a) prosta n:
| yC−yB | 5+1 | |||
aBC = | = | =−2 | ||
| xC−xB | 2−5 |
| xA+xB | −1+5 | |||
b) prosta k : S(xS,yS), xS = | = | = 2 | ||
| 2 | 2 |
| yA+yB | −2−1 | 3 | 1 | |||||
yS = | = | = − | = −1 | |||||
| 2 | 2 | 2 | 2 |
| 1 | ||
Czyli S(2,−1 | ) | |
| 2 |
| yC−yA | 5+2 | 7 | ||||
aAC= | = | = | ||||
| xC−xA | 2+1 | 3 |
| 7 | ||
k∥AC ⇒ ak = aAC = | ||
| 3 |
| 1 | 7 | 7 | 37 | |||||
y + 1 | = | (x−2) ⇒[C[y = | x − | |||||
| 2 | 3 | 3 | 6 |
| 7 | 1 | 3 | ||||
aAC = | , m⊥AC ⇒ am = − | = − | ||||
| 3 | aAC | 7 |
| xB+xC | 5+2 | 1 | ||||
K(xK,yK), xK = | = | =3 | ||||
| 2 | 2 | 2 |
| yB+yC | −1+5 | |||
yK = | = | = 2 | ||
| 2 | 2 |
| 1 | ||
Czyli: K(3 | ,2) | |
| 2 |
| 3 | 7 | 3 | 1 | |||||
y − 2 = − | (x − | ) ⇒ y=− | x + 3 | |||||
| 7 | 2 | 7 | 2 |