In the standard (x, y) coordinate plane, triangle XYZ has vertices X(1, 2), Y(4, 2), and Z… — ACT

In the standard $(x, y)$ coordinate plane, triangle $XYZ$ has vertices $X(1, 2)$ , $Y(4, 2)$ , and $Z(1, 6)$ . The triangle is reflected across the line $y = x$ , and then the resulting image is dilated by a scale factor of $3$ with the center of dilation at $(2, 1)$ to form triangle $X''Y''Z''$ . What are the coordinates of the vertex $Z''$ ?

A
$(12, 0)$
$(14, 1)$ Answer
C
$(16, -1)$
D
$(18, 3)$
E
$(-22, -5)$

Answer

The correct coordinates of the vertex

Z''

are

(14, 1)

The correct coordinate pair is

(14, 1)

. Swapping the coordinates of

Z(1, 6)

across the reflection line

y = x

gives

Z'(6, 1)

. The dilation of

Z'(6, 1)

by a scale factor of 3 relative to the center

(2, 1)

involves scaling the displacement vector

(4, 0)

(12, 0)

and adding the center back, resulting in

(14, 1)

Step-by-Step Solution

Reflect the point

Z(1, 6)

across the line

y = x

The intermediate image is

Z'(6, 1)

Reflecting a point across the line

y = x

swaps its

x

- and

y

-coordinates.

Calculate the displacement of

Z'(6, 1)

relative to the center of dilation

(2, 1)

(6 - 2, 1 - 1) = (4, 0)

To perform a dilation centered at a point other than the origin, the point's coordinate must first be measured relative to the center.

Multiply the displacement by the scale factor of

3

3 \times (4, 0) = (12, 0)

Dilation scales the distance from the center of dilation by the given factor.

Add the scaled displacement back to the center of dilation

(2, 1)

to find the final absolute coordinates.

(2 + 12, 1 + 0) = (14, 1)

This translates the relative coordinates back into the standard coordinate plane.

Key Concept

Applying composite transformations in the coordinate plane, specifically combining a line reflection and a dilation about a non-origin center.

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