F.5 Maths Equation of st. line

Let (0, a) be the coordinates of A, and (0, q)be the coordinates of Q, and (n, m) be the coordinates of B.

A(0, a) lies on L­1: 4x + 3y - 9 = 0
4(0) + 3a - 9 = 0
a = 3
Hence, the coordinates of A = (0, 3)

Q(0, q) lies on L3 : 5x - 6y - 21 = 0
5(0) - 6q - 21 = 0
q = -7/2
Hence, the coordinates of Q = (0, -7/2)

B(n, m) lies on both L1 and L2.
4n + 3m - 9 = 0 ...... [1]
5n - 6m = 0 ...... [2]

[1]*2 :
8n + 6m - 18 = 0 ...... [3]

[2] + [3] :
13n - 18 = 0
n = 18/13

Put n = 18/13 into [2] :
5(18/13) - 6m = 0
m = 15/13

Hence, the coordinates of B = (18/13, 15/13)

Area of ΔAPQ
= (1/2) * AQ * (x-coordinate of P)
= (1/2) * [3 + (7/2)] * 3
= 39/4

Area of ΔABO
= (1/2) * AO * (x-coordinate of B)
= (1/2) * 3 * (18/13)
= 27/13

Area of quad. BOQP
= (39/4) - (27/13)
= (507/52) - (108/52)
= 399/52

Area of ΔAPQ : Area of quad. BOQP
= (39/4) : (399/52)
= (507/52) : (399/52)
= 507 : 399
= 169: 133

2015-03-15 20:34:23 補充：
Alternative method to calculate the area of quad. BOQP, where quad. BOQP is a trapezium :

OB = √{[(18/13) - 0]² + [(15/13) - 0]²} = (3/13)√61
PQ = √{(3 - 0)² + [-1 - (-7/2)]² = √(117/4) = (1/2)√61
BP = |5(3) - 6(-1)| / √(5² + 6²) = 21/√61

2015-03-15 20:34:39 補充：
Area of quad. BOQP
= (1/2) * (OB + PQ) * BP
= (1/2) * {[(3/13)√61] + [(1/2)√61]} * (21/√61)
= (1/2) * [(3/13) + (1/2)] * 21
= (1/2) * [(6/26) + (13/26)] * 21
= 399/52

2015-03-15 20:35:56 補充：
You have not shown your answer, and thus no one knows what your mistake is.

2015-03-15 20:39:49 補充：
Sorry. In the above alternative method, "BP" should be "The height of the trapezium" instead.

YTC，你沒有交代你的各個數字，所以很難指出你哪裏錯了。
而且，我「懷疑」你「誤會」了那些三角形是直角三角形，但其實沒有證據指出它們是，（其實它們也不是）。

但是，我應該教你一個簡單一點的方法去計算本題。

你留意到 △AOB 和 △AQP 是 similar 的嗎？
所以你可以利用 AO 和 AQ 的長度去計算比例關係。

AO = 3
AQ = 3 + 3.5 = 6.5

AO : AQ = 3 : 6.5 = 6 : 13

△AOB ~ △AQP，因此，面積比例 = 6² : 13² = 36 : 169

2015-03-15 18:01:08 補充：
所以，所求的比例 = 169 : (169 - 36) = 169 : 133

2015-03-15 18:07:42 補充：
題目在以上的講解已經解決了，但仍幫你檢查一下吧。

你說「AQ is greater than the sum of AP and PQ」
那應該是不對的。

A = (0, 3)
P = (3, -1)
Q = (0, -3.5)

AQ = 6.5
AP = √(9 + 16) = 5
PQ = √(9 + 6.25) = √15.25 ≈ 3.9

AQ < AP + PQ 是必然的。

2015-03-16 23:58:33 補充：
YTC：
你再看我的意見 001，我的懷疑是對的。
你真的誤會了以為 L₁ ⊥ L₂ 及 L₁ ⊥ L₃。
其實它們不是垂直的！！！

如果你真的要計△AQP 的面積，
你要考慮AQ為底 (6.5)
而 P 的 x-coordinate 為高，
即 Area = (3)(6.5)/2 = 9.75

同樣地，Area of △AOB
= (3)(18/13)/2 = 27/13

因此，Area of BOQP = 27/13 - 9.75 = 399/52

所求的比是 9.75 : 399/52 = 169 : 133