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Geometria Analítica (Resolução) – Prova 1 – 28/03/2015 Carlos Alberto 
 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟏. 𝑆𝑒𝑗𝑎𝑚 �⃗� 𝑒 𝑣 𝑣𝑒𝑡𝑜𝑟𝑒𝑠 𝑒𝑚 ℝ2 𝑡𝑎𝑖𝑠 𝑞𝑢𝑒 ‖�⃗� ‖ = 2; ‖𝑣 ‖ = 3 𝑒 𝑜 â𝑛𝑔𝑢𝑙𝑜 
𝑒𝑛𝑡𝑟𝑒 �⃗� 𝑒 𝑣 é 
3𝜋
4
𝑟𝑎𝑑 . 𝐶𝑎𝑙𝑐𝑢𝑙𝑒 |< 2�⃗� − 𝑣 , �⃗� − 2𝑣 >| . 
 
∗ 𝐴𝑝𝑙𝑖𝑐𝑎𝑛𝑑𝑜 𝑎𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑝𝑟𝑜𝑝𝑟𝑖𝑒𝑑𝑎𝑑𝑒𝑠 𝑑𝑜 𝑝𝑟𝑜𝑑𝑢𝑡𝑜 𝑒𝑠𝑐𝑎𝑙𝑎𝑟 (𝑜𝑢 𝑖𝑛𝑡𝑒𝑟𝑛𝑜) 𝑒𝑛𝑡𝑟𝑒 
𝑑𝑜𝑖𝑠 𝑣𝑒𝑡𝑜𝑟𝑒𝑠 𝑞𝑢𝑎𝑖𝑠𝑞𝑢𝑒𝑟, 𝑡𝑒𝑚𝑜𝑠: 
 
< �⃗� + 𝑣 , �⃗⃗� > = < �⃗� , �⃗⃗� > +< 𝑣 , �⃗⃗� > 
< 𝜆�⃗� , 𝑣 > = 𝜆 < �⃗� , 𝑣 > 
< �⃗� , �⃗� ≥ ‖�⃗� ‖2 
 
∗ 𝐴𝑝𝑙𝑖𝑐𝑎𝑛𝑑𝑜 𝑒𝑠𝑠𝑎𝑠 𝑝𝑟𝑜𝑝𝑟𝑖𝑒𝑑𝑎𝑑𝑒𝑠 à 𝑛𝑜𝑠𝑠𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜, 𝑜𝑏𝑡𝑒𝑚𝑜𝑠: 
 
< 2�⃗� − 𝑣 , �⃗� − 2𝑣 > = < 2�⃗� − 𝑣 , �⃗� > −< 2�⃗� − 𝑣 , 2𝑣 > 
 = < 2�⃗� , �⃗� > −< 𝑣 , �⃗� > −< 2�⃗� , 2𝑣 > +< 𝑣 , 2𝑣 > 
 = 2 < �⃗� , �⃗� > −< �⃗� , 𝑣 > −4 < �⃗� , 𝑣 > +2 < 𝑣 , 𝑣 > 
 = 2‖�⃗� ‖2 − 5 < �⃗� , 𝑣 > +2‖𝑣 ‖2 
 
∗ 𝐷𝑒𝑣𝑒𝑚𝑜𝑠 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜 < �⃗� , 𝑣 >. 𝑆𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑜 𝑝𝑟𝑜𝑑𝑢𝑡𝑜 
𝑒𝑠𝑐𝑎𝑙𝑎𝑟 𝑒𝑛𝑡𝑟𝑒 𝑑𝑜𝑖𝑠 𝑣𝑒𝑡𝑜𝑟𝑒𝑠 é 𝑛𝑢𝑚𝑒𝑟𝑖𝑐𝑎𝑚𝑒𝑛𝑡𝑒 𝑖𝑔𝑢𝑎𝑙 𝑎𝑜 𝑝𝑟𝑜𝑑𝑢𝑡𝑜 𝑑𝑎𝑠 𝑛𝑜𝑟𝑚𝑎𝑠 𝑑𝑒 
𝑎𝑚𝑏𝑜𝑠 𝑝𝑒𝑙𝑜 𝑐𝑜𝑠𝑠𝑒𝑛𝑜 𝑑𝑜 â𝑛𝑔𝑢𝑙𝑜 𝑓𝑜𝑟𝑚𝑎𝑑𝑜 𝑒𝑛𝑡𝑟𝑒 𝑒𝑙𝑒𝑠. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 
 
< �⃗� , 𝑣 > = ‖�⃗� ‖. ‖𝑣 ‖. cos
3𝜋
4
 
 
→ 𝐶𝑜𝑚 𝑖𝑠𝑠𝑜, 𝑡𝑒𝑚𝑜𝑠: 
 
 < 2�⃗� − 𝑣 , �⃗� − 2𝑣 > = 2‖�⃗� ‖2 − 5‖�⃗� ‖. ‖𝑣 ‖. cos
3𝜋
4
+ 2‖𝑣 ‖2 
 = 2(2)2 − 5. (2). (3). (−
√2
2
) + 2. (3)2 
 = 8 + 15√2 + 18 
 = 26 + 15√2 
 
𝐸𝑛𝑡ã𝑜, |< 2�⃗� − 𝑣 , �⃗� − 2𝑣 > | = |26 + 15√2| = 26 + 15√2. 
 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟐.𝐷𝑒𝑚𝑜𝑛𝑠𝑡𝑟𝑒 𝑞𝑢𝑒 𝑜 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜 𝑞𝑢𝑒 𝑢𝑛𝑒 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑚é𝑑𝑖𝑜𝑠 𝑑𝑜𝑠 𝑙𝑎𝑑𝑜𝑠 𝑛ã𝑜 
𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑠 𝑑𝑒 𝑢𝑚 𝑡𝑟𝑎𝑝é𝑧𝑖𝑜 é 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜 à𝑠 𝑏𝑎𝑠𝑒𝑠, 𝑒 𝑠𝑢𝑎 𝑚𝑒𝑑𝑖𝑑𝑎 é 𝑎 𝑚é𝑑𝑖𝑎 𝑎𝑟𝑖𝑡𝑚é𝑡𝑖𝑐𝑎 
𝑑𝑎𝑠 𝑚𝑒𝑑𝑖𝑑𝑎𝑠 𝑑𝑎𝑠 𝑏𝑎𝑠𝑒𝑠. 
 
 
Geometria Analítica (Resolução) – Prova 1 – 28/03/2015 Carlos Alberto 
𝑆𝑒𝑗𝑎𝑚 𝐴𝐵̅̅ ̅̅ 𝑒 𝐶𝐷̅̅ ̅̅ 𝑜𝑠 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜𝑠 𝑞𝑢𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑚 𝑎𝑠 𝑏𝑎𝑠𝑒𝑠 𝑑𝑜 𝑡𝑟𝑎𝑝é𝑧𝑖𝑜 𝑡𝑎𝑖𝑠 𝑞𝑢𝑒 
𝐴 = (𝑎1, 𝑎2) , 𝐵 = (𝑏1, 𝑏2), 𝐶 = (𝑐1, 𝑐2) 𝑒 𝐷 = (𝑑1, 𝑑2). 
 
∗ 𝐸𝑚 𝑢𝑚 𝑡𝑟𝑎𝑝é𝑧𝑖𝑜, 𝑎𝑠 𝑏𝑎𝑠𝑒𝑠 𝑠ã𝑜 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜𝑠 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑠 𝑒𝑛𝑡𝑟𝑒 𝑠𝑖 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 
𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑛𝑑𝑜 𝑝𝑜𝑟 �⃗� = 𝐴𝐵⃗⃗⃗⃗ ⃗ 𝑒 𝑣 = 𝐶𝐷⃗⃗⃗⃗ ⃗ , 𝑡𝑒𝑚𝑜𝑠 �⃗� = 𝜆𝑣 . 
 
�⃗� = (𝑏1 − 𝑎1, 𝑏2 − 𝑎2) 𝑒 𝑣 = (𝑑1 − 𝑐1, 𝑑2 − 𝑐2) 
 
∗ 𝑆𝑒𝑗𝑎𝑚 𝑀1 𝑒 𝑀2 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑚é𝑑𝑖𝑜𝑠 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑜𝑠 𝑎𝑜𝑠 𝑙𝑎𝑑𝑜𝑠 𝑛ã𝑜 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑠 𝐴𝐶̅̅ ̅̅ 𝑒 𝐷𝐵̅̅ ̅̅ 
𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑎𝑚𝑒𝑛𝑡𝑒. 𝐿𝑜𝑔𝑜, 
 
 𝑀1 = (
𝑎1 + 𝑐1
2
,
𝑎2 + 𝑐2
2
) 𝑒 𝑀2 = (
𝑏1 + 𝑑1
2
,
𝑏2 + 𝑑2
2
) 
 
∗ 𝑂 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜 𝑀1𝑀2̅̅ ̅̅ ̅̅ ̅̅ 𝑝𝑜𝑑𝑒 𝑠𝑒𝑟 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑑𝑜 𝑝𝑒𝑙𝑜 𝑣𝑒𝑡𝑜𝑟 �⃗⃗� , 𝑑𝑎𝑑𝑜 𝑝𝑜𝑟: 
�⃗⃗� = (
𝑏1 − 𝑎1 + 𝑑1 − 𝑐1
2
,
𝑏2 − 𝑎2 + 𝑑2 − 𝑐2
2
) 
�⃗⃗� =
1
2
[(𝑏1 − 𝑎1, 𝑏2 − 𝑎2) + (𝑑1 − 𝑐1, 𝑑2 − 𝑐2)] 
�⃗⃗� =
1
2
(�⃗� + 𝑣 ) 
∗ 𝐶𝑜𝑚𝑜 �⃗� = 𝜆𝑣 , 𝑜𝑢 𝑠𝑒𝑗𝑎, 𝑠ã𝑜 𝑣𝑒𝑡𝑜𝑟𝑒𝑠 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑠, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑟𝑒𝑒𝑠𝑐𝑟𝑒𝑣𝑒𝑟 𝑜 𝑣𝑒𝑡𝑜𝑟 �⃗⃗� 
𝑑𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑓𝑜𝑟𝑚𝑎: 
�⃗⃗� =
1
2
(𝜆𝑣 + 𝑣 ) =
1
2
(𝜆 + 1). 𝑣 
𝐸𝑥𝑝𝑟𝑒𝑠𝑠𝑎𝑛𝑑𝑜 𝑜 𝑣𝑒𝑡𝑜𝑟 �⃗⃗� 𝑑𝑒𝑠𝑠𝑎 𝑓𝑜𝑟𝑚𝑎, 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠 𝑞𝑢𝑒 �⃗⃗� é 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜 𝑎𝑜 𝑣𝑒𝑡𝑜𝑟 𝑣 𝑒, 
𝑐𝑜𝑛𝑠𝑒𝑞𝑢𝑒𝑛𝑡𝑒𝑚𝑒𝑛𝑡𝑒, �⃗⃗� 𝑡𝑎𝑚𝑏é𝑚 é 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜 𝑎𝑜 𝑣𝑒𝑡𝑜𝑟 𝑣 . 
 
→ 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑜 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜 𝑞𝑢𝑒 𝑢𝑛𝑒 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑚é𝑑𝑖𝑜𝑠 𝑑𝑜𝑠 𝑙𝑎𝑑𝑜𝑠 𝑛ã𝑜 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑠 𝑑𝑒 𝑢𝑚 
𝑡𝑟𝑎𝑝é𝑧𝑖𝑜 é 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜 à𝑠 𝑏𝑎𝑠𝑒𝑠! 
 
∗ 𝐷𝑒𝑣𝑒𝑚𝑜𝑠 𝑚𝑜𝑠𝑡𝑟𝑎𝑟 𝑞𝑢𝑒 ‖�⃗⃗� ‖ =
1
2
(‖�⃗� ‖ + ‖𝑣 ‖) 
→ 𝑃𝑒𝑙𝑎 𝑝𝑟𝑜𝑝𝑟𝑖𝑒𝑑𝑎𝑑𝑒 𝑑𝑎 𝑛𝑜𝑟𝑚𝑎 𝑑𝑒 𝑢𝑚 𝑣𝑒𝑡𝑜𝑟, 𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 ‖𝜆�⃗� ‖ = |𝜆|. ‖�⃗� ‖. 𝐿𝑜𝑔𝑜, 
 
�⃗⃗� =
1
2
(�⃗� + 𝑣 ) ⇒ ‖�⃗⃗� ‖ =
1
2
‖�⃗� + 𝑣 ‖ =
1
2
√‖�⃗� ‖2 + 2 < �⃗� , 𝑣 > +‖𝑣‖2 = 
1
2
√‖�⃗� ‖2 + 2‖�⃗� ‖‖𝑣 ‖. cos 0° + ‖𝑣 ‖2 =
1
2
√‖�⃗� ‖2 + 2‖�⃗� ‖‖𝑣 ‖ + ‖𝑣 ‖2 = 
1
2
√(‖�⃗� ‖ + ‖𝑣 ‖)2 =
1
2
(‖�⃗� ‖ + ‖𝑣 ‖). 
 
∗ 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑎 𝑚𝑒𝑑𝑖𝑑𝑎 𝑑𝑜 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜 𝑀1𝑀2̅̅ ̅̅ ̅̅ ̅̅ é 𝑎 𝑚é𝑑𝑖𝑎 𝑎𝑟𝑖𝑡𝑚é𝑡𝑖𝑐𝑎 𝑒𝑛𝑡𝑟𝑒 𝑎𝑠 𝑏𝑎𝑠𝑒𝑠 
𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑑𝑎𝑠 𝑝𝑒𝑙𝑜𝑠 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜𝑠 𝐴𝐵̅̅ ̅̅ 𝑒 𝐶𝐷̅̅ ̅̅ . 
Geometria Analítica (Resolução) – Prova 1 – 28/03/2015 Carlos Alberto 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟑. 
 
𝑎) 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑎𝑠 𝑒𝑞𝑢𝑎çõ𝑒𝑠 𝑑𝑎𝑠 𝑐𝑖𝑟𝑐𝑢𝑛𝑓𝑒𝑟ê𝑛𝑐𝑖𝑎𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 à𝑠 𝑟𝑒𝑡𝑎𝑠 𝑟1 𝑒 𝑟2, 𝑑𝑎𝑑𝑎𝑠 
𝑝𝑜𝑟 𝑟1: 4𝑥 + 3𝑦 − 1 = 0 𝑒 𝑟2: 3𝑥 − 4𝑦 − 2 = 0 𝑐𝑜𝑚 𝑐𝑒𝑛𝑡𝑟𝑜𝑠 𝑠𝑜𝑏𝑟𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑠 𝑑𝑎𝑑𝑎 
𝑝𝑜𝑟 𝑠: 2𝑥 + 𝑦 − 2 = 0. 
 
∗ 𝑆𝑒𝑗𝑎 𝐶 = (𝑥, 𝑦) 𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑎 𝑐𝑖𝑟𝑐𝑢𝑛𝑓𝑒𝑟ê𝑛𝑐𝑖𝑎 𝜆1, 𝑐𝑜𝑚𝑜 𝐶 ∈ 𝑠, 𝑖𝑚𝑝𝑙𝑖𝑐𝑎 𝑑𝑖𝑧𝑒𝑟 𝑞𝑢𝑒 
𝐶 = (𝑥,−2𝑥 + 2). 
∗ 𝐶𝑜𝑚𝑜 𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑟1 𝑒 𝑟2 𝑠ã𝑜 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 à 𝑐𝑖𝑟𝑐𝑢𝑛𝑓𝑒𝑟ê𝑛𝑐𝑖𝑎, 𝑒𝑛𝑡ã𝑜 
 𝑑(𝐶, 𝑟1) = 𝑑(𝐶, 𝑟2) = 𝑅 
 
𝑑(𝐶, 𝑟1) =
|4𝑥 + 3(−2𝑥 + 2) − 1|
√42 + 32
=
|4𝑥 − 6𝑥 + 6 − 1|
√25
=
|−2𝑥 + 5|
5
 
 
𝑑(𝐶, 𝑟2) =
|3𝑥 − 4(−2𝑥 + 2) − 2|
√32 + (−4)2
=
|3𝑥 + 8𝑥 − 8 − 2|
√25
=
|11𝑥 − 10|
5
 
 
∗ 𝐶𝑜𝑚𝑜 𝑑(𝐶, 𝑟1) = 𝑑(𝐶, 𝑟2), 𝑡𝑒𝑚𝑜𝑠 … 
|−2𝑥 + 5|
5
=
|11𝑥 − 10|
5
 
|
11𝑥 − 10
−2𝑥 + 5
| = 1 
11𝑥 − 10
−2𝑥 + 5
= ±1 
 
∗ 𝐴𝑠𝑠𝑖𝑚, 𝑜𝑏𝑡𝑒𝑚𝑜𝑠 𝑜𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑥 𝑑𝑜 𝑝𝑜𝑛𝑡𝑜 𝐶: 
 
1) 11𝑥 − 10 = −2𝑥 + 5 2) 11𝑥 − 10 = 2𝑥 − 5 
 13𝑥 = 15 9𝑥 = 5 
 𝑥1 =
15
13
 𝑥2 =
5
9
 
 
→ 𝑃𝑎𝑟𝑎 𝑥 =
15
13
 , 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑠, 𝑜𝑏𝑡𝑒𝑚𝑜𝑠 𝑎𝑠 𝑐𝑜𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑎𝑠 𝑑𝑜 
𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑎 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝑐𝑖𝑟𝑐𝑢𝑛𝑓𝑒𝑟ê𝑛𝑐𝑖𝑎. 
𝑦1 = −2𝑥1 + 2 → 𝑦1 = −2(
15
13
) + 2 = −
30
13
+ 2 = −
4
13
 . 𝑃𝑜𝑛𝑡𝑜 𝐶1 = (
15
13
,−
4
13
) 
𝑦2 = −2𝑥2 + 2 → 𝑦2 = −2(
5
9
) + 2 = −
10
9
+ 2 =
8
9
 . 𝑃𝑜𝑛𝑡𝑜 𝐶2 = (
5
9
,
8
9
) 
 
∗ 𝑃𝑎𝑟𝑎 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑜 𝑟𝑎𝑖𝑜 𝑑𝑎𝑠 𝑐𝑖𝑟𝑐𝑢𝑛𝑓𝑒𝑟ê𝑛𝑐𝑖𝑎𝑠 𝜆1 𝑒 𝜆2 𝑟𝑒𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑚𝑜𝑠 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 
𝑑𝑖𝑠𝑡â𝑛𝑐𝑖𝑎 𝑑𝑒 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 à 𝑟𝑒𝑡𝑎. 𝐴𝑠𝑠𝑖𝑚, 
 
Geometria Analítica (Resolução) – Prova 1 – 28/03/2015 Carlos Alberto 
𝑑(𝐶1, 𝑟1) =
|−2𝑥1 + 5|
5
=
|−2 (
15
13) + 5|
5
=
|−
30
13 + 5|
5
=
|
35
13|
5
=
35
13
5
=
7
13
= 𝑅1 
 
𝑑(𝐶2, 𝑟1) =
|−2𝑥2 + 5|
5
=
|−2(
5
9) + 5|
5
=
|−
10
9 + 5|
5
=
|
35
9 |
5
=
35
9
5
=
7
9
= 𝑅2 
 
→ 𝐷𝑎𝑑𝑜𝑠 𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 𝐶(𝑎, 𝑏) 𝑒 𝑜 𝑟𝑎𝑖𝑜 𝑅 𝑑𝑒 𝑢𝑚𝑎 𝑐𝑖𝑟𝑐𝑢𝑛𝑓𝑒𝑟ê𝑛𝑐𝑖𝑎 𝑠𝑢𝑎 𝑒𝑞𝑢𝑎çã𝑜 é 𝑑𝑎𝑑𝑎 𝑝𝑜𝑟: 
(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑅2 
 
∗ 𝑃𝑎𝑟𝑎 𝜆1 𝑡𝑒𝑚𝑜𝑠 𝐶1 = (
15
13
,−
4
13
) 𝑒 𝑅1 =
7
13
 , 𝑙𝑜𝑔𝑜: 
 
𝜆1: (𝑥 −
15
13
)
2
+ (𝑦 +
4
13
)
2
=
49
169
 
 
∗ 𝑃𝑎𝑟𝑎 𝜆2 𝑡𝑒𝑚𝑜𝑠 𝐶2 = (
5
9
,
8
9
) 𝑒 𝑅2 =
7
9
 , 𝑙𝑜𝑔𝑜:𝜆2: (𝑥 −
5
9
)
2
+ (𝑦 −
8
9
)
2
=
49
81
 
 
𝑏) 𝐸𝑠𝑏𝑜ç𝑎𝑟 𝑜 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑑𝑜 𝑑𝑜 𝑖𝑡𝑒𝑚 𝑎. 
 
 
Geometria Analítica (Resolução) – Prova 1 – 28/03/2015 Carlos Alberto 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟒 
 
𝑎) 𝐷𝑎𝑑𝑜𝑠 𝑜𝑠 𝑣é𝑟𝑡𝑖𝑐𝑒𝑠 𝐴 = (−3,5), 𝐵 = (1,7) 𝑑𝑒 𝑢𝑚 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚𝑜 𝑒 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑑𝑒 
𝑖𝑛𝑡𝑒𝑟𝑠𝑒çã𝑜 𝑀 = (1,1) 𝑑𝑒 𝑠𝑢𝑎𝑠 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑖𝑠, 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑒 𝑜𝑠 𝑜𝑢𝑡𝑟𝑜𝑠 𝑑𝑜𝑖𝑠 𝑣é𝑟𝑡𝑖𝑐𝑒𝑠. 
 
∗ 𝑆𝑜𝑏𝑟𝑒 𝑢𝑚 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚𝑜 𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑖𝑛𝑡𝑒𝑟𝑠𝑒çã𝑜 𝑒𝑛𝑡𝑟𝑒 𝑎𝑠 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑖𝑠 𝑑𝑜 
𝑚𝑒𝑠𝑚𝑜 𝑑𝑖𝑣𝑖𝑑𝑒 𝑐𝑎𝑑𝑎 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑎𝑜 𝑚𝑒𝑖𝑜, 𝑜𝑢 𝑠𝑒𝑗𝑎,𝑀 é 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑚é𝑑𝑖𝑜 𝑒𝑛𝑡𝑟𝑒 𝑐𝑎𝑑𝑎 
𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑑𝑜 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚𝑜 𝐴𝐵𝐶𝐷.𝐷𝑒 𝑡𝑎𝑙 𝑓𝑜𝑟𝑚𝑎 𝑞𝑢𝑒 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟 
𝑓𝑎𝑐𝑖𝑙𝑚𝑒𝑛𝑡𝑒 𝑜𝑠 𝑜𝑢𝑡𝑟𝑜𝑠 𝑣é𝑟𝑡𝑖𝑐𝑒𝑠, 𝑢𝑚𝑎 𝑣𝑒𝑧 𝑞𝑢𝑒 𝐴 𝑒 𝐵 𝑛ã𝑜 𝑠ã𝑜 𝑜𝑠 𝑣é𝑟𝑡𝑖𝑐𝑒𝑠 𝑞𝑢𝑒 
𝑐𝑜𝑚𝑝õ𝑒𝑚 𝑢𝑚𝑎 𝑑𝑎𝑠 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑖𝑠, 𝑝𝑜𝑜𝑞𝑢𝑒 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑚é𝑑𝑖𝑜 𝑒𝑛𝑡𝑟𝑒 𝐴 𝑒 𝐵 é 𝑃 = (−1,4). 
 
→ 𝐷𝑒𝑠𝑠𝑎 𝑓𝑜𝑟𝑚𝑎, 𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑀 é 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑒𝑛𝑡𝑟𝑒 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝐴 𝑒 𝐶, 𝑒 𝑒𝑛𝑡𝑟𝑒 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 
𝐵 𝑒 𝐷. 𝐷𝑎𝑑𝑜: 𝐶 = (𝑥𝐶 , 𝑦𝐶) 𝑒 𝐷 = (𝑥𝐷 , 𝑦𝐷) 
 
𝑥𝐴 + 𝑥𝐶
2
= 𝑥𝑀 →
−3 + 𝑥𝐶
2
= 1 → −3 + 𝑥𝐶 = 2 ∴ 𝑥𝐶 = 5 
 
𝑦𝐴 + 𝑦𝐶
2
= 𝑦𝑀 →
5 + 𝑦𝐶
2
= 1 → 5 + 𝑦𝐶 = 2 ∴ 𝑦𝐶 = −3 
 
𝑃𝑜𝑛𝑡𝑜 𝐶 = (5,−3) 
 
𝑥𝐵 + 𝑥𝐷
2
= 𝑥𝑀 →
1 + 𝑥𝐷
2
= 1 → 1 + 𝑥𝐷 = 2 ∴ 𝑥𝐷 = 1 
 
𝑦𝐵 + 𝑦𝐷
2
= 𝑦𝑀 →
7 + 𝑦𝐷
2
= 1 → 7 + 𝑦𝐷 = 2 ∴ 𝑦𝐷 = −5 
 
∗ 𝑉é𝑟𝑡𝑖𝑐𝑒𝑠 𝐴 = (−3,5), 𝐵 = (1,7), 𝐶 = (5,−3) 𝑒 𝐷 = (1,−5) 
 
Geometria Analítica (Resolução) – Prova 1 – 28/03/2015 Carlos Alberto 
𝑆𝑒𝑗𝑎 �⃗� = 𝐴𝐷⃗⃗ ⃗⃗ ⃗ = (4, −10) 𝑒 𝑣 = 𝐴𝐵⃗⃗⃗⃗ ⃗ = (4,2) , 𝑒𝑛𝑡ã𝑜 ‖�⃗� ‖ = √116 𝑒 ‖𝑣 ‖ = √20 ; 
→ 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑛𝑑𝑜 𝑜 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜 𝐴𝐵̅̅ ̅̅ 𝑐𝑜𝑚𝑜 𝑎 𝑏𝑎𝑠𝑒 𝑑𝑜 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚𝑜, 𝑒𝑛𝑡ã𝑜 𝑎 
𝑎𝑙𝑡𝑢𝑟𝑎 𝑑𝑒𝑠𝑠𝑒 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚𝑜 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑒 𝑎𝑜 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 à 𝐴𝐵̅̅ ̅̅ , 𝑑𝑎𝑑𝑜 
𝑝𝑒𝑙𝑎 𝑛𝑜𝑟𝑚𝑎 𝑑𝑜 𝑣𝑒𝑡𝑜𝑟 𝑜𝑟𝑡𝑜𝑔𝑜𝑛𝑎𝑙 à 𝑝𝑟𝑜𝑗𝑒çã𝑜 𝑑𝑜 𝑣𝑒𝑡𝑜𝑟 𝑣 𝑠𝑜𝑏𝑟𝑒 𝑜 𝑜 𝑣𝑒𝑡𝑜𝑟 �⃗� . 𝑂𝑢 𝑠𝑒𝑗𝑎, 
ℎ = ‖𝑣 − 𝑃�⃗⃗� 
�⃗� ‖. 
 
𝑃�⃗⃗� 
�⃗� =
< �⃗� , 𝑣 >
‖�⃗� ‖2
. �⃗� =
4.4 + (−10). 2
116
. (4, −10) = −
4
116
(4,−10) = (−
16
116
,
40
116
) 
 
𝑣 − 𝑃�⃗⃗� 
�⃗� = (4,2) − (−
16
116
,
40
116
) = (
480
116
,
192
116
) =
1
116
(480,192) =
96
116
(5,2) 
 
‖𝑣 − 𝑃�⃗⃗� 
�⃗� ‖ =
96
116
. ‖(5,2)‖ =
96
116
√29 
 
∗ 𝑀𝑒𝑑𝑖𝑑𝑎 𝑑𝑎 𝑏𝑎𝑠𝑒 = ‖�⃗� ‖ = √116 
 
𝑆𝐴𝐵𝐶𝐷 = 𝑏 × ℎ = √116 ×
96
116
√29 =
96√116√29
116
=
96√4√29√29
116
=
96(2)(29)
116
= 
96(58)
2(58)
=
96
2
= 48𝑢. 𝐴