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The coordinates of trapezoid EFGH are E(4, 6), F(2, 3), G(4, 2), and H(8, 4). The image of EFGH under dilation is E'F'G'H'.

If the coordinates of vertex G' are (2, 1), what are the coordinates of vertex H'?

Respuesta :

DeanR
When it doesn't say, we assume the dilation is around the origin, so we just multiply all the coordinates by the dilation factor [tex]r[/tex].

I'll include more detail than I would in my own homework; if you're comfortable skipping a step, feel free.

[tex]G' = r G[/tex]

[tex](2,1) = r (4,2)[/tex]

That's two equations, [tex]2=4r[/tex] and [tex]1=2r[/tex].  Fortunately they have a common root,

[tex]r=\frac 1 2[/tex]

[tex]H' = rH[/tex]

[tex]H' = \frac 1 2(8,4)=(4,2)[/tex]



[tex]H'(4,2)[/tex]

Answer:  The required co-ordinates of vertex H' are (4, 2).

Step-by-step explanation:  Given that the co-ordinates of trapezoid EFGH are  E(4, 6), F(2, 3), G(4, 2), and H(8, 4) and its image EFGH under dilation is E'F'G'H'.

The co-ordinates of vertex G' are (2, 1).

We are to find the co-ordinates of vertex GH.

Let d denote the dilation factor of trapezoid EFGH to E'F'G'H'.

Then, according to the given information, we must have

[tex]d\times \textup{co-ordinates of G}=\textup{co-ordinates of G'}\\\\\\\Rightarrow d(4, 2)=(2,1)\\\\\\\Rightarrow d\times2(2, 1)=(2,1)\\\\\Rightarrow 2d=1\\\\\Rightarrow d=\dfrac{1}{2}.[/tex]

Therefore, the co-ordinates of vertex H' are given by

[tex]d\times\textup{co-ordinates of H}\\\\=\dfrac{1}{2}(8,4)\\\\=(4,2).[/tex]

Thus, the required co-ordinates of vertex H' are (4, 2).

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