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ra1l [238]
3 years ago
5

Write a sequence of transformation that maps quadrilateral ABCD onto quadrilateral A. B. C. D IN THE PICTURE

Mathematics
1 answer:
Mice21 [21]3 years ago
5 0

Answer:

The sequence of transformation is reflected across the y-axis and translated 2 units down

Step-by-step explanation:

Lets revise some transformation

- If point (x , y) reflected across the x-axis

∴ Its image is (x , -y)

- If point (x , y) reflected across the y-axis

∴ Its image is (-x , y)

- If point (x , y) translate h units to the right

∴ Its image is (x + h , y)

- If point (x , y) translate h units to the left

∴ Its image is (x - h , y)

- If point (x , y) translate k units up

∴ Its image is (x , y + k)

- If point (x , y) translate k units down

∴ Its image is (x , y - k)

* Now lets solve the problem

∵ The vertices of figure ABCD are:

  A (-1 , 3) , B (1 , 0) , C (2 , 3) , D (1 , 4)

∵ The vertices of figure A"B"C"D" are:

  A" (1 , 1) , B" (-1 , -2) , C" (-2 , 1) , D" (-1 , 2)

* Lets compare between ABCD and A"B"C"D"

∵ All x-coordinates has opposite signs

  -1 ⇒ 1 , 1 ⇒ -1 , 2 ⇒ -2 , 1 ⇒ -1

∴ The ABCD is reflected across the y-axis

∵ All y-coordinates subtracted by 2

 3 ⇒ 1 , 0 ⇒ -2 , 3 ⇒ 1 , 4 ⇒ 2

∴ The ABCD is translated 2 units down

* The sequence of transformation is reflected across the y-axis

  and translated 2 units down

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14.       1.5, 10    <- Answer
15.       5,1          <- Answer

Proof 14 

Solve the following system:
{2 x - y = -7 | (equation 1)
4 x - y = -4 | (equation 2)
Swap equation 1 with equation 2:
{4 x - y = -4 | (equation 1)
2 x - y = -7 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{4 x - y = -4 | (equation 1)
0 x - y/2 = -5 | (equation 2)
Multiply equation 2 by -2:
{4 x - y = -4 | (equation 1)
0 x+y = 10 | (equation 2)
Add equation 2 to equation 1:
{4 x+0 y = 6 | (equation 1)
0 x+y = 10 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 3/2 | (equation 1)
0 x+y = 10 | (equation 2)
Collect results:
Answer: {x = 1.5               
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Proof 15. 

Solve the following system:
{5 x + 7 y = 32 | (equation 1)
8 x + 6 y = 46 | (equation 2)
Swap equation 1 with equation 2:
{8 x + 6 y = 46 | (equation 1)
5 x + 7 y = 32 | (equation 2)
Subtract 5/8 × (equation 1) from equation 2:{8 x + 6 y = 46 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Divide equation 1 by 2:
{4 x + 3 y = 23 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Multiply equation 2 by 4/13:
{4 x + 3 y = 23 | (equation 1)
0 x+y = 1 | (equation 2)
Subtract 3 × (equation 2) from equation 1:
{4 x+0 y = 20 | (equation 1)
0 x+y = 1 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 5 | (equation 1)
0 x+y = 1 | (equation 2)
Collect results:
Answer:  {x = 5                           y = 1
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