1.求過原點o ,且與已知直線y = kx + b 的夾角為銳角A的直線方程
2.已知直線(2m^2 + m - 3)x + (m^2 - m)y - 4m + 1 = 0 與 直線 x- 2y+ 6 = 0 的夾角為arctan3 ,求 m
F5的解幾
2007-11-04 4:54 am
回答 (2)
2007-11-04 5:59 am
✔ 最佳答案
1)設直線為y = mx, tan A = l ( m1 – m2 ) / ( 1 + m1m2 ) l
= l ( k – m ) / ( 1 + km ) l
tan A = ( k – m ) / ( 1 + km ) 或 ( m – k ) / ( 1 + km )
m = ( k – tan A ) / ( k tan A + 1 ) 或 ( tan A + k ) / ( 1 – k tan A )
所以y = [( k – tan A )/( k tan A + 1 )]x或 y = [( tan A + k )/( 1 – k tan A )]x。
2)(2m2 + m - 3)x + (m2 - m)y - 4m + 1 = 0的斜率:( 2m2 + m – 3 ) / ( m – m2 )
x- 2y+ 6 = 0的斜率: 1/2
於是,
3 = l ( m1 – m2 ) / ( 1 + m1m2 ) l
3 = l [( 2m2 + m – 3 ) / ( m – m2 ) – 1/2 ] / [ 1 + ( 2m2 + m – 3 ) / 2( m – m2 )]
3 = l ( 5m2 + m – 6 ) / ( 3m – 3 ) l
3 = l ( m – 1 )( 5m + 6 ) / 3 ( m – 1 ) l
9 = l 5m + 6 l
m = 3 / 5 或 – 3
參考: My Maths Knowledge
2007-11-08 4:56 am
2)m = 3 / 5 或 – 3
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