Maths(Coordinate Treatment of Simple Locus Problems)

( 1 ) Do u mean A ( 4 , 5 )?

From the information given,

AC : CD = 5 : 1

So let C ( x , y ).

x = { ( 5 )( 10 ) + ( 1 )( 4 ) } / ( 1 + 5 )

= 9

y = { ( 5 )( 11 ) + ( 1 )( 5 ) } / ( 1 + 5 )

= 10

So C ( 9 , 10 ) which is ans e.

( 2 ) The slope of AB = ( b - 0 ) / ( a - 0 ) = b /a

As AB // DC, the slope of DC is also b / a.

Let C be ( x , y ).

( y - a ) / ( x + b ) = b / a

So y - a = b, x + b = a

x = a - b , y = a + b

Hence the ans is d.

( 4 ) Let C be ( 0 , y ).

AC = CB ( given )

√( -4 - 0 )^2 + ( 2 - y )^2 = √ ( 1 - 0 )^2 + ( - 3 - y )^2

16 + ( 2 - y )^2 = 1 + ( 3 + y )^2

16 + 4 - 4y + y^2 = 1 + 9 + 6y + y^2

10 y = 10

y = 1

So C ( 0 , 1 ) which is ans e

1. AB:BC:CD = 3:2:1
Hence AC:CD = 5:1
Let C(x,y).By section formula,
x=[ 1(10) + 5(4) ]/6 = 5
y=[ 1(11) + 5(5) ]/6 = 6

So the correct option is a.

2.The correct option is d.
I don't know how to explain ... sorry ...

3.Let C(0, y) (because any points on the y-axis has x-coordinate equal to 0)
Now only option d or e is correct. Substitute y=1,
AC = sqrt(16 + 1) = sqrt(17). CB=sqrt(1+16)=sqrt(17).
So d is correct.

2007-07-13 17:16:07 補充：
No.4 (no.3 above) : it is e not d ... I type it wrong