Question

如图，AB解析式为y=1/3x+1,C(0,-1),直线MN交AB于M点，交AC与N点,MH⊥BC，交y轴于H点，若BM=CN，求PH的值

Answer 1

似乎少一个条件。

A(-3, 0), B(0,1), C(0. -1)
MN: y = kx + b
P(0. b)

AC: x/(-3) + y/(-1) = 1, y = -x/3 - 1
与MN联立: N(-3(b+ 1)/(3k + 1), (b - 3k)/(3k + 1))
MN与AB联立: M(3(1 - b)/(3k - 1), (3k - b)/(3k - 1)), H(0, (3k - b)/(3k - 1))

BM² = CN²
BM² = 10(b-1)²/(3k -1)²
CN² = 10(b + 1)²/(3k + 1)²
(b + 1)/(3k + 1) = ±(b - 1)/(3k - 1)
(i)(b + 1)/(3k + 1) = (b - 1)/(3k - 1)
b = 3k
P(0, 3k), H(0, 0)
PH = 3|k|

(ii)(b + 1)/(3k + 1) = -(b - 1)/(3k - 1)
b = 1/(3k)
P(0, 1/(3k)), H(0, (3k + 1)/(3k))
PH = (3k + 1)/(3k) - 1/(3k) = 1