我想問下 Straight lines and rectilinear Figures 呢課
裡面有條式 d= l Ax+ By + C / √ A^2 + B^2 l
咁 d 呢個distance 係唔係一定要同某條線成直角 ga??
咁如果有2條線佢地唔係parallel ,有交點
咁可唔可以代入上面個個公式??
咁佢地distance 係指邊點?
1. Find the points on the line 2x-y-3=0 such that their disatnces to the line 3x+y-2 =0 are √10 units.
呢題我試個畫個2條lines 出黎, 但佢地有交點wor
咁以呢題為例 √10 units 係指邊到 同邊到distance??
佢地唔係perpenticular ,y 都係用d= l Ax+ By + C / √ A^2 + B^2 l ??
如果可以ge 請畫埋圖 thz~
plz explain ~ thz!!!!!
A.maths 高手請答
2007-04-30 11:17 pm
回答 (2)
2007-05-01 2:26 am
✔ 最佳答案
we apply this formula when we want to find the distance between a line and a pointthe equation of given line, L : Ax + By +C = 0
the given point, K ( x1 , y1)
d = abs[ ( Ax1 + By1 + C ) / root ( A^2 + B^2) ]
the distance is the line from K ┴ to L
i.e. the line with distance, d is ┴ to L
if there are two non // lines,
we never say the distance between them
the distance between two lines is only occured in // lines
if we want to find the distance between two // lines,
d = abs[ ( C - C' ) / root ( A^2 + B^2 ) ]
where the eqt of the two // lines are
Ax + By +C = 0 and Ax + By + C' = 0
For the question given,
Let the coordinate of the required point be ( x' , y' )
Substituting x = x' into 2x-y-3=0
y = 2x' -3
i.e. coordinate of the required pt. is ( x' , 2x' -3)
root 10 = ( 3x' + 2x' -3 -2 )/ root( 3^2 + 1^2)
10 = 5x' - 5
x' = 3
substituting x' = 3 into y' = 2x' -3
y' = 3
i.e. the required pt is ( 3 , 3)
參考: Keith ^^
2007-04-30 11:40 pm
這是 Additional Maths 的題目吧？ 我還以為是 Apply Maths.......
在直線"2x-y-3=0"上，找出(一或多)點，與直線"3x+y-2 =0"的距離為 √10 units
Set 2x-y-3=0 --- (1)
3x+y-2 =0 --- (2)
set root of (1) is (a,b) & root of (2) is (c,d)
√10 = │d-b│/│c-a│ --- (3)
2a-b-3=0 --- (4)
3c+d-2 =0 --- (5)
在直線"2x-y-3=0"上，找出(一或多)點，與直線"3x+y-2 =0"的距離為 √10 units
Set 2x-y-3=0 --- (1)
3x+y-2 =0 --- (2)
set root of (1) is (a,b) & root of (2) is (c,d)
√10 = │d-b│/│c-a│ --- (3)
2a-b-3=0 --- (4)
3c+d-2 =0 --- (5)
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