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triangle qrs is dilated according to the rule ( d_{o,2}(x,y) ). what is…

Question

triangle qrs is dilated according to the rule ( d_{o,2}(x,y) ). what is true about the image ( \triangle qrs )? select three options. which statements are true? ( square d_{o,2}(x,y)=(2x,2y) ) ( square ) side ( qs ) lies on a line with a slope of -1. ( square overline{qr} ) is longer than ( overline{qr} ). ( square ) the vertices of the image are closer to the origin than those of the pre - image. ( square ) the distance from ( q ) to the origin is twice the distance from ( q ) to the origin.

Explanation:

Step1: Analyze the dilation rule

The dilation rule \(D_{O,2}(x,y)\) means that each coordinate of the original point \((x,y)\) is multiplied by \(2\). So \(D_{O,2}(x,y)=(2x,2y)\) is correct.

Step2: Calculate the slope of \(Q'S'\)

First, find \(Q(-3,3)\) and \(S(-1,1)\). After dilation, \(Q'(-6,6)\) and \(S'(-2,2)\). The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{2 - 6}{-2+6}=\frac{-4}{4}=-1\). So side \(Q'S'\) lies on a line with a slope of \(-1\) is correct.

Step3: Compare the lengths of \(QR\) and \(Q'R'\)

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(QR\): \(Q(-3,3)\), \(R(2,4)\), \(d_{QR}=\sqrt{(2 + 3)^2+(4 - 3)^2}=\sqrt{25 + 1}=\sqrt{26}\)
For \(Q'R'\): \(Q'(-6,6)\), \(R'(4,8)\), \(d_{Q'R'}=\sqrt{(4 + 6)^2+(8 - 6)^2}=\sqrt{100+4}=\sqrt{104} = 2\sqrt{26}\). So \(\overline{QR}\) is longer than \(\overline{Q'R'}\) is incorrect.

Step4: Check the distance from vertices to the origin

The distance from a point \((x,y)\) to the origin is \(d=\sqrt{x^{2}+y^{2}}\). Since we are dilating by a factor of \(2\), the vertices of the image are farther from the origin than those of the pre - image. So the vertices of the image are closer to the origin than those of the pre - image is incorrect.

Step5: Check the distance from \(Q\) and \(Q'\) to the origin

The distance from \(Q(-3,3)\) to the origin \(d_Q=\sqrt{(-3)^{2}+3^{2}}=\sqrt{9 + 9}=\sqrt{18}\)
The distance from \(Q'(-6,6)\) to the origin \(d_{Q'}=\sqrt{(-6)^{2}+6^{2}}=\sqrt{36+36}=\sqrt{72}=2\sqrt{18}\). So the distance from \(Q'\) to the origin is twice the distance from \(Q\) to the origin is incorrect.

Answer:

A. \(D_{O,2}(x,y)=(2x,2y)\), B. Side \(Q'S'\) lies on a line with a slope of \(-1\), D. The vertices of the image are closer to the origin than those of the pre - image.