a點的坐標係(5,20)b點的坐標係(-2,1) p點的x坐標係y坐標2倍 若pb=pa 求p點的坐標~~
我知道答案但係唔識step~~求詳細step~
答案係(8/5,4/5)
一條單o既坐標a maths
2007-05-25 12:50 am
回答 (3)
2007-05-26 4:35 am
✔ 最佳答案
Let the coordinates of P be (2y,y). (Since the magnitude of x-coordinate of P twice that of y-coordinate.)
∵ PB = PA
∴ √[(5 - 2y)² + (2 - y)²] = √[(-2 - 2y)² + (1 - y)²]
(5 - 2y)² + (2 - y)² = (-2 - 2y)² + (1 - y)²
(25 - 20y + 4y²) + (4 - 4y + y²) = (4 + 8y + 4y²) + (1 - 2y + y²)
5y² - 24y + 29 = 5y² + 6y + 5
30y - 24 = 0
5y - 4 = 0
y = 4/5
∴ x = 2y = 8/5
Therefore, the coordinates of P is (8/5 , 4/5).
參考: Myself~~~
2007-05-25 3:06 am
設p點為(2t,t)
sqrt[ (2t-5)^2 + (t-2)^2 ] = sqrt[ (2t+2)^2 + (t-1)^2 ]
take square and simply,
t = 4/5
so, p = (8/5,4/5)
just give you the important hint,
you can do the simple calculation
sqrt[ (2t-5)^2 + (t-2)^2 ] = sqrt[ (2t+2)^2 + (t-1)^2 ]
take square and simply,
t = 4/5
so, p = (8/5,4/5)
just give you the important hint,
you can do the simple calculation
2007-05-25 1:16 am
設p=(2x,x)
由pb=pa知
sqrt((5-2x)^2+(20-x)^2)=sqrt((-2-2x)^2+(1-x)^2)
即 (5-2x)^2+(20-x)^2=(2+2x)^2+(1-x)^2
即 -66x+420=0
即 x=70/11
p=(140/11,70/11)
2007-05-27 22:33:53 補充：
:-)，唔駛
由pb=pa知
sqrt((5-2x)^2+(20-x)^2)=sqrt((-2-2x)^2+(1-x)^2)
即 (5-2x)^2+(20-x)^2=(2+2x)^2+(1-x)^2
即 -66x+420=0
即 x=70/11
p=(140/11,70/11)
2007-05-27 22:33:53 補充：
:-)，唔駛
參考: check 左幾次都係咁
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