數學~分點座標

Question

1.設A(8,-3),B(-7,7),若P點在AB上,且AP:PB=3:2,求P點座標。

2.設A(-2,4),B(7,-8),若P點在AB之延長線上,且AP:PB=3:2,求P點座標。

最好能夠一步步解說 謝~
+15

Answer 1

1.用分點公式求，如下圖：
圖片參考：http://img157.imageshack.us/img157/4964/09092vg7.jpg

Answer 2

Use vector.
1. V(A,B)=(-7,7)-(8,-3)=(-15,10). Since AP:PB=3:2, so V(A,P)=(3/5)*V(A,B)=(-9,6)
Let P=(x,y), then V(A,P)=(x,y)-(8,-3)=(x-8,y+3)=(-9,6), Thus P=(x,y)=(-1,3)#

2. V(A,B)=(7,-8)-(-2,4)=(9,-12). Since AP:PB=3:2, so AB:PB=1:2=> PB=2AB
Let P=(x,y), then V(B,P)=(x,y)-(7,-8)=(x-7,y+8)=2V(A,B)=(18,-24)
SO P=(x,y)=(25,-32)#