[問題] 高一數學

(short)(-15074)

2009-08-22 05:30:51 UTC

Post by ...
想請問一題數學：
過原點之直線將頂點A(1,0) B(0,1) C(3/2,0)之三角形面積平分
求此直線方程式? Ans：x-3y=0
把圖畫出來後，過原點的線似乎會把三角形切成一個小的三角形
和一個四邊形，這樣要怎麼知道面積有沒有平分呢?
而且高一還沒教行列式，該怎麼解呢?

首先 △ABC的面積是 (1/2)*1/2=1/4

設所求的線是 y=kx

易知直線 AB 為 x+y=1
直線 BC 為 2x+3y=3

若設所求線與 AB BC 的交點分別為 D E 原點為 O

則 △ADE = △AOE - △AOD
= (AO * E的x座標)/2 + (AO * D的x座標)/2
= AO * (E的x座標 - D的x座標) /2

計算可得 D的x座標為 1/(k+1) E的x座標為 3/(3k+2)

故可列式: 1/8 = 1 * (3/(3k+2)-1/(k+1))/2

計算略可解得 k = 1/3 or -2 (不合)

即得所求線為 y=x/3 => x-3y=0

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