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F.4 MATHS..................................... 圖片參考：http://imgcld.yimg.com/8/n/HA00002921/o/701007160100213873369590.jpg
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15.(a) ΔOAB is an Isosceles triangle ∠ABO = ∠AOB OA = AB, OP = BP ΔAOP & ΔABP are congruent triangles ∠APB + ∠APO = 180° ∠APB = ∠APO = 90° AP⊥OB (b)(i) Slope of OB = -2 Slope AP = (-1) / (-2) = 1/2 By point slope form, AP: y - 0 = (1/2)[x - (-5)] AP: 2y = x + 5 Solve AP and OB: 2y = x + 5 ...(1) y = -2x ... (2) (1) * 2 + (2): 4y + y = 2x + 10 -2x 5y = 10 y = 2 x = -1 P (-1, 2) sub x = 0 into (1) y = 5/2 Q (0, 5/2) (b)(ii) B (-2, 4) (c) By two point form, equation of line BQ: (y - 4)/(5/2 - 4) = [x - (-2)] / [0 - (-2)] 2(y - 4)/(-3) = (x + 2) / 2 4y - 16 = -3x - 6 4y = -3x + 10 when y = 0 x = 10/3 R (10/3, 0) Length of BQ = √[(-2 - 0)^2 + (4 - 5/2)^2] = 2.5 Length of QR = √[(0 - 10/3)^2 + (5/2 - 0)^2] = 25/6 BQ / QR = 2.5 / (25/6) = 3 / 5 BQ : QR = 3 : 5
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F.4 MATHS發問:
F.4 MATHS..................................... 圖片參考：http://imgcld.yimg.com/8/n/HA00002921/o/701007160100213873369590.jpg
最佳解答:
15.(a) ΔOAB is an Isosceles triangle ∠ABO = ∠AOB OA = AB, OP = BP ΔAOP & ΔABP are congruent triangles ∠APB + ∠APO = 180° ∠APB = ∠APO = 90° AP⊥OB (b)(i) Slope of OB = -2 Slope AP = (-1) / (-2) = 1/2 By point slope form, AP: y - 0 = (1/2)[x - (-5)] AP: 2y = x + 5 Solve AP and OB: 2y = x + 5 ...(1) y = -2x ... (2) (1) * 2 + (2): 4y + y = 2x + 10 -2x 5y = 10 y = 2 x = -1 P (-1, 2) sub x = 0 into (1) y = 5/2 Q (0, 5/2) (b)(ii) B (-2, 4) (c) By two point form, equation of line BQ: (y - 4)/(5/2 - 4) = [x - (-2)] / [0 - (-2)] 2(y - 4)/(-3) = (x + 2) / 2 4y - 16 = -3x - 6 4y = -3x + 10 when y = 0 x = 10/3 R (10/3, 0) Length of BQ = √[(-2 - 0)^2 + (4 - 5/2)^2] = 2.5 Length of QR = √[(0 - 10/3)^2 + (5/2 - 0)^2] = 25/6 BQ / QR = 2.5 / (25/6) = 3 / 5 BQ : QR = 3 : 5
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