Picture: http://postimg.org/image/c9eep9isr/
In the figure, the straight line L1: 4x+3y-9=0 passes through P(3,-1). It cuts the
positive y-axis and the straight line L2: 5x-6y=0 at A and B respectively. The
straight line L3: 5x-6y-21=0 passes through P and parallel to L2.
If L3 cuts the y-axis at Q, find the ratio of the area of APQ to the area of the
quadrilateral BOQP
I found that the figure is just like this:
http://postimg.org/image/nybllszd5/
And I found that A=(0,3) , Q=(0, -7/2)
Since BOQP should be a trapezium, I tried to find the length of OB, QP and BP in order
to find the area of BOQP.
However, the ratio I found(1.269) has a little bit difference with the correct answer
169/133 (=1.271). And I found that I should make a mistake in calculating the area of △APQ as the length AQ is greater than the sum of AP and PQ. But I still cannot calculate
the correct answer.
I want to ask what is/are the mistake(s) that I make, Thank you!!
F.5 Maths Equation of st. line
2015-03-16 1:37 am
回答 (2)
2015-03-16 4:13 am
✔ 最佳答案
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.
2015-03-16 2:00 am
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
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
收錄日期: 2021-04-15 18:47:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150315000051KK00089