Question

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)

Answer 1

Answer:

The y-coordinate of B' is equal to 2

Step-by-step explanation:

step 1

Find the distance of AB

$AB=10-2=8\ units$

step 2

Find the length of A'B'

Multiply the length of AB by the scale factor

$A'B'=8*\frac{2}{5}= 3.2\ units$

step 3

Find the midpoint AB

$M=(\frac{2+10}{2},\frac{2+2}{2} )$

$M=(6,2)$

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'

$6+\frac{3.2}{2}=6+1.6=7.6$

Find the y-coordinate of B'

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

Answer 2

Answer:

The y-coordinate of point B' is $\frac{4}{5}$ or 0.8.

Step-by-step explanation:

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

$(x,y)\rightarrow(kx,ky)$

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

The scale factor is $\frac{2}{5}$, so the coordinates of image are

$A(2,2)\rightarrowA'(\frac{2}{5}(2),\frac{2}{5}(2))\Rightarrow A'(\frac{4}{5},\frac{4}{5})=A'(0.8,0.8)$

$B(10,2)\rightarrowB'(\frac{2}{5}(10),\frac{2}{5}(2))\Rightarrow B'(4,\frac{4}{5})=B'(4,0.8)$

Therefore the y-coordinate of point B' is $\frac{4}{5}$ or 0.8.