Kegiyatan pentas seni kasebut kanggoMahagya apa?

A n y p o in t o n t h e y a x i s c an b e s t a t e d a s C ( 0 , y ) D i s t an ce F or m u l a : G i v e n t h e tw o p o in t s ( x 1 ​ , y 1 ​ ) an d ( x 2 ​ , y 2 ​ ) , t h e d i s t an ce b e tw ee n t h ese p o in t s i s g i v e n b y t h e f or m u l a : d = ( x 2 ​ − x 1 ​ ) 2 + ( y 2 ​ − y 1 ​ ) 2 ​ C A = ( − 2 − 0 ) 2 + ( − 3 − y ) 2 ​ = ( − 2 ) 2 + ( − 3 − y ) 2 ​ = 4 + 9 + 6 y + y 2 ​ = − y 2 + 6 y + 13 ​
CB = ( 1 − y ) 2 + ( 6 − 0 ) 2 ​ = 1 − 2 y + y 2 + 36 ​ = y 2 − 2 y + 37 ​ C A = CB y 2 + 6 y + 13 ​ = y 2 − 2 y + 37 ​ ∣ 2 y 2 + 6 y + 13 = y 2 − 2 y + 37 y 2 + 6 y − y 2 + 2 y = 37 − 13 8 y = 24 / : 8 y = 3 C = ( 0 , 3 )

Answered by Lilith | 2024-06-10

Looking the centre of the circle on the y-axis passing through the points (6;1) and (-2,-3). The coordinates of the center of this a circle are equal y-intercept of line segment bisector.
( x − 6 ) 2 + ( y − 1 ) 2 = ( x + 2 ) 2 + ( y + 3 ) 2 x 2 − 12 x + 36 + y 2 − 2 y + 1 = x 2 + 4 x + 4 + y 2 + 6 y + 9 − 12 x − 2 y + 37 = 4 x + 6 y + 13 − 2 y − 6 y = 4 x + 12 x + 13 − 37 − 8 y = 16 x − 24 ∣ : ( − 8 ) y = − 2 x + 3 A n s w er : ( 0 ; 3 )

Answered by Anonymous | 2024-06-10

The point on the y-axis that is equidistant from the points (6, 1) and (-2, -3) is (0, 3). This was determined using the distance formula and setting the distances equal to find the value of y. After solving the equation, we found that y = 3, resulting in the point on the y-axis being C(0, 3).
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Answered by Lilith | 2024-12-24

Jawaban:Biasane, pentas seni kuwi digelar kanggo:°Nampilaké bakat lan kreativitas siswa•Sarana hiburan kanggo warga sekolah/masyarakat°Nguri-uri kabudayan (melestarikan budaya)•Ajang apresiasi seni

Answered by eckobachir1 | 2025-08-22