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Part D
Complete the table to find the rule for the dilation, the coordinates of trapezoid P′Q′R′S′, and the coordinates of the dilated image, trapezoid P″Q″R″S″.

Part D Complete The Table To Find The Rule For The Dilation The Coordinates Of Trapezoid PQRS And The Coordinates Of The Dilated Image Trapezoid PQRS class=
Part D Complete The Table To Find The Rule For The Dilation The Coordinates Of Trapezoid PQRS And The Coordinates Of The Dilated Image Trapezoid PQRS class=

Sagot :

The shrink or stretch of the image of PQRS which is the dilation of PQRS is given as follows;

  • The rule of the dilation is [tex]\underline{D_{\frac{1}{2} }}[/tex], followed by a reflection across the x-axis

Reason:

The given completed  table is presented as follows;

[tex]\begin{array}{|cc|c|} \mathbf{Original \ Coordinates}&&\mathbf{New \ Coordinates}\\(x, \ y)&&(,)\\K(3, \, 2)&&P(6, \, -4)\\L(1, \, 1)&&Q(2, \, -2)\\M(1, \, 3)&&R(2, \, -6)\\N(3, \, 3)&&S(6, \, -6)\end{array}[/tex]

Length of the side [tex]\overline{LM}[/tex] = 3 - 1 = 2

Length of the side [tex]\overline{MN}[/tex] = 3 - 1 = 2

Length of the side [tex]\overline{NK}[/tex] = 3 - 2 = 1

Length of the side [tex]\overline{LK}[/tex] = [tex]\sqrt{(2 - 1)^2 + (3 - 1)^2} = \sqrt{5}[/tex]

Length of the side [tex]\overline{RQ}[/tex] = -2 - (-6) = 4

Length of the side [tex]\overline{RS}[/tex] = 6 - 2 = 4

Length of the side [tex]\overline{SP}[/tex] = -4 - (-6) = 2

Length of the side [tex]\overline{QP}[/tex] = [tex]\sqrt{(6 - 2)^2 + (-4 - (-2))^2} = 2\cdot \sqrt{5}[/tex]

[tex]\dfrac{\overline{LM}}{\overline{RQ}} = \dfrac{\overline{MN}}{\overline{RS}} = \dfrac{\overline{NK}}{\overline{SP}} = \dfrac{\overline{LK}}{\overline{QP}} = \dfrac{1}{2}[/tex]

However, the coordinates of the image of PQRS, which is P'Q'R'S', following a dilation are'

[tex]P' \left(\dfrac{1}{2} \times 6, \, \dfrac{1}{2} \times (-4) \right) = P'(3, \, -2)[/tex]

[tex]Q' \left(\dfrac{1}{2} \times 2, \, \dfrac{1}{2} \times (-2) \right) = Q'(1, \, -1)[/tex]

[tex]R' \left(\dfrac{1}{2} \times 2, \, \dfrac{1}{2} \times (-6) \right) = R'(1, \, -3)[/tex]

[tex]S' \left(\dfrac{1}{2} \times 6, \, \dfrac{1}{2} \times (-6) \right) = S'(3, \, -3)[/tex]  

The transformation of the coordinates of P'Q'R'S', to KLMN are;

  • (x, y) → (x, -y)

which is equivalent to a reflection across the x-axis

Therefore, the transformation that gives the coordinate of KLMN from PQRS is a dilation by a scale factor of [tex]\underline{\dfrac{1}{2}}[/tex], which is a rule of dilation of [tex]\underline{D_{\frac{1}{2} }}[/tex], followed by a reflection across the x-axis

Learn more here:

https://brainly.com/question/12382913

Answer:

Original / New

P′(4 , 6)  P″(2 , 3)

Q′(2 , 2) Q″(1 , 1)

R′(6 , 2)  R″(3 , 1)

S′(6 , 6) S″(3 , 3)

I hope this helps :)

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