## Trending News

# 求 New Progress in certifficate mathatics revised edition 20點

求 New Progress in certifficate mathatics revised edition as Re.ex 9

Q18,19, answer 超急,快,要有solution

### 1 Answer

- So MuchLv 71 decade agoFavorite Answer
let the slope of L = m ,

then the equation of L is :

(y-3)=m(x+3)

y=mx+3(m+1)

when y=0,

x= -3(m+1)/m

let the angle = x

then m= tan(90+x)=-cot(x)

slope of R= tan(90-x)=cot (x)=-m

so..

the equation of R =

(y-0)=-m[x+3(m+1)/m]

y=-mx-3(m+1)

put it into C,

x^2+[-mx-3(m+1)]^2-4x-4[-mx-3(m+1)]+7=0

Dlata=0

then we get m=-3/4 or -4/3..

so the equation is

3x+4y-3=0 and 4x+3y+3=0

2008-05-26 11:39:44 補充：

19.

let the center is (x, 1-2x),

abs[(x+2-4x+5)/√5]=√[(x-1)^2+(2-2x)^2]

(-3x+7)^2=5(5x^2-10x+5)

9x^2-42x+49=25x^2-50x+25

16x^2-8x-24=0

2x^2-x-3=0

(2x-3)(x+1)=0

x=3/2 or x=-1

y=-2 , y=3,

radius = √[(-1-1)^2+(2+2)^2]=√20

so the equation are :

C1:(x+1)^2+(y-3)^2=20

and C2: (x-3/2)^2+(y+2)^2=20