Given that the x-intercept of a straight line L is 12 and its y-intercept is-8, find the equation of a straight line that is perpendicular to L and passaes through the point(3,-1)
Anwer: 3x+2y-7 =0
請詳細解釋!!
MATHS.. 20分
2007-11-09 3:46 am
回答 (3)
2007-11-09 3:52 am
✔ 最佳答案
The equation of L:x / a + y / b = 1
x / 12 + y / ( - 8 ) = 1
8x - 12y = 96
2x - 3y - 24 = 0
Slope of L = 2/3
So the slope pf the line perpendicular to L and passaes through the point(3,-1): -3/2
Then the equation:
( y + 1 ) / ( x - 3 ) = - 3 / 2
3x + 2y - 7 = 0
參考: My Maths Knowledge
2007-11-09 3:57 am
Because (12,0) and (0,-8) lie on L,
slope of L = (0-(-8))/(12-0)=2/3
By point-slope from,
(y-(-1))/(x-3)=-1/(2/3) (Because the line is perpendicular to L)
y+1=(-3/2)(x-3)
2y+2=-3x+9
3x-2y-7=0
slope of L = (0-(-8))/(12-0)=2/3
By point-slope from,
(y-(-1))/(x-3)=-1/(2/3) (Because the line is perpendicular to L)
y+1=(-3/2)(x-3)
2y+2=-3x+9
3x-2y-7=0
2007-11-09 3:56 am
that means L passes through the point (12,0) and (0,-8).
Slope of L: (8/12) = 2/3
the slope of a straight line that is perpendicular to L = -3/2
as it passes through the point (3,-1),equation of a straight line : y+1= (-3/2)(x-3)
2y+2 = -3x +9
3x +2y -7 = 0
Slope of L: (8/12) = 2/3
the slope of a straight line that is perpendicular to L = -3/2
as it passes through the point (3,-1),equation of a straight line : y+1= (-3/2)(x-3)
2y+2 = -3x +9
3x +2y -7 = 0
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