Question

The rule is applied to ΔABC. On a coordinate plane, 5 triangles are shown. Triangle A B C has points (2, negative 4), (4, negative 4), (4, negative 2). Triangle 1 has points (negative 4, 2), (negative 4, 4), (negative 2, 4). Triangle 2 has points (negative 3, 3), (negative 1, 3), (negative 1, 1). Triangle 3 has points (1, 1), (1, 3), (3, 3). Triangle 4 has points (2, 4), (4, 2), (4, 4). Which triangle shows the final image? 1 2 3 4

Answer 1

Answer:

Consider triangle ABC with vertices at points A(2,-4), B(4,-4) and C(4,-2).

1. The rotation  acts with the rule:

Then:

2. The reflection across the y-axis has a rule:

So,

Triangle A''B''C'' is exactly the same as tiangle from figure 1.

Answer: correct choice is 1.

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Step-by-step explanation:

Answer 2

Geometric transformations includes rotation, translation, reflection, and scaling or resizing

The correct option for the triangle that shows the final image is triangle 1

The reason triangle 1 gives the correct final image is as follows;

The given coordinates of triangle ABC are;

A(2, -4), B(4, -4), C(4, -2)

The required transformation is $r_{y -axis} \circ R_{O, \, 90 ^{\circ}} (x, y)$

The transformation on the right of a composite transformation is carried out first as follows;

Rotation 90° about the origin:

The rotation of a point (x, y) by 90 degrees about the origin gives (-y, x)

Therefore, for the triangle, ΔABC, with point (2, -4), (4, -4), (4, -2), we have;

$R_{O, \, 90 ^{\circ}} (2, -4) = (4, \, 2)$

$R_{O, \, 90 ^{\circ}} (4, -4) = (4, \, 4)$

$R_{O, \, 90 ^{\circ}} (4, -2) = (2, \, 4)$

The coordinates of the image of triangle ABC, following a rotation of 90° about the origin are (4, 2), (4, 4), and (2, 4)

Reflection across the y-axis:

The coordinates of the image of a point (x, y) following a reflection across the y-axis is (-x, y)

Therefore, we have;

$r_{y -axis}(4, \, 2) = (-4, \, 2)$

$r_{y -axis}(4, \, 4) = (-4, \, 4)$

$r_{y -axis}(2, \, 4) = (-2, \, 4)$

Therefore, the rule, $r_{y -axis} \circ R_{O, \, 90 ^{\circ}} (x, y)$, applied to triangle ΔABC, gives the final image of the triangle with coordinates (-4, 2), (-4, 4), and (-2, 4), which is triangle 1

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