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L is a family of lines perpendicular to the line 2x-6y+3=0
a.)Write down an equation of L
b.)Hence find the equation of L1 , which is a line of L passing through ( 2, -1/2 )
A-math
2007-05-25 4:03 am
回答 (2)
2007-05-25 4:14 am
✔ 最佳答案
( a )2x - 6y + 3 = 0
6y = 2x + 3
y = x/3 + 1/2
As L is perpendicular to the line, the product of their slopes = -1
Let the equation of L be y = mx + c
m( 1 / 3 ) = -1
m = -3
Equation of L is y = -3x + c, where c is a constant
An equation of L is y = -3x
____________________________
( b )
Put ( 2 , -1/2 ) into y = -3x + c
-1/2 = -6 + c
c = 11 / 2
Equation of L1 is y = -3x + 11/2
2y = -6x + 11
6x + 2y - 11 = 0
2007-05-25 11:51 pm
a.)2x-6y+3=0
6y=2x+3
y=1/3x+1/2
as L is perpendicular to "2x-6y+3=0"
so, slope of L=-3
The required equation is
y=-3x+k where k is any real number
b.)Putting(2, -1/2) into"y=-3x+k"
-1/2=(-3)(2)+k
1=12-2k
k=11/2
so, the required equation is y=-3x+11/2
6y=2x+3
y=1/3x+1/2
as L is perpendicular to "2x-6y+3=0"
so, slope of L=-3
The required equation is
y=-3x+k where k is any real number
b.)Putting(2, -1/2) into"y=-3x+k"
-1/2=(-3)(2)+k
1=12-2k
k=11/2
so, the required equation is y=-3x+11/2
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