Math-坐標幾何

(a)
BC的斜率
=(8-6)/(4-0)
=1/2
(b)
AB┴BC
AB斜率
=(8-x)/4
[(8-x)/4]*1/2=-1
x=16
A=(0,16)
(c)
AB=√[(4-0)^2+(16-8)^2]=4√5
BC=√[(8-6)^2+(4-0)^2]=2√5
(d)
AREA
=(16)*4/2
=32 平方單位

Somuch ...]] @ ktss: 你又出矛招霸頭位? c part 都未做就post答案= =

1) AB┴BC，而C點坐標為(0,6)B點為(4,8)A點為(0,x)



(a) 求BC的斜率。
Ans:


slope of BC=(8-6)/(4-0)=1/2//


(b) 求A點的坐標。
Ans:
Since AB┴BC ,

So slope of AB * slope of BC = -1

1/2*(8-x)/(4-0)=-1
8-x=-8
x=16//


So the 坐標 of A is (0,16)


d)


Area of OABC=(1/2)(6*4)=12//







2008-07-30 07:42:27 補充：
1) AB┴BC，而C點坐標為(0,6)B點為(4,8)A點為(0,x)
(a) 求BC的斜率。
Ans:
slope of BC=(8-6)/(4-0)=1/2//
(b) 求A點的坐標。
Ans:
Since AB┴BC ,
So slope of AB * slope of BC = -1

1/2*(8-0)/(4-x)=-1
4=-4+x
x=8//

So the 坐標 of A is (8,0)

d)

Area of OABC=(1/2)(16*6-6*4)=36//