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Pleaseee help me....what will the y-coordinate of b' if ab is dilated by a factor of 2/5, centered at the origin. A=(2,2) B=(10,2)

Pleaseee help mewhat will the ycoordinate of b if ab is dilated by a factor of 25 centered at the origin A22 B102 class=

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calculista

Answer:

The y-coordinate of B' is equal to 2

Step-by-step explanation:

step 1

Find the distance of AB

[tex]AB=10-2=8\ units[/tex]

step 2

Find the length of A'B'

Multiply the length of AB by the scale factor

[tex]A'B'=8*\frac{2}{5}= 3.2\ units[/tex]

step 3

Find the midpoint AB

[tex]M=(\frac{2+10}{2},\frac{2+2}{2} )[/tex]

[tex]M=(6,2)[/tex]

step 4

Find the x-coordinate of B'

The x-coordinate of B' is equal to the x-coordinate of midpoint plus the semi distance A'B'

[tex]6+\frac{3.2}{2}=6+1.6=7.6[/tex]

Find the y-coordinate of B'

The y-coordinate of B' is equal to 2 (because is a horizontal line)

Answer:

The y-coordinate of point B' is [tex]\frac{4}{5}[/tex] or 0.8.

Step-by-step explanation:

If a figure dilated by scale factor k, centered at the origin, then

[tex](x,y)\rightarrow(kx,ky)[/tex]

It is given that the coordinates of A are (2,2) and coordinates of B are (10,2).

The scale factor is [tex]\frac{2}{5}[/tex], so the coordinates of image are

[tex]A(2,2)\rightarrowA'(\frac{2}{5}(2),\frac{2}{5}(2))\Rightarrow A'(\frac{4}{5},\frac{4}{5})=A'(0.8,0.8)[/tex]

[tex]B(10,2)\rightarrowB'(\frac{2}{5}(10),\frac{2}{5}(2))\Rightarrow B'(4,\frac{4}{5})=B'(4,0.8)[/tex]

Therefore the y-coordinate of point B' is [tex]\frac{4}{5}[/tex] or 0.8.

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