Consider ΔABC , with vertices A(0,3) , B(0,1) , and C(−4,2) , and ΔDEF , with vertices D(0,6) , E(0,2) , and F(8,4) . Which transformations could be used to show that the triangles are similar?(1 point) Responses ΔABC is dilated by a scale factor of 14 and is reflected across the y-axis. triangle upper A upper B upper C is dilated by a scale factor of Start Fraction 1 over 4 End Fraction and is reflected across the y -axis. ΔABC is dilated by a scale factor of −2 and is reflected across the y-axis. triangle upper A upper B upper C is dilated by a scale factor of negative 2 and is reflected across the y -axis. ΔABC is dilated by a scale factor of 2 and is reflected across the y-axis. triangle upper A upper B upper C is dilated by a scale factor of 2 and is reflected across the y -axis. ΔABC is dilated by a scale factor of 12 and is reflected across the x-axis.

1 answer

To determine whether triangles \( \Delta ABC \) and \( \Delta DEF \) are similar, we first need to analyze their corresponding vertices and the ratios of their side lengths.

Vertices of the triangles:

  • \( \Delta ABC \) has vertices \( A(0, 3) \), \( B(0, 1) \), and \( C(-4, 2) \).
  • \( \Delta DEF \) has vertices \( D(0, 6) \), \( E(0, 2) \), and \( F(8, 4) \).

Finding the lengths of the sides:

  1. In \( \Delta ABC \):

    • \( AB = \sqrt{(0 - 0)^2 + (3 - 1)^2} = \sqrt{0 + 4} = 2 \)
    • \( BC = \sqrt{(0 - (-4))^2 + (1 - 2)^2} = \sqrt{(4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17} \)
    • \( AC = \sqrt{(0 - (-4))^2 + (3 - 2)^2} = \sqrt{(4)^2 + (1)^2} = \sqrt{16 + 1} = \sqrt{17} \)

    So the side lengths of \( \Delta ABC \) are:

    • \( AB = 2 \)
    • \( BC = \sqrt{17} \)
    • \( AC = \sqrt{17} \)
  2. In \( \Delta DEF \):

    • \( DE = \sqrt{(0 - 0)^2 + (6 - 2)^2} = \sqrt{0 + 16} = 4 \)
    • \( EF = \sqrt{(0 - 8)^2 + (2 - 4)^2} = \sqrt{(8)^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68} \)
    • \( DF = \sqrt{(0 - 8)^2 + (6 - 4)^2} = \sqrt{(8)^2 + (2)^2} = \sqrt{64 + 4} = \sqrt{68} \)

    So the side lengths of \( \Delta DEF \) are:

    • \( DE = 4 \)
    • \( EF = \sqrt{68} \)
    • \( DF = \sqrt{68} \)

Finding the ratio of the corresponding sides:

  • The ratio of \( AB \) to \( DE \) is \( \frac{2}{4} = \frac{1}{2} \).
  • The ratio of \( BC \) to \( EF \) is \( \frac{\sqrt{17}}{\sqrt{68}} = \frac{\sqrt{17}}{2\sqrt{17}} = \frac{1}{2} \).
  • The ratio of \( AC \) to \( DF \) is \( \frac{\sqrt{17}}{\sqrt{68}} = \frac{\sqrt{17}}{2\sqrt{17}} = \frac{1}{2} \).

Since all corresponding side lengths have the same ratio of \( \frac{1}{2} \), the triangles are similar.

Transformation to show similarity: Given the similarity ratios, the transformation involves dilation by the scale factor \( \frac{1}{2} \). Since the triangles are in the same orientation across the y-axis, a reflection across the y-axis can be included, but the essential transformation is the dilation.

Thus, the transformation that shows that the triangles are similar is:

  • \( \Delta ABC \) is dilated by a scale factor of \( \frac{1}{2} \) and could also be reflected across the y-axis.

Correct option:

  • \( \Delta ABC \) is dilated by a scale factor of \( \frac{1}{2} \) and is reflected across the y-axis.