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

By the definition of dilation with a scale factor \(k = 2\) centered at the origin \(O\), the rule \(D_{O,k}(x,y)=(kx,ky)\). So for \(k = 2\), \(D_{O,2}(x,y)=(2x,2y)\)

Step2: Find the coordinates of \(Q'\) and \(S'\)

If \(Q(-3,3)\), then \(Q'=D_{O,2}(-3,3)=(-6,6)\). If \(S(-1,1)\), then \(S'=D_{O,2}(-1,1)=(-2,2)\)
The slope of the line passing through \(Q'(-6,6)\) and \(S'(-2,2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{2 - 6}{-2+6}=\frac{-4}{4}=-1\)

Step3: Analyze the distance from a point to the origin

The distance from a point \(P(x,y)\) to the origin \(O(0,0)\) is \(d=\sqrt{x^{2}+y^{2}}\). For \(Q(-3,3)\), \(d_Q=\sqrt{(-3)^{2}+3^{2}}=\sqrt{9 + 9}=\sqrt{18}\). For \(Q'(-6,6)\), \(d_{Q'}=\sqrt{(-6)^{2}+6^{2}}=\sqrt{36+36}=\sqrt{72}\), and \(\sqrt{72}=2\sqrt{18}\)

For the option \(C\): Since the scale factor \(k = 2>1\), \(\overline{Q'R'}=2\overline{QR}\), so \(\overline{QR}\) is shorter than \(\overline{Q'R'}\)
For the option \(D\): Since \(k = 2>1\), the vertices of the image \(\triangle Q'R'S'\) are farther from the origin than those of the pre - image \(\triangle QRS\)

Answer:

A. \(D_{O,2}(x,y)=(2x,2y)\)
B. Side \(Q'S'\) lies on a line with a slope of \(- 1\)
E. The distance from \(Q'\) to the origin is twice the distance from \(Q\) to the origin