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直线L经过点p(-4,3)与x轴,y轴分别交与AB俩点,且AP绝对值:PB绝对值=3:5,求直线l的方程

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直线L经过点p(-4,3)与x轴,y轴分别交与AB俩点,且AP绝对值:PB绝对值=3:5,求直线l的方程
let L be
(y-3)/(x+4) = m
put x =0
y= 4m+3
put y=0
-3 = (x+4)m
x = -3/m -4
ie A(0,4m+3),B( -3/m -4,0)
|PA| = (16+16m^2)^(1/2)
|PB| = (9/m^2 + 9)^(1/2)
|PA|/|PB| = 3/5
(16+16m^2)^(1/2)/( (9/m^2 + 9)^(1/2)) = 3/5
(16+16m^2)/(9/m^2 + 9) = 9/25
25(16+16m^2) = 9((9/m^2 + 9)
400m^2+319 - 81/m^2 =0
400m^4 +319m -81 =0
m^2 = (-319+481)/800 or (-319-481)/800 (rejected)
= 162/400
m = √162/20 or -√162/20
L:
(y-3)/(x+4) = √162/20 or (y-3)/(x+4) = -√162/20