{2}}=0.75\sqrt{x^{2}+y^{2}}\lt\sqrt{x^{2}+y^{2}}\). So the vertices of the image are closer to the origin than those of the pre - image.

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

Let \(K(-4,4)\) and \(M(-2,2)\). The length of \(KM=\sqrt{(-4 + 2)^{2}+(4 - 2)^{2}}=\sqrt{4 + 4}=\sqrt{8}\). After dilation \(K'(-4\times0.75,4\times0.75)=(-3,3)\) and \(M'(-2\times0.75,2\times0.75)=(-1.5,1.5)\). The length of \(K'M'=\sqrt{(-3+1.5)^{2}+(3 - 1.5)^{2}}=\sqrt{2.25 + 2.25}=\sqrt{4.5}=0.75\sqrt{8}\lt\sqrt{8}\), so \(KM\) is longer than \(K'M'\) (not shorter).

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

Let \(M(-2,2)\), the distance from \(M\) to the origin \(d_M=\sqrt{(-2)^{2}+2^{2}}=\sqrt{4 + 4}=\sqrt{8}\). After dilation \(M'(-1.5,1.5)\), the distance from \(M'\) to the origin \(d_{M'}=\sqrt{(-1.5)^{2}+1.5^{2}}=\sqrt{2.25+2.25}=\sqrt{4.5}=0.75\sqrt{8}
eq0.5\sqrt{8}\) (not half).

Answer:

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