Question

cari jawaban yg paling tepat dan benar pakai cara nya ​

Answer 1

Verified answer

Materi : Translasi Geometri

P( x,y ) => T[ a,b ] => P'( x+a, y+b )

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Bagian A

T[ 2,-3 ]

x' = x + 2 [ x = x' - 2 ]

y' = y - 3 [ y = y' + 3 ]

---

( x - 2 )² + ( y + 4 )² = 16

( x' - 4 )² + ( y' + 7 )² = 16

x'² - 8x' + 16 + y'² + 14y' + 49 = 16

x'² + y'² - 8x' + 14y' + 49 = 0

Bagian B

T[ -2,4 ]

x' = x - 2 [ x = x' + 2 ]

y' = y + 4 [ y = y' - 4 ]

---

( x - 2 )² + ( y + 4 )² = 16

( x' )² + ( y' )² = 16

x'² + y'² = 16

Bagian C

T[ 3,5 ]

x' = x + 3 [ x = x' - 3 ]

y' = y + 5 [ y = y' - 5 ]

---

( x - 2 )² + ( y + 4 )² = 16

( x' - 5 )² + ( y' - 1 )² = 16

x'² + y'² - 10x' - 2y' + 10 = 0

Semoga bisa membantu

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Answer 2

Jawaban:

x'=x+a

y'=y+b

a.

(x-2)²+(y+4)²=16

Translasin(2,-3) maka bentuknya

x=x'-2

y=y'+3

(x'-4)²+(y'+7)²=16

b.

(x-2)²+(y+4)²=16

Translasi(-2,4) maka bentuknya

x=x'+2

y=y'-4

(x')²+(y')²=16

x'²+y'²=16

c.

(x-2)²+(y+4)²=16

Translasi(3,5) maka bentuknya

x=x'-3

y=y'-5

(x'-5)²+(y'-1)²=16

Semoga bermanfaat