Question

triangle qrs is dilated according to the rule $d_{o,2}(x,y)$. what is true about the image $\delta qrs$? 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 Dilation Rule

Dilation with rule \( D_{O,2}(x,y) \) means scaling by factor 2 from origin. So \( (x,y) \to (2x,2y) \). This makes first option true.

Step2: Check Slope of \( Q'S' \)

Original \( Q(-3,3) \), \( S(-1,1) \). After dilation, \( Q'(-6,6) \), \( S'(-2,2) \). Slope \( m=\frac{2 - 6}{-2 - (-6)}=\frac{-4}{4}=-1 \). So second option true.

Step3: Compare Lengths \( QR \) and \( Q'R' \)

Dilation by factor 2 increases length, so \( Q'R' = 2QR \). Thus \( QR \) is shorter than \( Q'R' \), third option false.

Step4: Distance from Origin

Dilation by 2 moves points farther from origin (since scale factor >1). So fourth option false.

Step5: Distance from \( Q' \) to Origin

Original \( Q(-3,3) \): distance \( \sqrt{(-3)^2 + 3^2}=\sqrt{18} \). \( Q'(-6,6) \): distance \( \sqrt{(-6)^2 + 6^2}=\sqrt{72}=2\sqrt{18} \). So fifth option true.

Answer:

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