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Kita teruskan dgn bab akhir 
dlm Geometri Koordinat.

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Kali ini mengenai Lokus.

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Cth soalan: Titik bergerak, P (x,y)

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bergerak pd jarak malar

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dari titik A dan B.

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Jika ditafsir kpd persamaan Matematik,

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PA = PB

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Jarak PA = jarak PB

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Sekarang kita guna formula jarak,

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x2 = x, x1 = -1,

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maka, PA: 
√ [(x - (-1))2 + (y - 1)2] =

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PB: 
√ [(x - 5)2 + (y - (-1))2]

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Jadi,


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√ [(x +1)2 + (y - 1)2] = √ [(x - 5)2 + (y +1)2]

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Bagaimana nak hapuskan kesemua √ ?

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Caranya dgn kuasa-dua kan 
persamaan kiri dan kanan.

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Maka tiada lagi √ .

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(x + 1)2 + (y - 1)2 = (x-5)2 + (y + 1 )2

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Kembangkan persamaan ini.

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PA: 
x2 + 2x + 1 + y2 - 2y + 1

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PB:
x2 - 10x + 25 + y2 + 2y + 1

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Sekarang, jadikan persamaan ini = 0

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Kumpulkan semua anu ke sebelah kiri.

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x2 & y2 ada di kedua-dua belah.

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Jadi ia boleh terus dihapuskan.

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Maka,
2x - 2y + 2 +10x -25 -2y - 1= 0

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Ringkaskan persamaan ini.

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12x - 4y - 24 = 0

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÷ persamaan ini dgn 4

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3x - y - 6 = 0

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Inilah cara utk dapatkan persamaan Lokus.

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Anda cuma tafsirkan ayat 
ke dlm bentuk persamaan Matematik.