这道题怎么做?在线等急!!
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(1)
令其斜率为k (k < 0), y - 1 = k(x - 2)
分别令y = 0和x = 0, 得B(2 - 1/k, 0), C(0, 1 - 2k)
△OBC的面积 = (1/2)*OB*OC = (1/2)(2 - 1/k)(1 - 2k)=(1/2)(4 - 4k - 1/k)
-4k = -1/k时, 面积最小, 此时4k = 1/k, k = -1/2 (舍去k = 1/2 > 0)
y = 1 - (x - 2)/2 = -x/2 + 2
(2)
|OB| + |OC| = |2 - 1/k| + |1 - 2k| = 2 - 1/k + 1 - 2k = 3 - 2k - 1/k
-2k = -1/k时, 和最小, 此时k = -1/√2 (舍去k = 1/√2 > 0)
y = 1 - (x - 2)/√2 = -x/√2 + 1 + √2
(3)
相当于|AB|²|AC|²最小
|AB|² = (2 - 1/k - 2)² + (0 - 1)² = 1/k² + 1
|AC|² = (2 - 0)² + (1 - 1 + 2k)² = 4k² + 4
其积为p = 4(1/k² + 1)(k² + 1) = 4(2 + k² + 1/k²)
k² = 1/k²时, p最小, 此时k² = 1, k = -1 (舍去k = 1>0)
y = 1 - (x - 2) = -x + 3
令其斜率为k (k < 0), y - 1 = k(x - 2)
分别令y = 0和x = 0, 得B(2 - 1/k, 0), C(0, 1 - 2k)
△OBC的面积 = (1/2)*OB*OC = (1/2)(2 - 1/k)(1 - 2k)=(1/2)(4 - 4k - 1/k)
-4k = -1/k时, 面积最小, 此时4k = 1/k, k = -1/2 (舍去k = 1/2 > 0)
y = 1 - (x - 2)/2 = -x/2 + 2
(2)
|OB| + |OC| = |2 - 1/k| + |1 - 2k| = 2 - 1/k + 1 - 2k = 3 - 2k - 1/k
-2k = -1/k时, 和最小, 此时k = -1/√2 (舍去k = 1/√2 > 0)
y = 1 - (x - 2)/√2 = -x/√2 + 1 + √2
(3)
相当于|AB|²|AC|²最小
|AB|² = (2 - 1/k - 2)² + (0 - 1)² = 1/k² + 1
|AC|² = (2 - 0)² + (1 - 1 + 2k)² = 4k² + 4
其积为p = 4(1/k² + 1)(k² + 1) = 4(2 + k² + 1/k²)
k² = 1/k²时, p最小, 此时k² = 1, k = -1 (舍去k = 1>0)
y = 1 - (x - 2) = -x + 3
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