11.4 1) k−AB−−−→k=
5−2 5−1
=
3 4
=√
32+ 42 =√
9 + 16 =√ 25 = 5
k−AC−−−→k=
0−2
−7−1
=
−2
−8
= −2
1 4
=| −2|
1 4
=
= 2√
12+ 42 = 2√
1 + 16 = 2√ 17 k−BC−−−→k=
0−5
−7−5
=
−5
−12
=p
(−5)2+ (−12)2 =√
25 + 144 =
=√
169 = 13
Le périmètre du triangleABC vaut donc5 + 2√
17 + 13 = 18 + 2√ 17.
2) k−AB−−−→k=
4−2 3−1 4−3
=
2 2 1
=√
22+ 22+ 12 =√ 9 = 3
k−AC−−−→k=
2−2 6−1
−9−3
=
0 5
−12
=p
02+ 52 + (−12)2 =
=√
0 + 25 + 144 =√
169 = 13 k−BC−−−→k=
2−4 6−3
−9−4
=
−2 3
−13
=p
(−2)2+ 32+ (−13)2 =
=√
4 + 9 + 169 =√ 182
Ainsi le périmètre du triangle ABC est3 + 13 +√
182 = 16 +√ 182.
Géométrie : norme Corrigé 11.4
