Question

triangle qrs is dilated according to the rule $d_{o,2}(x, y)$. which statements are true? $\overline{qr}$ is longer than $\overline{qr}$. side $qs$ lies on a line with a slope of $-1$. the distance from $q$ to the origin is twice the distance from q to the origin. $d_{o,2}(x, y) = (2x, 2y)$

Explanation:

Step1: Analyze Dilation Rule

Dilation rule \( D_{O,2}(x,y) \) means scaling by factor 2 from origin, so \( (x,y) \to (2x,2y) \). So the fourth statement \( D_{O,2}(x,y)=(2x,2y) \) is true.

Step2: Check Length of \( \overline{QR} \) and \( \overline{Q'R'} \)

Dilation with scale factor >1 increases length. So \( \overline{Q'R'} = 2 \times \overline{QR} \), so \( \overline{QR} \) is shorter than \( \overline{Q'R'} \). First statement is false.

Step3: Find Slope of \( \overline{Q'S'} \)

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

Step4: Distance from \( Q \) and \( Q' \) to Origin

Distance from \( Q(-3,3) \) to origin: \( \sqrt{(-3)^2 + 3^2} = \sqrt{18} \). Distance from \( Q'(-6,6) \) to origin: \( \sqrt{(-6)^2 + 6^2} = \sqrt{72} = 2\sqrt{18} \). So third statement is true.

Answer:

The true statements are:

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