Question

triangle klm was dilated according to the rule ( d_{0.075}(x,y) ) what is true about the image ( \triangle klm )? select three options ( square d_{0.075}(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 with a scale factor \(k = 0.75\) centered at the origin \(O\), the rule \(D_{O,k}(x,y)=(kx,ky)\). So \(D_{O,0.75}(x,y)=(0.75x,0.75y)\) is correct.

Step2: Check parallelism

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

Step3: Compare distances 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 \(d'=\sqrt{(0.75x)^{2}+(0.75y)^{2}}=0.75\sqrt{x^{2}+y^{2}}\). Since \(0.75\lt1\), the vertices of the image are closer to the origin than those of the pre - image.

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

The length of a line segment with endpoints \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is \(l=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\). For \(KM\) with \(K(-4,4)\) and \(M(-2,2)\), \(l_{KM}=\sqrt{(-2 + 4)^{2}+(2 - 4)^{2}}=\sqrt{4 + 4}=\sqrt{8}\). After dilation \(K'(-3,3)\) and \(M'(-1.5,1.5)\), \(l_{K'M'}=\sqrt{(-1.5+3)^{2}+(1.5 - 3)^{2}}=\sqrt{2.25 + 2.25}=\sqrt{4.5}=0.75\sqrt{8}\). So \(KM\) is longer than \(K'M'\), not shorter.

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 \(M'(-1.5,1.5)\) to the origin \(d_{M'}=\sqrt{(-1.5)^{2}+1.5^{2}}=\sqrt{2.25+2.25}=\sqrt{4.5}=0.75\sqrt{8}
eq0.5\sqrt{8}\)

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