F.3 Coordinate Geometry (15 分)

25.
(a) T is the mid-point of PR, so T = (5/2, 3)
(b) QS = 4, QT = 2.
PR = sqrt[3^2+4^2] = 5, so PT = 5/2
For a rhombus, QS ans PR are perpendicular, so PQ = sqrt[2^2 + (5/2)^2] = sqrt(41)/2

26.
(a) First find that the equation of BC is x/(-2)+y/3=1, that is, 3x-2y+6=0, the slope is 3/2. Therefore, the slope of AP is -2/3. Consider the equation of AP as (y-0)/(x-1) = -2/3, that is 3y = 2-2x. that is 2x+3y-2=0. When x=0, y=2/3, so H=(0,2/3)

(b) When AB is the base, OC is an altitude, which passes through H. Since two altitudes of triangle ABC pass through H, H is the orthocentre of triangle ABC.

2013-08-19 22:33:46 補充：
Re: 皎潔明月

是用了畢氏定理 (Pythagoras Theorem)。
也用了菱形特性, 兩對角線互相垂直, 互相平分。

Masterijk,
25(b) 我唔明你個方法. 我只諗到用畢氏定理......