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Triangle XYZ where X(7, 2), Y(-3, 8), and Z(-4, 6) is dilated with a scale factor of one-fourth . Where is the dilation located?

AX’the coordinate (negative seven-fourths, one-half), Y’the coordinate (three-fourths, negative 2and Z’he coordinate (1, negative three-halves)
BX’the coordinate (one-half, seven-fourths), Y’the coordinate (2, negative three-fourthsand Z’the coordinate (three-halves, negative 1)
CX’the coordinate (seven-fourths, one-half), Y’the coordinate (negative three-fourths, 2and Z’the coordinate (negative 1, three-halves)
DX’the coordinate (negative one-half, seven-fourths), Y’the coordinate (negative 2, three-fourthsand Z’the coordinate (negative three-halves, 1)

Verified Answer :

queenzy

The answer to the question is C. the simplest method to ever solve for a dilation is by multiplying the axis by the factor. like in your question, the X(7,2), should be multiplied by the scale factor 1/4. You, first of all, multiply the x-axis(7) by 1/4=71/4=7/4. then you multiply the y-axis(2) by 1/4=21/4=1/2.After multiplying, the X coordinate would be(7/4,1/2). then solve for the other coordinates and the answer would be C. This is the simplest way to solve for dilation.

Answer:  The correct option is

(C) X coordinate (seven-fourths, one-half), Y coordinate (negative three-fourths, 2) and Z coordinate (negative 1, three-halves).

Step-by-step explanation:  Given that the triangle  XYZ where X(7, 2), Y(-3, 8), and Z(-4, 6) is dilated with a scale factor of one-fourth.

We are to select the correct location of the dilation.

We know that

if a point (x, y) is a vertex of a triangle, which is dilated with a scale factor of s, then the new co-ordinates of the vertex will be (sx, sy).

According to the given information, the scale factor of dilation is

\(s=\dfrac{1}{4}.\)

So, the new co-ordinates of the image triangle after dilation are

\(X(7,2)~~~\Rightarrow~~~\left(7\times\dfrac{1}{4},2\times\dfrac{1}{4}\right)=\left(\dfrac{7}{4},\dfrac{1}{2}\right),\\\\\\Y(-3,8)~~~\Rightarrow~~~\left((-3)\times\dfrac{1}{4},8\times\dfrac{1}{4}\right)=\left(-\dfrac{3}{4},2\right),\\\\\\Z(-4,6)~~~\Rightarrow~~~\left((-4)\times\dfrac{1}{4},6\times\dfrac{1}{4}\right)=\left(-1,\dfrac{3}{2}\right).\)

Thus, the required location of the dilation is the new triangle with vertices \(\left(\dfrac{7}{4},\dfrac{1}{2}\right),\) \(\left(-\dfrac{3}{4},2\right)\) and \(\left(-1,\dfrac{3}{2}\right).\)

Option (C) is CORRECT.

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