(((20點!!!急求救))求P和Q的長度
已知直線A與B相交於A(-2,-5),而它們的斜率分別為4和-2。若A和B分別與y軸相交於P和Q,求P和Q的長度。
(((20點!!!急求救))求P和Q的長度
2008-10-12 10:41 pm
回答 (2)
2008-10-12 11:37 pm
直線A 有 點 A(-2,-5)與P(0,p) , 斜率為4
4 = (p-(-5))/(0-(-2))
p = 4*(0-(-2))+(-5) = 8-5 = 3
=> P(0,3)
直線B 有 點 A(-2,-5)與Q(0,q), 斜率為 -2
-2 = (q-(-5))/(0-(-2))
q = (-2)*(0-(-2))+(-5) = -4-5 = -9
=> Q(0,-9)
Distance of PQ = p-q = 3-(-9) = 12
.---------/+P(0,p)
.--------/-|
(-2,-5)/--|
.--------\-|
.---------\+Q(0,q)
4 = (p-(-5))/(0-(-2))
p = 4*(0-(-2))+(-5) = 8-5 = 3
=> P(0,3)
直線B 有 點 A(-2,-5)與Q(0,q), 斜率為 -2
-2 = (q-(-5))/(0-(-2))
q = (-2)*(0-(-2))+(-5) = -4-5 = -9
=> Q(0,-9)
Distance of PQ = p-q = 3-(-9) = 12
.---------/+P(0,p)
.--------/-|
(-2,-5)/--|
.--------\-|
.---------\+Q(0,q)
2008-10-12 10:47 pm
Line A's Y-intercept = 4*(0-(-2))+(-5) = 3
Therefor, P=(0,3)
Line B's Y-intercept = -2*(0-(-2))+(-5) = -9
Therefor, Q=(0,-9)
Distance of PQ = 3-(-9) = 12
Therefor, P=(0,3)
Line B's Y-intercept = -2*(0-(-2))+(-5) = -9
Therefor, Q=(0,-9)
Distance of PQ = 3-(-9) = 12
參考: me
收錄日期: 2021-04-13 16:09:19
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https://hk.answers.yahoo.com/question/index?qid=20081012000051KK01146