Two points A (-3, -2) and B (5, 1) are given . P is a point on the x-axis such that A, P and B are collinear.
Q is a point lying on the AB produced such that AP=PQ. Find the coordinates of Q.
A. Maths (F.4)
2007-05-21 3:43 am
回答 (3)
2007-05-21 3:52 am
✔ 最佳答案
Let P be (x, 0)SLOPE OF AP = SLOPE OF AB
[0 - (-2) ]/ [x - (-3)] = [1- (-2)] / 5- (-3)
2/ (x+3) = 3/ 8
x = 7/ 3
P = (7/3, 0)
Q = ( (-3 + 7/3) / 2 , (-2 + 0) / 2 )
= ( -1/3, -1 )
2007-05-31 00:18:16 補充：
但係我計錯ans 呀!
我睇錯題目
2007-05-25 10:36 am
Let P be (x,0)
-2/(-3-x)=1/(5-x)
-10+2x= -3-x
x=7/3
∴P = (7/3,0)
Let Q be (a,b)
∵AP:AQ=1:2
3/8=(-3+a)/2
6= -24++8a
a=30/8
0=(-2+b)/2
b=2
∴Q=(30/8,2)
2007-05-25 02:51:13 補充：
sor,我代錯左P...計a個到應該係...7/3=(-3+a)/214= -9+3a a=23/3你做lee d數個陣,畫圖就易好多,通常一題有好多資料,就咁唸會慢好多同好易錯,按比例畫圖就準好多!背左:1)line既長度=[(x-x1)^2+(y-y1)^2]^(1/2)2)line既mid-pt既coor.=( (x+x1)/2, (y+y1)/2 )3)slope-point form(m=slope): m=(y-y1)/(x-x1)等等既式...做起黎就易好多
-2/(-3-x)=1/(5-x)
-10+2x= -3-x
x=7/3
∴P = (7/3,0)
Let Q be (a,b)
∵AP:AQ=1:2
3/8=(-3+a)/2
6= -24++8a
a=30/8
0=(-2+b)/2
b=2
∴Q=(30/8,2)
2007-05-25 02:51:13 補充：
sor,我代錯左P...計a個到應該係...7/3=(-3+a)/214= -9+3a a=23/3你做lee d數個陣,畫圖就易好多,通常一題有好多資料,就咁唸會慢好多同好易錯,按比例畫圖就準好多!背左:1)line既長度=[(x-x1)^2+(y-y1)^2]^(1/2)2)line既mid-pt既coor.=( (x+x1)/2, (y+y1)/2 )3)slope-point form(m=slope): m=(y-y1)/(x-x1)等等既式...做起黎就易好多
2007-05-21 3:53 am
AP:PB = 2:1
x corrdinates of P is = 2*5+(-3)/(3) = 7/3
P = (7/3, 0)
P is the mid-point of AQ
therefore, (-2+y)/2 = 0 ==> y=2
(-3+x)/2 = 7/3
x = 23/3
Therefore, coordinates of Q are (23/3, 2).
x corrdinates of P is = 2*5+(-3)/(3) = 7/3
P = (7/3, 0)
P is the mid-point of AQ
therefore, (-2+y)/2 = 0 ==> y=2
(-3+x)/2 = 7/3
x = 23/3
Therefore, coordinates of Q are (23/3, 2).
參考: me
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