Anonymous
Anonymous asked in 科學數學 · 2 decades ago

一題直線方程式的問題

若一直線通過點(-2,2),且與兩座標軸所成之三角形的面積為1,則其方程式為何??

請告訴我解法,還有計算過程

4 Answers

Rating
  • 2 decades ago
    Favorite Answer

    設直線和兩座標軸交點(a,0),(0,b),則面積=1/2*ab=1所以ab=2 a=2/b..............1

    直線通過(a,0),(0,b)兩點,則直線方程式為y=-b/a*x+b

    因為直線通過點(-2,2),帶入直線方程式y=-b/a*x+b 得:2b+ab=2a................2

    1帶入2:得 b*b+b-2=0 為:(b+2)*(b-1)=0 所以b=-2or1 帶入1得a=-1or2

    所以直線方程式為y=-1/2x+1 or y=-2x-2

  • Anonymous
    2 decades ago

    假設斜率=m

    點斜式→y-2=m(x+2)

    與y軸交點→x=0

    y-2=2m→y=2m+2→交點(0,2m+2)

    與x軸交點→y=0

    -2=mx+2m→x=-(2m+2)/m→交點(-(2m+2)/m,0)

    │2m+2│*│-(2m+2)/m│=2

    (2m+2)^2=±2m

    (I)

    (2m+2)^2=2m

    →4m^2+6m+4=0

    →無實數解

    (II)

    (2m+2)^2=-2m

    →4m^2+10m+4=0

    →2m^2+5m+2=0

    →(2m+1)*(m+2)=0

    →m=-1/2,-2

    直線方程式

    當m=-1/2時

    →y-2=-(x+2)/2→x-2y=-6

    當m=-2時

    →y-2=-2(x+2)→2x+y=-2#

    Source(s): 自己
  • Anonymous
    2 decades ago

    (-2,2)在第二象限,所以假設線斜率m為正設直線方程為y=m(x+2)+2令x=0, y=2m+2 >0令y=0, x= (-2-2m)/m<0面積=1 = (2m+2)(2+2m)/(2m)整理得: 2m2+3m+2=0因32-4*2*2<0m無解----------------------------------(-2,2)在第二象限,所以假設線斜率m為負設直線方程為y=m(x+2)+2令x=0, y=2m+2 令y=0, x= (-2-2m)/m若2m+2>0  則(-2-2m)/m>0(2m+2)*(-2-2m)/m=2 ---->2m2+5m+2=0--->m=(-1/2) 或 (-2)方程式為x+2y-2=0  或 2x+y+2=0

  • 2 decades ago

    假設這條直線是y-2=m(x+2)與兩座標軸交於(0,2m+2),((-2m-2)/m,0)(1/2)(2m+2)(-2m-2)/m=1m=-1/2或-2y-2=(-1/2)(x+2)或y-2=(-2)(x+2)

Still have questions? Get your answers by asking now.