A=(1,2)
B=(3,−2)
Odcinek ma dwie osie symetrii
Pierwsza os symetrii to symetralna odcinka
(x−xA)2+(y−yA)2=(xB−x)2+(yB−y)2
(x−1)2+(y−2)2=(3−x)2+(−2−y)2
x2−2x+1+y2−4y+4=x2−6x+9+y2+4y+4
−4y−4y−2x+6x+5−13=0
−8y+4x−8=0
−8y=−4x+8
| 1 | ||
y= | x−1 (Czerwona | |
| 2 |
| −2−2 | ||
y= | (x−1)+2 | |
| 3−1 |
| −4 | ||
y= | (x−1)+2 | |
| 2 |
C = (x, y), A = (1, 2), B = (3, −2)
| 1 | ||
|AC|2 = |BC|2 ⇒ (x−1)2+(y−2)2=(x−3)2+(y+2)2 ⇒ ... ⇒ y= | x−1 | |
| 2 |