Question

Triangle A″B″C″ is formed by a reflection over x = 1 and dilation by a scale factor of 2 from the origin. Which equation shows the correct relationship between ΔABC and ΔA″B″C″?


coordinate plane with triangle ABC at A negative 3 comma 3, B 1 comma negative 3, and C negative 3 comma negative 3


segment AB over segment A double prime B double prime = the square root of 13 over 2 times the square root of 13

segment C double prime A double prime over segment CA = the square root of 16 over 2 times the square root of 16

segment AB = zero point two segment A double prime B double prime

segment C double prime A double prime = zero point two segment CA

Answer 1

Answer:

segment AB over segment A double prime B double prime = the square root of 13 over 2 times the square root of 13

Step-by-step explanation:

Triangle ABC has vertices at points A(-3,3), B(1,-3) and C(-3,-3).

1. Reflection over x = 1 maps vertices A, B and C as follows

• A(-3,3)→A'(5,3);
• B(1,-3)→B'(1,3);
• C(-3,-3)→C'(5,-3).

2. Dilation by a scale factor of 2 from the origin has the rule

(x,y)→(2x,2y)

So,

• A'(5,3)→A''(10,6);
• B'(1,3)→B''(2,6);
• C'(5,-3)→C''(10,-6)

See attached diagram for details

Note that

$A''B''=2AB\\ \\A''C''=2AC\\ \\B''C''=2BC,\\ \\AB=\sqrt{(-3-1)^2+(3-(-3))^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}$

so

$\dfrac{AB}{A''B''}=\dfrac{2\sqrt{13}}{2\cdot 2\sqrt{13}}=\dfrac{\sqrt{13}}{2\sqrt{13}}$