Question

Triangle A"B"C" is formed using the translation (x + 2 y + 0) and the dilation by a scale factor of from the origin. Which equation explains the relationship between AB and A"B"?

Answer 1

Answer:

segment a double prime b double prime = segment ab over 2

Step-by-step explanation:

Triangle A″B″C″ is formed using the translation (x + 2, y + 0) and the dilation by a scale factor of one half from the origin. which equation explains the relationship between segment ab and segment a double prime b double prime?

coordinate plane with triangle ABC at A (- 3,  3), B (1, - 3), and C (- 3, -3).

segment a double prime b double prime = segment ab over 2

segment ab = segment a double prime b double prime over 2

segment ab over segment a double prime b double prime = one half

segment a double prime b double prime over segment ab = 2

Answer:  If triangle ABC at A (- 3,  3), B (1, - 3), and c (- 3, -3) is translated using   (x + 2, y + 0), then the new coordinates of triangle ABC would be at A'(- 3 + 2,  3), B'(1 + 2, - 3), and C'(- 3 + 2, -3) Which is A'(- 1,  3), B'(3, - 3), and C'(- 1, -3). The dilation by a scale factor of from the origin gives A″ (-1/2, 3/2), B″ (3/2, -3/2) and C″(-1/2, -3/2) = (-0.5, 1.5), B″ (1.5, -1.5) and C″(-0.5, -1.5)

The distance AB and A"B" are given as:

$AB=\sqrt{(1-(-3))^2+(-3-3)^2} =\sqrt{5 2} \\\\A"B"=\sqrt{(1.5-(-0.5))^2+(-1.5-1.5)^2} =\sqrt{13}$

AB = √52 = √(4×52) = 2√13 = 2A"B"

A"B" = AB/2