座標不懂怎樣解?10點!!!!!!!

1)
AB = a-(-2),
AC=AB = a+2,
CA = √[(6-a)²+(4-0)²] = a+2
36-12a+a²+16 = a²+4a+4
16a = 48
a = 3
So, A: (0, -3)

2)
4a²-49b²+10a-35b
= (2a-7b)(2a+7b)+5(2a-7b)
= (2a-7b)(2a+7b+5)

3)
唔知...sor

2007-06-07 21:11:30 補充：
Mind that triangles with same height, the area are direct proportional to the base of the triangles.BCE=DCE =aBFE=FEA =b∵BCA=CDA∴a+DEA=a+2bDEA = 2b

2007-06-07 21:12:05 補充：
DEA:EFA = 2:1,They share the same height when DE and EF are base.So, DE:EF = 2:1Diagram: http://i169.photobucket.com/albums/u209/lautszki/math.jpg

2007-06-07 21:25:24 補充：
我第三條誤以為BF=FA，所以我錯了。

2007-06-07 21:42:18 補充：
我希望對樓下作少少更正：1) 應為intercept theorem2) 作步驟補充：Let G is a pt on DF such that CG//BA∵BC=CD(given)∴DG=GF(intercept theorem),∵∠CGE=∠AFE, ∠GCE = ∠EAF (alt. ∠s, CG//AB)and CE=EA (given)∴△CGE≡△AFE(A.A.S.)∴GE=EF(corr. sides, ≡△s)

2007-06-07 21:42:32 補充：
Let GE = EF = x.GD = GF = GE + EF = 2x,DE : EF = (GD + GE) : EF= (2x+x) : x= 3:1diagram: http://i169.photobucket.com/albums/u209/lautszki/math-1.jpg

1.AB=AC
LET A PT =(0,a)
AC=√(4-0)^2+(6-a)^2=AB=a+2
16+36+a^2-12a=a^2+4a+4
16a=48
a=3
A pt =(0,3)(B)

2.4a^2-49b^2+10a-35b
=(2a^2-7b^2)(2a^2+7b^2)+5(2a^2-7b^2)
=(2a^2+7b^2+5)(2a^2-7b^2)

3.在DF上加一點G使CG//BA
DG=GF(MID-PT.THEM)
又三角形CGE全等三角形AFE(AAS)
所以GE=EF
DE:EF=3:1

1)AB=AC
LET A PT =(0,a)
AC=√(4-0)^2+(6-a)^2=AB=a+2
16+36+a^2-12a=a^2+4a+4
16a=48
a=3
A pt =(0,3)(B)
2)4a^2-49b^2+10a-35b
=(2a^2-7b^2)(2a^2+7b^2)+5(2a^2-7b^2)
=(2a^2+7b^2+5)(2a^2-7b^2)
3)在DF上加一點G使CG//BA
DG=GF(MID-PT.THEM)
又三角形CGE全等三角形AFE(AAS)
所以GE=EF
DE:EF=3:1