Question

Parallelogram ABCD is translated (x + 3, y − 2) and then rotated 90° about the origin in the clockwise direction. Complete the table to show the locations of A″, B″, C″, and D″ after both transformations.
Parallelogram ABCD is shown. A is at negative 5, 1. B is at negative 4, 3. C is at negative 1, 3. D is at negative 2, 1.


A (− 5, 1) A″ ?
B (−4, 3) B″ ?
C (−2, 3) C″ ?
D (−1, 1) D″ ?

A″ (−1, 2), B″ (1, 1), C″ (1, −2), D″ (−1, −1)
A″ (1, −2), B″ (−1, −1), C″ (−1, 2), D″ (1, 1)
A″ (−2, −1), B″ (−1, 1), C″ (2, 1), D″ (1, −1)
A″ (1, 1), B″ (−1, 2), C″ (−1, −1), D″ (1, −2)

Answer 1

Answer: A″ (−1, 2), B″ (1, 1), C″ (1, −1), D″ (−1, −2)

Step-by-step explanation:

• there are basically 2 parts in the question.

• first one is the translation(easier one)

after translation, the corresponding points will be:

A (− 5, 1)   --> a(-5+3,1-2) = (-2,-1)

B (−4, 3)  --> b(-4+3,3-2) = (-1,1)

C (−2, 3)  --> c(-2+3,3-2) = (1,1)

D (−1, 1)    --> d(-1+3,1-2)  = (2,-1)

• now the second part:

let (x,y) be a point and if it is rotated by "$\alpha$"

then, the new coordinates (x',y') will be :

x'= $xcos\alpha +ysin\alpha$

y'=$-xsin\alpha +ycos\alpha$

here, $\alpha = 90 degrees$

so,

a(-2,-1)  --> A"$(-2cos[90]-1sin[90], -[-2]sin[90]+[-1]cos[90]) =(0-1, 2+0)=(-1,2)$

(since, cos[90]=0, sin[90]=1)

b(-1,1)    -->B"(-1cos[90]+1sin[90], -[-1]sin[90]+1cos[90])=(1,1)

c(1,1)     --> C"(1cos[90]+1sin[90],-[1]sin[90]+1cos[90]) = (1,-1)

d(2,-1)   --> D"(2cos[90]+[-1]sin[90],-2sin[90]+[-1]cos[90]) =(-1,-2)

these are the coordinates after both the transformations.

Answer 2

Answer:

The answer is A″ (−1, 2), B″ (1, 1), C″ (1, −2), D″ (−1, −1)

Step-by-step explanation:

I took the test and got the answer right