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F.5 Maths Equation of st. line
發問:
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... 顯示更多 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!! 更新: RE 知足常樂： 我是用altitude去找△AQP的面積， 答案亦和胡雪8°先生的一樣。 我想是BOQP計錯了@@ 我的計算； http://postimg.org/image/4jywrlrwz/ 應該是有星星的那張圖才有錯的地方。 雖然用similar triangle的確快很多， 但我想知道本來的方法錯了甚麼（不過我還是找不出BOQP在哪裏錯了@@） 請問可以幫我檢查一下嗎？ 更新 2: 另外「AQ is greater than the sum of AP and PQ」我打錯了， 應該是 「PQ」 is greater than the sum of AP and 「AQ」， 不過我知道這個計錯甚麼了， 我計到PQ=15.25，忘記了要開方=_=
最佳解答:
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 L-1: 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 L? ⊥ L? 及 L? ⊥ L?。 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
F.5 Maths Equation of st. line
發問:
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... 顯示更多 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!! 更新: RE 知足常樂： 我是用altitude去找△AQP的面積， 答案亦和胡雪8°先生的一樣。 我想是BOQP計錯了@@ 我的計算； http://postimg.org/image/4jywrlrwz/ 應該是有星星的那張圖才有錯的地方。 雖然用similar triangle的確快很多， 但我想知道本來的方法錯了甚麼（不過我還是找不出BOQP在哪裏錯了@@） 請問可以幫我檢查一下嗎？ 更新 2: 另外「AQ is greater than the sum of AP and PQ」我打錯了， 應該是 「PQ」 is greater than the sum of AP and 「AQ」， 不過我知道這個計錯甚麼了， 我計到PQ=15.25，忘記了要開方=_=
最佳解答:
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 L-1: 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.
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其他解答: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 L? ⊥ L? 及 L? ⊥ L?。 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
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