| 1 | ||
y=− | x+b | |
| 3 |
| 7 | ||
b= | ||
| 3 |
| 1 | 7 | |||
y=− | x+ | |||
| 3 | 3 |
| 1 | 7 | |||
{y=− | x+ | |||
| 3 | 3 |
| 1 | 7 | |||
3x+4=− | x+ | //*3 | ||
| 3 | 3 |
| 1 | ||
x=− | ||
| 2 |
| 5 | ||
y= | ||
| 2 |
| 1 | 5 | |||
P(− | ; | ) | ||
| 2 | 2 |
| x+2 | x−3 | |||
O=( | ; | ) | ||
| 2 | 2 |
| 4 | 2 | |||
Odpowiedzi: (x+5 | )2+(y+ | )2=25 | ||
| 5 | 5 |
, To zadanie numer 3
x2+y2−4x+6y−12=0 ⇔(x−2)2+(y+3)2=25, S=(2,−3), r=5
względem prostej
k: y=3x+4
Prostopadła do k
| −1 | −1 | 1 | ||||
m: y= | x+b i −3= | *2+b⇔b=−2 | ||||
| 3 | 3 | 3 |
| −1 | 7 | |||
m: y= | x− | |||
| 3 | 3 |
| −1 | 7 | |||
3x+4= | x− | /*3 | ||
| 3 | 3 |
| x+2 | y−3 | |||
−1,9= | i −1,7= | |||
| 2 | 2 |
| −58 | −29 | |||
x=−5,8= | = | |||
| 10 | 5 |
| −2 | ||
y=−0,4= | ||
| 5 |
| −29 | −2 | |||
S1=( | , | ) | ||
| 5 | 5 |
| 29 | 2 | |||
(x+ | )2+(y+ | )2=25 | ||
| 5 | 5 |
| 7 | 7 | |||
Wszystko zrozumiałem, zrobiłem błąd rachunkowy. b=− | a nie | . | ||
| 3 | 3 |