2題math =40分

1a) 3x - y - 2 = 0 --- ( 1 )

x - y - 4 = 0 --- ( 2 )

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( 1 ) - ( 2 ) : 2x + 2 = 0

x = - 1

-1 - y - 4 = 0

y = - 5

So A ( - 1 , - 5 ).

b) L : ( y - 0 ) / ( x - 2 ) = ( 0 + 5 ) / ( 2 + 1 )

5x - 3y - 10 = 0

2a) C: ( x - 1 )^2 + ( y + 2 )^2 = r^2

( 2 - 1 )^2 + ( 3 + 2 )^2 = r^2 [ Sub ( 2,3 ) ]

r^2 = 26

So C: ( x - 1 )^2 + ( y + 2 )^2 = 26

x^2 - 2x + 1 + y^2 + 4y + 4 = 26

x^2 + y^2 - 2x + 4y - 21 = 0

b) Sub y = 0 into the circle [ as the circle cuts the x-axis ] ,

x^2 + 0^2 - 2x + 4 ( 0 ) - 21 = 0

x^2 - 2x - 21 = 0

By quadratic formula,

x = 1 ±√22

Hence the pts. are ( 1 + √22 , 0 ) and ( 1 - √22 , 0 ) respectively.


2007-09-21 00:18:28 補充：
For 2b, apart from leaving the answers in surds, you may also give the answer in numerials, then x = 5.69 or -3.69 ( cor. to 3 s.f. ).

(a)L1:3x-y-2=0
L2:x-y-4=0
-y=4-x
sub -y=4-x into L1:
3x+(4-x)-2=0
2x+2=0
2x=-2
x=-1
Sub x=-1into L2:
-y=4-(-1)
-y=4+1
-y=5
y=-5
So the coordinate of A is (-1,-5).(only 1 coordinate)
(b)Eq of A &(2,0)
y=mx+c
y=[(y2-y1)/(x2-x1)]x+c
y=[(0-(-5))/(2-(-1))]x+c
y=(5/3)x+c
when x=2, y=0
so 0=(5/3)(2)+c
c=-(10/3)
the Eq of L is y=5/3x-10/3 OR 3y-5x=-10 Or 5x-3y=10

2a) (x-h)^2+(y-k)^2=r^2 (h,k)is the centre r is the radius of the circle.
(x-1)^2+(y-(-2))^2=((-2-3)^2+(1-2)^2)^1/2)^2
(x-1)^2+(y+2)^2=26
2b)the coordinates of x-ax y must be 0
(x-1)^2+(0+2)^2=26
(x-1)^2+4=26
(x-1)^2=22
(x-1)=22^(1/2)
x=1(+-)4.69041576
x=6.69041576 &-3.69041576
the coordinates of pts are (6.69041576,0)&(-3.69041576,0)