F.3 MATHS (1 question)

恕我冒昧,我不知道為甚麼要provide ABC is a right-angled triangle,所以,希望其他高手提供alternative method
AP = sq rt [(x-0)^2 + (y-3)^2] = sq rt (x^2 + y^2 -6y + 9)
BP = sq rt [(-1-x)^2 + y^2] = sq rt (1 + 2x + x^2 + y^2)
CP = sq rt [(9-x)^2 + y^2] = sq rt (81 - 18x + x^2 + y^2)
AP = BP
-6y+9 = 1 + 2x
2x + 6y = 8
x + 3y = 4...(1)

BP = CP
1+2x = 81 -18x
x = 4

Sub x = 4 into (1),we get y = 0
Thus,the cooordinates of P are (4,0)

2012-06-01 21:31:45 補充：
可能上面寫得不夠詳細
Sub x = 4 into (1),
4 + 3y = 4
y = 0
so,we get y = 0

TO 。文。
謝謝額外的知道講解,但以我的水平未學所說的圓形的特性
但也可作參考

right-angled triangle 的確係有幫助架!!
同以下有一個圓形的特性有關:

let P,Q be two points on a circle，if centre O lies on PQ
ie, PQ is the diameter, then ∠PRQ is a right angle,
where R is another point on the circle
ie, ΔPQR is right-angled triangle

Note:
1. OP,OQ,OR have the same length (radii)
2. PQ, the diameter, is the hypotenuse of the right-angled triangle
(我讀舊學制，以前F4先學這特性，唔知你地新制幾時教)


睇返你條題目:
ΔABC is right-angled triangle, sketch it out,
BC is hypotenuse obviously.

according to the mentioned properties of circle
we can draw a circle passing through all the points A, B, C,
and the centre of such circle should be the mid point of BC
(since BC is the hypotenuse, ie, the diameter)

the centre is a point equidistant from A, B and C
so the required point P is the mid point of BC
since both B, C lies on the line y=0 (the x-axis)
y coordinate of P will also be 0

2012-06-01 22:07:17 補充：
x-coordinate of P = (9+(-1))/2 = 4

當然，
F3可能未教個property...
你用返Ans001 個方法最安全(噓!! 你要知道...大部分學校老師都是忌才的...)
你睇下當消磨下時間啦......( /_>\ )

2012-06-01 22:14:08 補充：
y=0 是算出來的，不是絕對的
改一改A,B,C的coordinate的話，什麼都變了
P的y-coordinate都會變

2012-06-02 12:09:26 補充：
其實有個更加簡單的方法
記唔記得circumcentre of triangle?
circumcentre of a triangle is a point with equal distance from the vertices
circumcentre係三條邊的perpendicular bisectors的intersection

obviously, perpendicular bisector of BC is x=4
find the remaining perpendicular bisectors' eqt. and sub x =4
we get y=0

我覺得你應該將e個問題設為<快問快答>會比較好D