F.3,,,coordinates.

3a)the slope of AB
= [-2﹣(-3)] / (6﹣2)
= (-2 + 3) / 4
= 1 / 4
the slope of DB
= [2﹣(-2)] / (5﹣6)
= (2 + 2) / -1
= -4
Since the slope of AB X the slope of DB = -1
∴AB and DB are perpendicular to each other.

b)DB = √[(-2﹣2)2 + (6﹣5)2] = √17 unit.
AB = √{ [(-2﹣(-3)]2 + (6﹣2)2] }
AB = √[(-2 + 3)2 + (6﹣2)2] = √17 unit.
∴the area of triangle ABD
= (AB X DB) / 2
= (√17 X √17) / 2
= 17 / 2 sq. unit.

ci)the slope of AD
= (-3﹣2) / (2﹣5)
= -5 / -3
= 5 / 3
Since ABCD is a parallelogram,
i.e.,the slope of BC = the slope of AD
[k﹣(-2)] / (9﹣6) = 5 / 3
(k + 2) / 3 = 5 / 3
k + 2 = 5
k = 3

cii)CD = √[(3﹣2)2 + (9﹣5)2] = √17unit.
Area of BCD
= (BD X CD) / 2
= (√17 X √17) / 2
= 17 / 2 sq. unit.
CB = √[(-2﹣3)2 + (6﹣9)2] = √34 unit.
Area of BCD = (DE X BC) / 2
17 / 2 = (DE X √34) / 2
17 = DE X √34
DE = 17 / √34
DE = 2.92 (3 .f .s.)

2008-07-30 15:32:42 補充：
To zoe_tpm：你的part cii)的答案抄到十足!!!!

zoe_tpm你都幾好嘢wor, 1分幾鐘打到成11行字......如果真係你打嘅應該可以創出世界紀錄, 但睇真d你part cii)個答案同ncy_0916嘅一模一樣, 唔通你勁到完全知人地諗咩?!
根本就冇可能啦! 好心你唔識就唔好答啦, 抄咩鬼答案呢?

a)slopes of straight lineAB=-2-(-3)/6-2
=1/4
slopes of straight lineDB=-2-2/6-5
=-4
Because slopes of straight lineAB x slopes of straight lineDB =-1
soAB is perpendicular to DB
b)AB=square root ((-2-(-3) )square+(6-2)square)
=square root 17
BD=square root (( 2-(-2))square+(5-6)square)
=square root 17
Area of triangle ABD=square root 17
xsquare root 17 x0.5=8.5
ci)slope of AB=slope of DC
ie.1/4=k-2/9-5
k=3
ii)I think you should do one question by your own!
I hope I can help you!

2008-07-30 15:11:08 補充：
cii)CD = √[(3﹣2)2 + (9﹣5)2] = √17unit.

Area of BCD

= (BD X CD) / 2

= (√17 X √17) / 2

= 17 / 2 sq. unit.

CB = √[(-2﹣3)2 + (6﹣9)2] = √34 unit.

Area of BCD = (DE X BC) / 2

17 / 2 = (DE X √34) / 2

17 = DE X √34

DE = 17 / √34

DE = 2.92 (3 .f .s.)