Question

triangle qrs is dilated according to the rule ( d_{o,2}(x,y) ). which statements are true? ( d_{o,2}(x,y)=(2x,2y) ) side ( qs ) lies on a line with a slope of -1. ( overline{qr} ) is longer than ( overline{qr} ). the vertices of the image are closer to the origin than those of the pre - image. 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 pre - image is multiplied by 2. So \(D_{O,2}(x,y)=(2x,2y)\).

Step2: Find the new coordinates

If \(Q(-3,3)\), then \(Q'(-6,6)\); if \(R(2,4)\), then \(R'(4,8)\); if \(S(-1,1)\), then \(S'(-2,2)\).

Step3: Calculate the slope of \(Q'S'\)

The slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For \(Q'(-6,6)\) and \(S'(-2,2)\), \(m=\frac{2 - 6}{-2+6}=\frac{-4}{4}=-1\).

Step4: Compare the lengths

The distance formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(QR\) with \(Q(-3,3)\) and \(R(2,4)\), \(d_{QR}=\sqrt{(2 + 3)^2+(4 - 3)^2}=\sqrt{25 + 1}=\sqrt{26}\). For \(Q'R'\) with \(Q'(-6,6)\) and \(R'(4,8)\), \(d_{Q'R'}=\sqrt{(4 + 6)^2+(8 - 6)^2}=\sqrt{100+4}=\sqrt{104}\). So \(QR\) is shorter than \(Q'R'\).

Step5: Compare distances from the origin

The distance from a point \((x,y)\) to the origin \((0,0)\) 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 \(\sqrt{72}=2\sqrt{18}\). The vertices of the pre - image \((x,y)\) and image \((2x,2y)\): \(\sqrt{(2x)^{2}+(2y)^{2}} = 2\sqrt{x^{2}+y^{2}}\), so the vertices of the pre - image are closer to the origin.

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