On a coordinate plane, 3 triangles are shown. Triangle B C D has points (1, 4), (1, 2), (5, 3). Triangle B prime C prime D prime

Answer:

It's c

Explanation:

I got it correct on the exam

The transformation sequence that maps ΔBCD to ΔB"C"D" is:

Reflection across the y-axis combined with a translation of 6 units in the x direction and -5 units in the y direction.

Step-by-step explanation:

Let's break down the reflection across the y-axis, horizontal translation, and vertical translation:

1. For point (x, y), reflecting it across the y-axis gives the point (-x, y).

2. Translating point (x, y) h units to the right results in (x + h, y), and h units to the left results in (x - h, y).

3. If point (x, y) is moved k units up, it becomes (x, y + k), and if moved k units down, it is (x, y - k).

∵ The vertices of triangle BCD are (1, 4), (1, 2), (5, 3).

∵ The vertices of triangle B'C'D' are (-1, 4), (-1, 2), (-5, 3).

∵ The x-coordinates of ΔB'C'D' have the same absolute value as those of ΔBCD but with opposite signs, indicating that ΔB'C'D' results from reflecting ΔBCD across the y-axis.

∵ The vertices of triangle B'C'D' are (-1, 4), (-1, 2), (-5, 3).

∵ The vertices of triangle B''C''D'' are (5, -1), (5, -3), (1, -2).

∵ The reflected x-coordinate -1 becomes +5, and -5 becomes +1,

thus the x-coordinates of triangle B'C'D' are increased by 6.

∵ The images of 4, 2, and 3 yield -1, -3, and -2 respectively,

hence subtracting 5 from the y-coordinates of triangle B'C'D' leads us to ΔB"C"D" through a translation of 6 units right and 5 units down ⇒ (x + 6, y - 5).

The transformation sequence that maps ΔBCD to ΔB"C"D" is:

Reflection across the y-axis combined with a translation of 6 units in the x direction and -5 units in the y direction.

Learn more:

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