IME / ITA ⇒ (Nível-Ime/Ita)Geometria Analítica Tópico resolvido

[Flavio2020]

Dado o gráfico, N(-5;0) e NY=YL.Cakcule as coordenadas de P.

IMG-20240410-WA0016.jpg (29.33 KiB) Exibido 113 vezes

a)(1;3)
b)(2;3)
c)(1;2)
d)(1/2;2)
e)(2;4)

Resposta

---

c

[petras]

Flavio2020,

Gabarito errado

[tex3]\triangle AON_{(ret-isosc)}\implies A=(0,5).\\

Y_{(pto~médio)} \implies \triangle NYO_{(isosc)}\\

\angle POA=\angle YOL=90^o−\angle NOY=90^o−\frac{180^o−53^o}{2}=\frac{53^o}{2}\\

\triangle PCA: \sqrt5k = 5\implies k =\sqrt5 \implies PO= 2k=2\sqrt5\\
\triangle POD: \sqrt5k = PO=2\sqrt5 \implies k = 2 \implies OD = k =2\\
PD = 2k = 4\\
\therefore \boxed{P(2,4)}

[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\triangle AON_{(ret-isosc)}\implies A=(0,5).\\

Y_{(pto~médio)} \implies \triangle NYO_{(isosc)}\\

\angle POA=\angle YOL=90^o−\angle NOY=90^o−\frac{180^o−53^o}{2}=\frac{53^o}{2}\\

\triangle PCA: \sqrt5k = 5\implies k =\sqrt5 \implies PO= 2k=2\sqrt5\\
\triangle POD: \sqrt5k = PO=2\sqrt5 \implies k = 2 \implies OD = k =2\\
PD = 2k = 4\\
\therefore \boxed{P(2,4)}

[/tex3]