幾何圖形及坐標計算

1.
(a)
PQ的中點
= ((0+3)/2, (4 + 0)/2)
= (3/2, 2)

PQ 的斜率
= (4 - 0)/(0 - 3)
= -4/3

所求直線的斜率
= -1/(-4/3)
= 3/4

所求直線的方程：
y - 2 = (3/4)[x - (3/2)]
4(y - 2) = 3[x - (3/2)]
4y - 8 = 3x - (9/2)
8y - 16 = 6x - 9
6x - 8y + 7 = 0

(b)
把 x = 0 代入方程 6x - 8y + 7 = 0 中：
0 - 8y + 7 = 0
y = 7/8
R = (0, 7/8)

(c)
ΔPQR面積
= (1/2) x [4 - (7/8)] x (3/2)
= 75/32 平方單位


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2.
由於 DB = BC，故此 ΔBDC 為等腰三角形。
∠BDC = ∠BCD = 30°

三角形外角等於兩內對角之和：
∠ABC = ∠BDC + ∠BCD
∠ABC = 30° + 30°
∠ABC = 60°

對同弧圓心角為圓周角的兩倍：
∠AOD = 2∠ABD
∠AOD = 2 x 60°
∠AOD = 120°

周角：
y + ∠AOD = 360°
y + 120° = 360°
y = 240°


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3.
(a)
∠ADB = ∠ACB (對同弧圓周角)
∠ADB = 30°

由於 AD = BD，故此 ΔDAB 為等腰三角形。
∠DAB = ∠DBA (等腰三角形兩底角)

∠ADB + ∠DAB + ∠DBA = 180° (Δ內角和)
30° + ∠DBA + ∠DBA = 180°
∠DBA = 75°

由於 ∠ADE = ∠DBA = 75°
故此 EF 是圓於 D 的切線 (圓切角等於對所夾弧的圓周角)

(b)
∠ABC = 90° (半圓內圓周角)

∠ABC + ∠ACB + ∠CAB = 180° (Δ內角和)
90° + 30° + ∠CAB = 180°
∠CAB = 60°

由於 ∠CAD + ∠CAB = ∠DAB
故此 ∠CAD + 60° = 75°
∠CAD = 15°