In the standard (x,y) coordinate plane, a triangle has vertices at A(2, 4), B(6, 4), and C… — ACT

In the standard $(x,y)$ coordinate plane, a triangle has vertices at $A(2, 4)$ , $B(6, 4)$ , and $C(2, 10)$ . The triangle undergoes a series of transformations: first, it is reflected across the line $y = x$ ; next, the resulting image is dilated by a scale factor of $\frac{1}{2}$ with the center of dilation at $(2, 2)$ ; finally, this image is translated $1$ unit to the left and $3$ units down. What are the coordinates of the final image of vertex $C$ ?

A
$(4, -2)$
B
$(7, 5)$
$(5, -1)$ Answer
D
$(5, 5)$
E
$(-5, -3)$

Answer

(5, -1)

The correct answer is

(5, -1)

because applying the three transformations sequentially yields the correct final coordinates: first, reflecting

C(2, 10)

across the line

y = x

gives

C'(10, 2)

; second, dilating

C'(10, 2)

by a scale factor of

\frac{1}{2}

centered at

(2, 2)

gives

C''(6, 2)

; and finally, translating

C''(6, 2)

1

unit left and

3

units down results in

(5, -1)

Step-by-Step Solution

Reflect the initial coordinates of vertex

C(2, 10)

across the line

y = x

C'(10, 2)

Reflecting a point

(x, y)

across the line

y = x

interchanges the coordinates, mapping

(x, y)

(y, x)

Dilate the point

C'(10, 2)

by a scale factor of

k = \frac{1}{2}

centered at

P(2, 2)

C''(6, 2)

The formula for a dilation centered at

(a, b)

(x'', y'') = (a + k(x' - a), b + k(y' - b))

. Substituting

a=2, b=2, k=\frac{1}{2}, x'=10, y'=2

gives

(2 + \frac{1}{2}(10-2), 2 + \frac{1}{2}(2-2)) = (2+4, 2+0) = (6, 2)

Translate the point

C''(6, 2)

1

unit to the left and

3

units down.

C'''(5, -1)

Translating

1

unit left subtracts

1

from the

x

-coordinate (

6 - 1 = 5

), and translating

3

units down subtracts

3

from the

y

-coordinate (

2 - 3 = -1

Key Concept

Composite transformations in the coordinate plane involving reflections, dilations with non-origin centers, and translations.

Estimated Time:2m 0s

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