Mathematics: Can someone me ( picture below) : )

Can someone me ( picture below) : ) - id893690, gabriellabadon2

davidoj13

18.04.2021

Given:

The vertices of a parallelogram are A (-2,3), B (3,3), C (4,6), and D(-1,6).

It is first rotated 90 degrees clockwise and then translated 4 units left and 2 units down to form quadrilateral A'B'C'D.

To find:

The distance between C' and D'.

Solution:

If a figure rotated 90 degrees clockwise, then

$(x,y)\to (y,-x)$

$C(4,6)\to C_1(6,-4)$

$D(-1,6)\to D_1(6,-(-1))=D_1(6,1)$

If figure translated 4 units left and 2 units down, then

$(x,y)\to (x-4,y-2)$

$C_1(6,-4)\to C'(6-4,-4-2)=C'(2,-6)$

$D_1(6,1)\to D'(6-4,1-2)=D'(2,-1)$

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between C'(2,-6) and D'(2,-1) is

$C'D'=\sqrt{(2-2)^2+(-1-(-6))^2}$

$C'D'=\sqrt{(0)^2+(-1+6)^2}$

$C'D'=\sqrt{(5)^2}$

C'D'=5

Therefore, the distance between C'D' is 5 units.

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