標題:
發問:
Find the equation of the inscribed circle of triangle ABC if the sides of triangle are given as follows. AB: 3x+4y-19=0, BC:12x-5y+53=0, AC: y=1.
最佳解答:
AB : 3x + 4y - 19 = 0 ---------- (1) BC : 12x - 5y + 53 = 0 ---------- (2) AC : y = 1 ---------- (3) from (1) and (3), 3x + 4(1) - 19 = 0, x = 5, A = (5,1) from (2) and (3), 12x - 5(1) + 53 = 0, x = -4, C = (-4,1) from (1), 12x + 16y - 76 = 0 ---------- (4) (4) - (2), 21y - 129 = 0, y = 43/7 12x - 5(43/7) + 53 = 0, x = -13/7 B = (-13/7, 43/7) AC = 9 AB = sqrt[ (-13/7 - 5)^2 + (43/7 - 1)^2 ] = 60/7 BC = sqrt[ (-13/7 + 4)^2 + (43/7 - 1)^2 ] = 39/7 AB + BC + AC = 162/7 incenter (centre of circle) = [(39/7) / (162/7)] (5,1) + [9 / (162/7)](-13/7, 43/7) + [(60/7) / (162/7)](-4,1) = (13/54) (5,1) + (7/18) (-13/7, 43/7) + (10 / 27) (-4,1) = (-1,3) radius = 2 circle : (x + 1)^2 + (y - 3)^2 = 4
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AMath-Circle發問:
Find the equation of the inscribed circle of triangle ABC if the sides of triangle are given as follows. AB: 3x+4y-19=0, BC:12x-5y+53=0, AC: y=1.
最佳解答:
AB : 3x + 4y - 19 = 0 ---------- (1) BC : 12x - 5y + 53 = 0 ---------- (2) AC : y = 1 ---------- (3) from (1) and (3), 3x + 4(1) - 19 = 0, x = 5, A = (5,1) from (2) and (3), 12x - 5(1) + 53 = 0, x = -4, C = (-4,1) from (1), 12x + 16y - 76 = 0 ---------- (4) (4) - (2), 21y - 129 = 0, y = 43/7 12x - 5(43/7) + 53 = 0, x = -13/7 B = (-13/7, 43/7) AC = 9 AB = sqrt[ (-13/7 - 5)^2 + (43/7 - 1)^2 ] = 60/7 BC = sqrt[ (-13/7 + 4)^2 + (43/7 - 1)^2 ] = 39/7 AB + BC + AC = 162/7 incenter (centre of circle) = [(39/7) / (162/7)] (5,1) + [9 / (162/7)](-13/7, 43/7) + [(60/7) / (162/7)](-4,1) = (13/54) (5,1) + (7/18) (-13/7, 43/7) + (10 / 27) (-4,1) = (-1,3) radius = 2 circle : (x + 1)^2 + (y - 3)^2 = 4
其他解答:
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