Página08
0200
Determine z ∈ C z\in\mathbb{C} z ∈ C tal que z 2 = i z^2=i z 2 = i .
0200 - Solução
Determinando z ∈ C z\in\mathbb{C} z ∈ C tal que z 2 = i z^2=i z 2 = i :
z = a + b i z 2 = i ⇒ z 2 = ( a 2 − b 2 ) + 2 a b i = i \left.
\begin{array}{rcl}
z & = & a+bi\\
z^2 & = & i\\
\end{array}\right.\Rightarrow z^2=(a^2-b^2)+2abi=i z z 2 = = a + bi i ⇒ z 2 = ( a 2 − b 2 ) + 2 abi = i
Então: { a 2 − b 2 = 0 2 a b = 1 … e, resolvendo o sistema, teremos: \left\{
\begin{array}{rcl}
a^2-b^2 & = & 0\\
2ab & = & 1\\
\end{array}\right.\ldots\text{e, resolvendo o sistema, teremos:} { a 2 − b 2 2 ab = = 0 1 … e, resolvendo o sistema, teremos:
I ) Se a = 2 2 , b = 2 2 , e z = 2 2 + i 2 2 I I ) Se a = − 2 2 , b = − 2 2 , e z = − 2 2 − i 2 2 \left.
\begin{array}{rcl}
I)\,\,\text{Se}\quad a=\dfrac{\sqrt{2}}{2}, & b=\dfrac{\sqrt{2}}{2}, &\text{e}\quad \boxed{z=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2}} \\
& & \\
II)\,\,\text{Se}\quad a=-\dfrac{\sqrt{2}}{2}, & b=-\dfrac{\sqrt{2}}{2}, &\text{e}\quad \boxed{z=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}}\\
\end{array}\right. I ) Se a = 2 2 , II ) Se a = − 2 2 , b = 2 2 , b = − 2 2 , e z = 2 2 + i 2 2 e z = − 2 2 − i 2 2
0199
Determine z ∈ C z\in\mathbb{C} z ∈ C tal que z 2 = 1 + i 3 z^2=1+i\sqrt{3} z 2 = 1 + i 3
0199 - Solução
Determinando z ∈ C z\in\mathbb{C} z ∈ C tal que z 2 = 1 + i 3 z^2=1+i\sqrt{3} z 2 = 1 + i 3 :
z = a + b i z 2 = 1 + i 3 ⇒ z 2 = ( a 2 − b 2 ) + 2 a b i = 1 + i 3 \left.
\begin{array}{lcl}
z & = & a+bi \\
z^2 & = & 1+i\sqrt{3}\\
\end{array}\right.\Rightarrow z^2=(a^2-b^2)+2abi=1+i\sqrt{3} z z 2 = = a + bi 1 + i 3 ⇒ z 2 = ( a 2 − b 2 ) + 2 abi = 1 + i 3
Então: { a 2 − b 2 = 1 2 a b = 3 … e, resolvendo o sistema, teremos: \left\{
\begin{array}{rcr}
a^2-b^2 & = & 1 \\
2ab & = & \sqrt{3}\\
\end{array}\right.\ldots\text{e, resolvendo o sistema, teremos:} { a 2 − b 2 2 ab = = 1 3 … e, resolvendo o sistema, teremos:
I ) Se a = 6 2 , b = 2 2 , e z = 6 2 + i 2 2 I I ) Se a = − 6 2 , b = − 2 2 , e z = − 6 2 − i 2 2 \left.
\begin{array}{rcl}
I)\,\,\text{Se}\quad a=\dfrac{\sqrt{6}}{2}, & b=\dfrac{\sqrt{2}}{2}, &\text{e}\quad \boxed{z=\dfrac{\sqrt{6}}{2}+i\dfrac{\sqrt{2}}{2}} \\
& & \\
II)\,\,\text{Se}\quad a=-\dfrac{\sqrt{6}}{2}, & b=-\dfrac{\sqrt{2}}{2}, &\text{e}\quad \boxed{z=-\dfrac{\sqrt{6}}{2}-i\dfrac{\sqrt{2}}{2}}\\
\end{array}\right. I ) Se a = 2 6 , II ) Se a = − 2 6 , b = 2 2 , b = − 2 2 , e z = 2 6 + i 2 2 e z = − 2 6 − i 2 2
0198
Sendo x 2 + y 2 = 1 x^2+y^2=1 x 2 + y 2 = 1 , prove que 1 + x + i y 1 + x − i y = x + i y \dfrac{1+x+iy}{1+x-iy}=x+iy 1 + x − i y 1 + x + i y = x + i y
0198 - Prova
Vamos à prova:
x 2 + y 2 = 1 ( I ) ⇒ y 2 = x 2 − 1 ( I I ) x^2+y^2=1\,(I)\Rightarrow y^2=x^2-1\,(II) x 2 + y 2 = 1 ( I ) ⇒ y 2 = x 2 − 1 ( II )
1 + x + i y 1 + x − i y = ( 1 + x ) + i y ( 1 + x ) − i y ⋅ ( 1 + x ) + i y ( 1 + x ) + i y = = [ ( 1 + x ) 2 − y 2 ] + 2 ( 1 + x ) y i ( 1 + x ) 2 + y 2 = = ( I I ) [ 1 + 2 x + x 2 + ( x 2 − 1 ) ] + 2 ( 1 + x ) y i 1 + 2 x + ( x 2 + y 2 ) = = ( I ) 2 x ( 1 + x ) + 2 ( 1 + x ) y i 1 + 2 x + 1 = = 2 ( 1 + x ) ( x + y i ) 2 ( 1 + x ) → 1 + x + i y 1 + x − i y = x + y i \left.
\begin{array}{rcl}
\dfrac{1+x+iy}{1+x-iy} & = & \dfrac{(1+x)+iy}{(1+x)-iy}\cdot\dfrac{(1+x)+iy}{(1+x)+iy}=\\
& & \\
& = & \dfrac{[(1+x)^2-y^2]+2(1+x)yi}{(1+x)^2+y^2}=\\
& &\\
& \overset{(II)}{=} & \dfrac{[1+2x+x^2+(x^2-1)]+2(1+x)yi}{1+2x+(x^2+y^2)}=\\
& &\\
& \overset{(I)}{=} & \dfrac{2x(1+x)+2(1+x)yi}{1+2x+1}=\\
& &\\
& = & \dfrac{\cancel{2(1+x)}(x+yi)}{\cancel{2(1+x)}}\to\\
\dfrac{1+x+iy}{1+x-iy} & = & x+yi\\
\end{array}\right. 1 + x − i y 1 + x + i y 1 + x − i y 1 + x + i y = = = ( II ) = ( I ) = = ( 1 + x ) − i y ( 1 + x ) + i y ⋅ ( 1 + x ) + i y ( 1 + x ) + i y = ( 1 + x ) 2 + y 2 [( 1 + x ) 2 − y 2 ] + 2 ( 1 + x ) y i = 1 + 2 x + ( x 2 + y 2 ) [ 1 + 2 x + x 2 + ( x 2 − 1 )] + 2 ( 1 + x ) y i = 1 + 2 x + 1 2 x ( 1 + x ) + 2 ( 1 + x ) y i = 2 ( 1 + x ) 2 ( 1 + x ) ( x + y i ) → x + y i
0197
Sabemos que "Q Q Q " é o afixo de z 1 = − 2 + i z_{1}=-2+i z 1 = − 2 + i e "R R R " é o afixo de z 2 = 3 − 2 i z_{2}=3-2i z 2 = 3 − 2 i . Escreva o par ordenado que corresponde a:
a ) ( z 1 ) 2 b ) z 2 ‾ z 1 c ) z 2 × z 1 d ) z 2 − z 1 e ) z 1 ‾ \begin{array}{lllll}
a) (z_{1})^2 & b) \dfrac{\overline{z_{2}}}{z_{1}} & c) z_{2}\times z_{1} & d) z_{2}-z_{1} & e) \overline{z_{1}}
\end{array} a ) ( z 1 ) 2 b ) z 1 z 2 c ) z 2 × z 1 d ) z 2 − z 1 e ) z 1
0197 - Soluções
Resolvendo, item a item:
a) ( z 1 ) 2 = ( − 2 + i ) 2 = 4 − 4 i − 1 = 3 − 4 i → ( 3 ; − 4 ) (z_{1})^2=(-2+i)^2=4-4i-1=3-4i\to (3;\,-4) ( z 1 ) 2 = ( − 2 + i ) 2 = 4 − 4 i − 1 = 3 − 4 i → ( 3 ; − 4 )
b) z 2 ‾ z 1 = 3 + 2 i i − 2 × i + 2 i + 2 = 3 i + 6 + 2 i 2 + 4 i i 2 − 4 = − 4 5 − i 7 5 → ( − 4 5 ; − 7 5 ) \dfrac{\overline{z_{2}}}{z_{1}}=\dfrac{3+2i}{i-2}\times \dfrac{i+2}{i+2}=\dfrac{3i+6+2i^2+4i}{i^2-4}=-\dfrac{4}{5}-i\dfrac{7}{5}\to \left( -\dfrac{4}{5};\,-\dfrac{7}{5} \right) z 1 z 2 = i − 2 3 + 2 i × i + 2 i + 2 = i 2 − 4 3 i + 6 + 2 i 2 + 4 i = − 5 4 − i 5 7 → ( − 5 4 ; − 5 7 )
c) z 2 × z 1 = ( 3 − 2 i ) × ( i − 2 ) = 3 i − 6 − 2 i 2 + 4 i = − 4 + 7 i → ( − 4 ; 7 ) z_{2}\times z_{1}=(3-2i)\times (i-2)=3i-6-2i^2+4i=-4+7i\to (-4;\,7) z 2 × z 1 = ( 3 − 2 i ) × ( i − 2 ) = 3 i − 6 − 2 i 2 + 4 i = − 4 + 7 i → ( − 4 ; 7 )
d) z 2 − z 1 = 3 − 2 i − ( i − 2 ) = ( 5 − 3 i ) → ( 5 ; − 3 ) z_{2}-z_{1}=3-2i-(i-2)=(5-3i)\to (5;\,-3) z 2 − z 1 = 3 − 2 i − ( i − 2 ) = ( 5 − 3 i ) → ( 5 ; − 3 )
e) z 1 ‾ = − 2 − i → ( − 2 ; − 1 ) \overline{z_{1}}=-2-i\to (-2;\,-1) z 1 = − 2 − i → ( − 2 ; − 1 )
0196
Calcule a área do triângulo C D E CDE C D E , onde "C C C " representa o afixo de i + 4 i+4 i + 4 ; "D D D " representa o afixo de 5 i + 1 5i+1 5 i + 1 e "E E E " representa o afixo de i − 2 i-2 i − 2 .
0196 - Solução
Vamos iniciar pela análise da imagem gerada pelos afixos e tecer as observações:
É visível que:
a) O ponto médio do segmento E C ‾ ( M E C ) \overline{EC}\,(M_{EC}) EC ( M EC ) tem coordenadas M E C = ( 1 ; 1 ) M_{EC}=(1;\,1) M EC = ( 1 ; 1 ) , cuja abscissa "1 1 1 " coincide com a abscissa do ponto "D = ( 1 ; 5 ) D=(1;\,5) D = ( 1 ; 5 ) ". Por isso, a base E C ‾ \overline{EC} EC tem comprimento igual a 6 u 6u 6 u e, ainda, o Δ C D E \Delta CDE Δ C D E é, no mínimo, isósceles;
b) A altura "h h h " tem comprimento igual a 4 u 4u 4 u , pois tem início na ordenada 5 5 5 do ponto D D D e termina na ordenada 1 1 1 do ponto médio M E C M_{EC} M EC .
Com esses dados, já é possível calcular a área(A A A ) do Δ C D E \Delta CDE Δ C D E : A = 6 ⋅ 4 2 → A = 12 u . a A=\dfrac{6\cdot 4}{2}\to\boxed{A=12\,u.a} A = 2 6 ⋅ 4 → A = 12 u . a
0195
Sendo "M M M " o afixo de z = − 3 + 5 i z=-3+5i z = − 3 + 5 i , determine o par ordenado que corresponde ao:
a) ponto simétrico de "z z z " em relação ao eixo vertical I m ( z ) Im(z) I m ( z ) ;
b) ponto simétrico de "z z z " em relação à origem;
c) número complexo obtido de z × i z\times i z × i .
0195 - Soluções
Partindo das coordenadas do afixo, que são ( − 3 ; 5 ) (-3;\,5) ( − 3 ; 5 ) , teremos:
a) ( 3 ; 5 ) (3;\,5) ( 3 ; 5 )
b) ( 3 ; − 5 ) (3;\,-5) ( 3 ; − 5 )
c) z ⋅ i = i ( − 3 + 5 i ) = − 5 − 3 i → ( − 5 ; − 3 ) z\cdot i=i(-3+5i)=-5-3i\to (-5;\,-3) z ⋅ i = i ( − 3 + 5 i ) = − 5 − 3 i → ( − 5 ; − 3 )
0194
Represente no plano de Argand-Gauss todos os números complexos que satisfazem, concomitantemente as inequações I m ( z ) < 3 Im(z)<3 I m ( z ) < 3 e R e ( z ) ≤ 3 Re(z)\leq 3 R e ( z ) ≤ 3 .
0194 - Solução
Lembrando que o termo "concomitantemente", neste caso, significa a interseção dessas duas inequações. Vamos às representações gráficas, para visualizar melhor e, após, a descrição detalhada:
A resposta final é a área hachurada em cinza mais escuro que corresponde à interseção dos semi planos α : R e ( z ) ≤ 3 \alpha: Re(z)\leq 3 α : R e ( z ) ≤ 3 e β : I m ( z ) < 3 \beta: Im(z)<3 β : I m ( z ) < 3 , ou seja α ∩ β \boxed{\alpha\cap\beta} α ∩ β .
0193
Os números complexos z z z e z ‾ \overline{z} z tais que:
{ z + z ‾ = 4 z ⋅ z ‾ = 13 \left\{\begin{array}{rcrcrr}
z & + & \overline{z} & = & 4 & \\
z & \cdot & \overline{z} & = & 13 &
\end{array}\right. { z z + ⋅ z z = = 4 13
são representados no plano de Argand-Gauss pelos pontos A A A e B B B .
Qual é área do Δ A B O \Delta ABO Δ A BO , sendo O O O a origem do plano?
0193 - Solução
Resolvendo o sistema, onde trataremos z = a + b i z=a+bi z = a + bi e z ‾ = a − b i \overline{z}=a-bi z = a − bi , chegamos a solução que: a = 2 a=2 a = 2 e b = ± 3 b=\pm 3 b = ± 3 . Assim, A ( 2 ; 3 ) A(2;\,3) A ( 2 ; 3 ) e B ( 2 ; − 3 ) B(2;\,-3) B ( 2 ; − 3 ) ou vice-versa, conforme ilustra a imagem a seguir:
Da imagem, podemos obter a altura(h h h ) do Δ A B O \Delta ABO Δ A BO no eixo horizontal R e ( z ) Re(z) R e ( z ) , da origem até o ponto ( 2 , 0 ) (2,0) ( 2 , 0 ) , portanto, h = 2 h=2 h = 2 . A base(b b b ) deste triângulo pode ser obtida da distância(visual) entre os pontos A A A e B B B , ou seja, b = 6 b=6 b = 6 . Portanto, a área(A A A ) será dada por:
A = b × h 2 → A = 6 × 2 2 → A = 6 u . a A=\dfrac{b\times h}{2}\to A=\dfrac{6\times 2}{2}\to \boxed{A=6\,u.a} A = 2 b × h → A = 2 6 × 2 → A = 6 u . a
0192
Represente o conjunto A = { z ∈ C / ∣ z − 1 ∣ = ∣ z − 3 i ∣ } A=\left\{ z\in\mathbb{C}/\,|z-1|=|z-3i|\right\} A = { z ∈ C / ∣ z − 1∣ = ∣ z − 3 i ∣ } .
0192 - Solução
Adotando z = a + b i z=a+bi z = a + bi , com a ∈ R a\in\mathbb{R} a ∈ R e b ∈ R b\in\mathbb{R} b ∈ R , teremos:
∣ a + b i − 1 ∣ = ∣ a + b i − 3 i ∣ → ∣ ( a − 1 ) + b i ∣ = ∣ a + i ( b − 3 ) ∣ → \left|a+bi-1\right|=\left|a+bi-3i\right|\to \left|(a-1)+bi\right|=\left|a+i(b-3)\right|\to ∣ a + bi − 1 ∣ = ∣ a + bi − 3 i ∣ → ∣ ( a − 1 ) + bi ∣ = ∣ a + i ( b − 3 ) ∣ →
( a − 1 ) 2 + b 2 = a 2 + ( b − 3 ) 2 → \sqrt{(a-1)^2+b^2}=\sqrt{a^2+(b-3)^2}\to ( a − 1 ) 2 + b 2 = a 2 + ( b − 3 ) 2 →
a 2 − 2 b + 1 + b 2 = a 2 + b 2 − 6 b + 9 ⇒ a^2-2b+1+b^2=a^2+b^2-6b+9\Rightarrow a 2 − 2 b + 1 + b 2 = a 2 + b 2 − 6 b + 9 ⇒
a − 3 b + 4 = 0 \boxed{a-3b+4=0} a − 3 b + 4 = 0
Portanto, teremos a equação de uma reta mediatriz do segmento de pontos ( 1 ; 0 ) (1;\,0) ( 1 ; 0 ) e ( 0 ; 3 ) (0;\,3) ( 0 ; 3 ) .
0191
Interprete geometricamente ∣ z − 2 + i ∣ = 5 |z-2+i|=5 ∣ z − 2 + i ∣ = 5 , para o número complexo z = a + b i z=a+bi z = a + bi .
0191 - Solução
Adotando z = a + b i z=a+bi z = a + bi , teremos ∣ a + b i − ( 2 − i ) ∣ = 5 |a+bi-(2-i)|=5 ∣ a + bi − ( 2 − i ) ∣ = 5 . Essa equação representa todos os números complexos "z z z " que possuem a distância até o número 2 − i 2-i 2 − i constante e igual a 5 5 5 . Trata-se da circunferência de centro C ( 2 ; − 1 ) C(2;\,-1) C ( 2 ; − 1 ) e raio 5 5 5 . Com este exercício resolvido, podemos generalizar:
∣ z − z 0 ∣ = a \left|z-z_{0}\right|=a ∣ z − z 0 ∣ = a , tal que z 0 ∈ C z_{0}\in\mathbb{C} z 0 ∈ C e a ∈ R + ∗ a\in\mathbb{R}^*_{+} a ∈ R + ∗ representa uma circunferência de centro z 0 z_{0} z 0 e raio "a a a ":
Obs : Imagem meramente ilustrativa e sem escala.
0190
Determine os valores mínimo e máximo de ∣ z − 4 ∣ |z-4| ∣ z − 4∣ sendo ∣ z + 3 i ∣ ≤ 1 |z+3i|\leq1 ∣ z + 3 i ∣ ≤ 1 , para z = a + b i z=a+bi z = a + bi .
0190 - Solução
Sabe-se que ∣ z − 4 ∣ |z-4| ∣ z − 4∣ representa a distância do afixo de z z z ao ponto ( 4 ; 0 ) (4;\,0) ( 4 ; 0 ) . Vamos determinar a licalização de z z z . Pela interpretação geométrica, z z z pertence a uma circunferência de centro ( 0 ; − 3 ) (0;\,-3) ( 0 ; − 3 ) e raio unitário.
Traçando uma reta pelos pontos ( 4 ; 0 ) (4;\,0) ( 4 ; 0 ) e ( 0 ; − 3 ) (0;\,-3) ( 0 ; − 3 ) encontramos os pontos "A A A " e "B B B ". Até o ponto "A A A " temos a menor distância, e até o ponto "B B B " temos a maior distância. Assim: A C ‾ = 4 \overline{AC}=4 A C = 4 e B C ‾ = 6 \overline{BC}=6 BC = 6 .
0189
Calcule ( − 2 1 + i ) 93 \left(-\dfrac{\sqrt{2}}{1+i}\right)^{93} ( − 1 + i 2 ) 93
0189 - Solução
Primeiramente, vamos passar o número complexo dado, que está na forma algébrica, para sua forma trigonométrica; posteriormente, efetuaremos a potenciação pedida; assim:
− 2 1 + i ⋅ 1 − i 1 − i = … = 2 2 − i ⋅ 2 2 = 2 2 ⋅ ( 1 − i ) -\dfrac{\sqrt{2}}{1+i}\cdot\dfrac{1-i}{1-i}=\ldots=\dfrac{\sqrt{2}}{2}-i\cdot\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2}\cdot (1-i) − 1 + i 2 ⋅ 1 − i 1 − i = … = 2 2 − i ⋅ 2 2 = 2 2 ⋅ ( 1 − i ) .
Na forma trigonométrica, ( 1 − i ) (1-i) ( 1 − i ) é: 2 ⋅ c i s ( 7 π 4 ) \sqrt{2}\cdot cis\left(\dfrac{7\pi}{4}\right) 2 ⋅ c i s ( 4 7 π )
Aplicando a potenciação a toda a expressão:
[ 2 2 ⋅ 2 ⋅ c i s ( 7 π 4 ) ] 93 = [ c i s ( 7 π 4 ) ] 93 = c i s ( 93 ⋅ 7 π 4 ) = \left[\dfrac{\sqrt{2}}{2}\cdot\sqrt{2}\cdot cis\left(\dfrac{7\pi}{4}\right)\right]^{93}=\left[cis\left(\dfrac{7\pi}{4}\right)\right]^{93}=cis\left(\dfrac{93\cdot 7\pi}{4}\right)= [ 2 2 ⋅ 2 ⋅ c i s ( 4 7 π ) ] 93 = [ c i s ( 4 7 π ) ] 93 = c i s ( 4 93 ⋅ 7 π ) =
= c i s ( 3 π 4 ) = cos 3 π 4 + i ⋅ sen 3 π 4 = − 2 2 + i ⋅ 2 2 =cis\left(\dfrac{3\pi}{4}\right)=\text{cos}\,\dfrac{3\pi}{4}+i\cdot\text{sen}\,\dfrac{3\pi}{4}=-\dfrac{\sqrt{2}}{2}+i\cdot\dfrac{\sqrt{2}}{2} = c i s ( 4 3 π ) = cos 4 3 π + i ⋅ sen 4 3 π = − 2 2 + i ⋅ 2 2 (resposta final)
0188
Calcule ( − 1 2 − i ⋅ 3 2 ) 40 \left(-\dfrac{1}{2}-i\cdot\dfrac{\sqrt{3}}{2}\right)^{40} ( − 2 1 − i ⋅ 2 3 ) 40
0188 - Solução
Idem ao anterior, vamos passar o número complexo dado, que está na forma algébrica, para sua forma trigonométrica; posteriormente, efetuaremos a potenciação pedida; assim:
∣ z ∣ = ρ = ( − 1 2 ) 2 + ( − 3 2 ) 2 ⇒ ρ = 1 |z|=\rho=\sqrt{\left(-\dfrac{1}{2}\right)^2+\left(-\dfrac{\sqrt{3}}{2}\right)^2}\Rightarrow \boxed{\rho=1} ∣ z ∣ = ρ = ( − 2 1 ) 2 + ( − 2 3 ) 2 ⇒ ρ = 1
sen θ = 3 2 cos θ = − 1 2 ⇒ θ ∈ 3 o Q ∴ θ = 12 0 o = 4 π 3 r a d \left.\begin{array}{rcr}
\text{sen}\theta & = & \dfrac{\sqrt{3}}{2}\\
& & \\
\text{cos}\theta & = & -\dfrac{1}{2}
\end{array}\right.\Rightarrow \theta\in 3^{o}\,Q\,\,\therefore\,\,\boxed{\theta=120^{o}=\dfrac{4\pi}{3}\,rad} sen θ cos θ = = 2 3 − 2 1 ⇒ θ ∈ 3 o Q ∴ θ = 12 0 o = 3 4 π r a d
Portanto devemos efetuar a seguinte potenciação:
[ 1 ⋅ c i s ( 4 π 3 ) ] 40 = 1 40 ⋅ c i s ( 40 ⋅ 4 π 3 ) = \left[1\cdot cis\left(\dfrac{4\pi}{3}\right)\right]^{40}=1^{40}\cdot cis\left(40\cdot\dfrac{4\pi}{3}\right)= [ 1 ⋅ c i s ( 3 4 π ) ] 40 = 1 40 ⋅ c i s ( 40 ⋅ 3 4 π ) =
= 1 ⋅ c i s ( 160 π 3 ) = c i s ( 156 π 3 26 voltas de 6 π 3 cada + 4 π 3 ) = =1\cdot cis\left(\dfrac{160\pi}{3}\right)=cis\left(\cancel{\dfrac{156\pi}{3}}^{\,\,26\,\text{voltas de }\frac{6\pi}{3}\,\,\text{cada}}+\dfrac{4\pi}{3}\right)= = 1 ⋅ c i s ( 3 160 π ) = c i s ( 3 156 π 26 voltas de 3 6 π cada + 3 4 π ) =
= c i s ( 4 π 3 ) =cis\left(\dfrac{4\pi}{3}\right) = c i s ( 3 4 π ) ou ( − 1 2 − i ⋅ 3 2 ) \left(-\dfrac{1}{2}-i\cdot\dfrac{\sqrt{3}}{2}\right) ( − 2 1 − i ⋅ 2 3 )
0187
Dados z 1 = 3 ⋅ cis π 4 z_{1}=3\cdot \text{cis}\,\dfrac{\pi}{4} z 1 = 3 ⋅ cis 4 π e z 2 = 4 ⋅ cis π 2 z_{2}=4\cdot \text{cis}\,\dfrac{\pi}{2} z 2 = 4 ⋅ cis 2 π , calcule:
a ) z 1 ⋅ z 2 b ) z 1 2 c ) z 2 2 d ) z 1 3 ⋅ 3 e ) z 2 3 \begin{array}{lllll}
a)\,z_{1}\cdot z_{2} & b)\,z_{1}^2 & c)\,z_{2}^2 & d)\,z_{1}^3\cdot\sqrt{3} & e)\,z_{2}^3
\end{array} a ) z 1 ⋅ z 2 b ) z 1 2 c ) z 2 2 d ) z 1 3 ⋅ 3 e ) z 2 3
0187 - Soluções
Resolvendo uma a uma:
a) z 1 ⋅ z 2 = 3 ⋅ 4 ⋅ c i s ( π 4 + π 2 ) = 12 ⋅ c i s 3 π 4 z_{1}\cdot z_{2}=3\cdot 4\cdot cis\left(\dfrac{\pi}{4}+\dfrac{\pi}{2}\right)=12\cdot cis\dfrac{3\pi}{4} z 1 ⋅ z 2 = 3 ⋅ 4 ⋅ c i s ( 4 π + 2 π ) = 12 ⋅ c i s 4 3 π
b) z 1 2 = 3 ⋅ 3 ⋅ c i s ( π 4 + π 4 ) = 9 ⋅ c i s π 2 z_{1}^2=3\cdot 3\cdot cis\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}\right)=9\cdot cis\dfrac{\pi}{2} z 1 2 = 3 ⋅ 3 ⋅ c i s ( 4 π + 4 π ) = 9 ⋅ c i s 2 π
c) z 2 2 = 4 ⋅ 4 ⋅ c i s ( π 2 + π 2 ) = 16 ⋅ c i s π z_{2}^2=4\cdot 4\cdot cis\left(\dfrac{\pi}{2}+\dfrac{\pi}{2}\right)=16\cdot cis\pi z 2 2 = 4 ⋅ 4 ⋅ c i s ( 2 π + 2 π ) = 16 ⋅ c i s π
d) z 1 3 ⋅ 3 = 3 ⋅ 3 ⋅ 3 ⋅ c i s ( π 4 + π 4 + π 4 ) ⋅ 3 = z_{1}^3\cdot\sqrt{3}=3\cdot 3\cdot 3\cdot cis\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}+\dfrac{\pi}{4}\right)\cdot\sqrt{3}= z 1 3 ⋅ 3 = 3 ⋅ 3 ⋅ 3 ⋅ c i s ( 4 π + 4 π + 4 π ) ⋅ 3 =
= 3 ⋅ [ 27 ⋅ c i s 3 π 4 ] = … = 3 ⋅ [ 27 ⋅ ( − 2 2 + i ⋅ 2 2 ) ] = =\sqrt{3}\cdot \left[27\cdot cis\dfrac{3\pi}{4}\right]=\ldots=\sqrt{3}\cdot \left[27\cdot\left(-\dfrac{\sqrt{2}}{2}+i\cdot\dfrac{\sqrt{2}}{2}\right)\right]= = 3 ⋅ [ 27 ⋅ c i s 4 3 π ] = … = 3 ⋅ [ 27 ⋅ ( − 2 2 + i ⋅ 2 2 ) ] =
3 ⋅ ( − 27 2 2 + i ⋅ 27 2 2 ) = − 27 6 2 + i ⋅ 27 6 2 \sqrt{3}\cdot \left(-\dfrac{27\sqrt{2}}{2}+i\cdot\dfrac{27\sqrt{2}}{2}\right)=\boxed{-\dfrac{27\sqrt{6}}{2}+i\cdot\dfrac{27\sqrt{6}}{2}} 3 ⋅ ( − 2 27 2 + i ⋅ 2 27 2 ) = − 2 27 6 + i ⋅ 2 27 6
e) z 2 3 = 4 ⋅ 4 ⋅ 4 ⋅ c i s ( π 2 + π 2 + π 2 ) = 64 ⋅ c i s 3 π 2 z_{2}^3=4\cdot 4\cdot 4\cdot cis\left(\dfrac{\pi}{2}+\dfrac{\pi}{2}+\dfrac{\pi}{2}\right)=64\cdot cis\dfrac{3\pi}{2} z 2 3 = 4 ⋅ 4 ⋅ 4 ⋅ c i s ( 2 π + 2 π + 2 π ) = 64 ⋅ c i s 2 3 π
0186
Disserte sobre "O Plano Cartesiano "
0186 - Dissertação
Assim como você pode representar números reais por pontos em uma reta de números reais, você pode representar pares de números reais por pontos em um plano chamado sistema de coordenadas retangulares ou plano cartesiano , assim nomeado pelo matemático francês René Descartes (1596-1650).
Duas retas de números reais intersectam-se em ângulos retos formando o plano Cartesiano. A reta horizontal é chamada de eixo "x" ou eixo das abscissas, e a reta vertical é chamada de eixo "y" ou eixo das ordenadas. O ponto de intersecção dos dois eixos é chamado de origem e estes dois eixos dividem o plano em quatro quartos chamados de quadrantes .
A cada ponto no plano associa-se um par ordenado ( x , y ) (x,\,y) ( x , y ) de números reais "x x x " e "y y y " chamados de coordenadas do ponto. A coordenada "x x x " representa a menor distância entre o eixo das ordenadas (eixo "y y y ") até o ponto. A coordenada y y y representa a menor distância entre o eixo das abscissas (eixo "x x x ") até o ponto.
Exemplo Coloque no plano cartesiano os pontos ( − 1 ; 2 ) , ( 3 ; 4 ) , ( 0 ; 0 ) , ( 3 ; 0 ) , ( − 2 ; − 3 ) (-1;\,2),(3;\,4),(0;\,0),(3;\,0),(-2;\,-3) ( − 1 ; 2 ) , ( 3 ; 4 ) , ( 0 ; 0 ) , ( 3 ; 0 ) , ( − 2 ; − 3 ) :
0185
A beleza do sistema de coordenadas retangulares ou plano Cartesiano é que ele nos permite observar relações entre as coordenadas dos pares ordenadas que indicam os pontos desse plano. Disserte a respeito de 3(três) aplicações matemáticas.
0185 - Dissertação
Vamos às três aplicações matemáticas:
1ª) O Teorema de Pitágoras
Para um triângulo retângulo com hipotenusa "a a a " e catetos "b b b " e "c c c " vale a relação(ou Teorema de Pitágoras) a 2 = b 2 + c 2 a^2=b^2+c^2 a 2 = b 2 + c 2 . Veja:
Observação : Não podemos esquecer das condições de existência, válidas para quaisquer tipos de triângulos, isto é, para um triângulo qualquer, de lados "a a a ", "b b b " e "c c c ", devemos observar a veracidade dessas afirmações:
∣ b − c ∣ < a < b + c | b - c | < a < b + c ∣ b − c ∣ < a < b + c
∣ a − c ∣ < b < a + c | a - c | < b < a + c ∣ a − c ∣ < b < a + c
∣ a − b ∣ < c < a + b | a - b | < c < a + b ∣ a − b ∣ < c < a + b
2ª) A distância entre dois pontos
Suponha que se queira determinar a distância "d d d " entre dois pontos ( x 1 ; y 1 ) (x_{1};\,y_{1}) ( x 1 ; y 1 ) e ( x 2 ; y 2 ) (x_{2};\,y_{2}) ( x 2 ; y 2 ) no plano. Esses dois pontos podem formar um triângulo retângulo, como mostra a imagem:
O comprimento do cateto vertical é ∣ y 2 − y 1 ∣ |y_{2}-y_{1}| ∣ y 2 − y 1 ∣ e o comprimento do cateto horizontal é ∣ x 2 − x 1 ∣ |x_{2}-x_{1}| ∣ x 2 − x 1 ∣ . Pelo Teorema de Pitágoras, teremos:
d 2 = ∣ x 2 − x 1 ∣ 2 + ∣ y 2 − y 1 ∣ 2 d = ∣ x 2 − x 1 ∣ 2 + ∣ y 2 − y 1 ∣ 2 d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \begin{array}{lcr}
d^2 & = & |x_{2}-x_{1}|^2+|y_{2}-y_{1}|^2\\&&\\
d & = & \sqrt{|x_{2}-x_{1}|^2+|y_{2}-y_{1}|^2}\\&&\\
d & = & \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}
\end{array} d 2 d d = = = ∣ x 2 − x 1 ∣ 2 + ∣ y 2 − y 1 ∣ 2 ∣ x 2 − x 1 ∣ 2 + ∣ y 2 − y 1 ∣ 2 ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
Assim, podemos definir que a distância "d d d " entre dois pontos ( x 1 ; y 1 ) (x_{1};\,y_{1}) ( x 1 ; y 1 ) e ( x 2 ; y 2 ) (x_{2};\,y_{2}) ( x 2 ; y 2 ) no plano cartesiano é dada por
d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
Exemplo01 : Obtenha a distância entre os pontos ( − 2 , 1 ) (-2,\,1) ( − 2 , 1 ) e ( 3 ; 4 ) (3;\,4) ( 3 ; 4 )
Fazendo ( x 1 ; y 1 ) = ( − 2 ; 1 ) (x_{1};\,y_{1})=(-2;\,1) ( x 1 ; y 1 ) = ( − 2 ; 1 ) e ( x 2 ; y 2 ) = ( 3 ; 4 ) (x_{2};\,y_{2})=(3;\,4) ( x 2 ; y 2 ) = ( 3 ; 4 ) , teremos:
d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = [ 3 − ( − 2 ) ] 2 + ( 4 − 1 ) 2 = ( 5 ) 2 + ( 3 ) 2 = 34 d ≈ 5 , 83 \begin{array}{lcl}
d & = & \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}\\&&\\
& = & \sqrt{[3-(-2)]^2+(4-1)^2}\\&&\\
& = & \sqrt{(5)^2+(3)^2}\\&&\\
& = & \sqrt{34}\\\\
d & \approx & 5,83
\end{array} d d = = = = ≈ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 [ 3 − ( − 2 ) ] 2 + ( 4 − 1 ) 2 ( 5 ) 2 + ( 3 ) 2 34 5 , 83
Exemplo02 : Obtenha a distância entre os pontos ( 3 , 1 ) (3,\,1) ( 3 , 1 ) e ( − 3 ; 0 ) (-3;\,0) ( − 3 ; 0 )
Fazendo ( x 1 ; y 1 ) = ( 3 ; 1 ) (x_{1};\,y_{1})=(3;\,1) ( x 1 ; y 1 ) = ( 3 ; 1 ) e ( x 2 ; y 2 ) = ( − 3 ; 0 ) (x_{2};\,y_{2})=(-3;\,0) ( x 2 ; y 2 ) = ( − 3 ; 0 ) , teremos:
d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( − 3 − 3 ) 2 + ( 0 − 1 ) 2 = ( − 9 ) 2 + ( − 1 ) 2 = 37 d ≈ 6 , 08 \begin{array}{lcl}
d & = & \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}\\&&\\
& = & \sqrt{(-3-3)^2+(0-1)^2}\\&&\\
& = & \sqrt{(-9)^2+(-1)^2}\\&&\\
& = & \sqrt{37}\\\\
d & \approx & 6,08
\end{array} d d = = = = ≈ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 ( − 3 − 3 ) 2 + ( 0 − 1 ) 2 ( − 9 ) 2 + ( − 1 ) 2 37 6 , 08
Exemplo03 : Mostre que os pontos ( 2 ; 1 ) (2;\,1) ( 2 ; 1 ) , ( 4 ; 0 ) (4;\,0) ( 4 ; 0 ) e ( 5 ; 7 ) (5;\,7) ( 5 ; 7 ) são vértices de um triângulo retângulo
Vamos encontrar as distâncias d 1 d_{1} d 1 , d 2 d_{2} d 2 e d 3 d_{3} d 3 ; após, vamos aplicar os resultados obtidos diretamente no Teorema de Pitágoras e, finalmente, construir a figura para comprovação visual:
d 1 = ( 5 − 2 ) 2 + ( 7 − 1 ) 2 = 9 + 36 = 45 d 2 = ( 4 − 2 ) 2 + ( 0 − 1 ) 2 = 4 + 1 = 5 d 3 = ( 5 − 4 ) 2 + ( 7 − 0 ) 2 = 1 + 49 = 50 \begin{array}{lcrcrcr}
d_{1} & = & \sqrt{(5-2)^2+(7-1)^2} & = \sqrt{9+36} & = & \sqrt{45} &\\\\
d_{2} & = & \sqrt{(4-2)^2+(0-1)^2} & = \sqrt{4+1} & = & \sqrt{5} &\\\\
d_{3} & = & \sqrt{(5-4)^2+(7-0)^2} & = \sqrt{1+49} & = & \sqrt{50} &
\end{array} d 1 d 2 d 3 = = = ( 5 − 2 ) 2 + ( 7 − 1 ) 2 ( 4 − 2 ) 2 + ( 0 − 1 ) 2 ( 5 − 4 ) 2 + ( 7 − 0 ) 2 = 9 + 36 = 4 + 1 = 1 + 49 = = = 45 5 50
Aplicando os valores encontrados no Teorema de Pitágoras:
( d 1 ) 2 + ( d 2 ) 2 = ( d 3 ) 2 → (d_{1})^2+(d_{2})^2=(d_{3})^2\to ( d 1 ) 2 + ( d 2 ) 2 = ( d 3 ) 2 →
( 45 ) 2 + ( 5 ) 2 = ( 50 ) 2 (\sqrt{45})^2+(\sqrt{5})^2=(\sqrt{50})^2\quad ( 45 ) 2 + ( 5 ) 2 = ( 50 ) 2
3ª) O ponto médio entre dois pontos
O ponto médio M ( x M ; y M ) M(x_{M};\,y_{M}) M ( x M ; y M ) do segmento de extremos ( x 1 ; y 1 ) (x_{1};\,y_{1}) ( x 1 ; y 1 ) e ( x 2 ; y 2 ) (x_{2};\,y_{2}) ( x 2 ; y 2 ) é dado por:
M ( x M ; y M ) = ( x 1 + x 2 2 ; y 1 + y 2 2 ) M(x_{M};\,y_{M})=\left(\dfrac{x_{1}+x_{2}}{2};\,\dfrac{y_{1}+y_{2}}{2}\right) M ( x M ; y M ) = ( 2 x 1 + x 2 ; 2 y 1 + y 2 )
Exemplo01 : Encontre o ponto médio do segmento de reta com extremos em ( − 5 ; − 3 ) (-5;\,-3) ( − 5 ; − 3 ) e ( 9 ; 3 ) (9;\,3) ( 9 ; 3 )
M ( x M ; y M ) = ( x 1 + x 2 2 ; y 1 + y 2 2 ) → M(x_{M};\,y_{M})=\left(\dfrac{x_{1}+x_{2}}{2};\,\dfrac{y_{1}+y_{2}}{2}\right)\to M ( x M ; y M ) = ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) →
M ( x M ; y M ) = ( − 5 + 9 2 ; − 3 + 3 2 ) → M ( 2 ; 0 ) M(x_{M};\,y_{M})=\left(\dfrac{-5+9}{2};\,\dfrac{-3+3}{2}\right)\to M(2;\,0) M ( x M ; y M ) = ( 2 − 5 + 9 ; 2 − 3 + 3 ) → M ( 2 ; 0 )
Observe a imagem ilustrativa:
Exemplo02 : Encontre o ponto médio do segmento de reta com extremos em ( − 2 ; 8 ) (-2;\,8) ( − 2 ; 8 ) e ( 4 ; − 10 ) (4;\,-10) ( 4 ; − 10 )
M ( x M ; y M ) = ( x 1 + x 2 2 ; y 1 + y 2 2 ) → M(x_{M};\,y_{M})=\left(\dfrac{x_{1}+x_{2}}{2};\,\dfrac{y_{1}+y_{2}}{2}\right)\to M ( x M ; y M ) = ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) →
M ( x M ; y M ) = ( − 2 + 4 2 ; 8 − 10 2 ) → M ( 1 ; − 1 ) M(x_{M};\,y_{M})=\left(\dfrac{-2+4}{2};\,\dfrac{8-10}{2}\right)\to M(1;\,-1) M ( x M ; y M ) = ( 2 − 2 + 4 ; 2 8 − 10 ) → M ( 1 ; − 1 )
Exemplo03 : A empresa "A" vendeu aproximadamente R$ 9,8 milhões em 2016 e R$ 11,7 milhões em 2018. Sem qualquer outra informação, estime as vendas de 2017.
Sem qualquer outra informação, vamos assumir uma estimativa linear. Dessa forma, a estimativa para 2017 é o ponto médio entre os pontos ( 2016 ; 9 , 8 ) (2016;\,9,8) ( 2016 ; 9 , 8 ) e ( 2018 ; 11 , 7 ) (2018;\,11,7) ( 2018 ; 11 , 7 )
M ( x M ; y M ) = ( x 1 + x 2 2 ; y 1 + y 2 2 ) → M(x_{M};\,y_{M})=\left(\dfrac{x_{1}+x_{2}}{2};\,\dfrac{y_{1}+y_{2}}{2}\right)\to M ( x M ; y M ) = ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) →
M ( x M ; y M ) = ( 2016 + 2018 2 ; 9 , 8 + 11 , 7 2 ) → M ( 2017 ; 10 , 75 ) M(x_{M};\,y_{M})=\left(\dfrac{2016+2018}{2};\,\dfrac{9,8+11,7}{2}\right)\to M(2017;\,10,75) M ( x M ; y M ) = ( 2 2016 + 2018 ; 2 9 , 8 + 11 , 7 ) → M ( 2017 ; 10 , 75 )
Portanto, a estimativa de venda em 2017 é de R$ 10,75 milhões.
0184
Encontre a distância entre os pontos:
a) ( − 2 ; 6 ) (-2;\,6) ( − 2 ; 6 ) e ( 3 ; − 6 ) (3;\,-6) ( 3 ; − 6 )
b) ( 8 ; 5 ) (8;\,5) ( 8 ; 5 ) e ( 0 ; 20 ) (0;\,20) ( 0 ; 20 )
c) ( 1 ; 4 ) (1;\,4) ( 1 ; 4 ) e ( − 5 ; − 1 ) (-5;\,-1) ( − 5 ; − 1 )
d) ( 1 ; 3 ) (1;\,3) ( 1 ; 3 ) e ( 3 ; − 2 ) (3;\,-2) ( 3 ; − 2 )
e) ( 1 2 ; 4 3 ) \left(\frac{1}{2};\,\frac{4}{3}\right) ( 2 1 ; 3 4 ) e ( 2 ; − 1 ) (2;\,-1) ( 2 ; − 1 )
f) ( 9 , 5 ; − 2 , 6 ) (9,5;\,-2,6) ( 9 , 5 ; − 2 , 6 ) e ( − 3 , 9 ; 8 , 2 ) (-3,9;\,8,2) ( − 3 , 9 ; 8 , 2 )
0184 - Soluções
Utilizando a fórmula da distância entre dois pontos: d = ( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2} d = ( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2
a) ( − 2 ; 6 ) (-2;\,6) ( − 2 ; 6 ) e ( 3 ; − 6 ) (3;\,-6) ( 3 ; − 6 )
→ d = [ 3 − ( − 2 ) ] 2 + ( − 6 − 6 ) 2 → d = 169 → d = 13 \to d=\sqrt{[3-(-2)]^2+(-6-6)^2}\to d=\sqrt{169}\to\boxed{d=13} → d = [ 3 − ( − 2 ) ] 2 + ( − 6 − 6 ) 2 → d = 169 → d = 13
b) ( 8 ; 5 ) (8;\,5) ( 8 ; 5 ) e ( 0 ; 20 ) (0;\,20) ( 0 ; 20 )
→ d = ( 0 − 8 ) 2 + ( 20 − 5 ) 2 → d = 289 → d = 17 \to d=\sqrt{(0-8)^2+(20-5)^2}\to d=\sqrt{289}\to\boxed{d=17} → d = ( 0 − 8 ) 2 + ( 20 − 5 ) 2 → d = 289 → d = 17
c) ( 1 ; 4 ) (1;\,4) ( 1 ; 4 ) e ( − 5 ; − 1 ) (-5;\,-1) ( − 5 ; − 1 )
→ d = ( − 5 − 1 ) 2 + ( − 1 − 4 ) 2 → d = 61 \to d=\sqrt{(-5-1)^2+(-1-4)^2}\to\boxed{d=\sqrt{61}} → d = ( − 5 − 1 ) 2 + ( − 1 − 4 ) 2 → d = 61
d) ( 1 ; 3 ) (1;\,3) ( 1 ; 3 ) e ( 3 ; − 2 ) (3;\,-2) ( 3 ; − 2 )
→ d = ( 3 − 1 ) 2 + ( − 2 − 3 ) 2 → d = 29 \to d=\sqrt{(3-1)^2+(-2-3)^2}\to\boxed{d=\sqrt{29}} → d = ( 3 − 1 ) 2 + ( − 2 − 3 ) 2 → d = 29
e) ( 1 2 ; 4 3 ) \left(\frac{1}{2};\,\frac{4}{3}\right) ( 2 1 ; 3 4 ) e ( 2 ; − 1 ) (2;\,-1) ( 2 ; − 1 )
→ d = ( 2 − 1 2 ) 2 + ( − 1 − 4 3 ) 2 → d = 9 4 + 49 9 → d = 277 36 → d = 277 6 \to d=\sqrt{\left(2-\frac{1}{2}\right)^2+\left(-1-\frac{4}{3}\right)^2}\to d=\sqrt{\frac{9}{4}+\frac{49}{9}}\to d=\sqrt{\frac{277}{36}} \to\boxed{d=\frac{\sqrt{277}}{6}} → d = ( 2 − 2 1 ) 2 + ( − 1 − 3 4 ) 2 → d = 4 9 + 9 49 → d = 36 277 → d = 6 277
f) ( 9 , 5 ; − 2 , 6 ) (9,5;\,-2,6) ( 9 , 5 ; − 2 , 6 ) e ( − 3 , 9 ; 8 , 2 ) (-3,9;\,8,2) ( − 3 , 9 ; 8 , 2 )
→ d = ( − 3 , 9 − 9 , 5 ) 2 + ( 8 , 2 + 2 , 6 ) 2 → d = ( 13 , 4 ) 2 + ( 10 , 8 ) 2 → \to d=\sqrt{(-3,9-9,5)^2+(8,2+2,6)^2}\to d=\sqrt{(13,4)^2+(10,8)^2}\to → d = ( − 3 , 9 − 9 , 5 ) 2 + ( 8 , 2 + 2 , 6 ) 2 → d = ( 13 , 4 ) 2 + ( 10 , 8 ) 2 →
→ d = 179 , 56 + 116 , 64 → d = 296 , 2 → d ≈ 17 , 21 \to d=\sqrt{179,56+116,64}\to d=\sqrt{296,2}\to\boxed{d\approx 17,21} → d = 179 , 56 + 116 , 64 → d = 296 , 2 → d ≈ 17 , 21
0183
Encontre o valor dos lados dos triângulos retângulos a seguir e mostre que esses valores satisfazem o Teorema de Pitágoras
a)
b)
0183 - Soluções
a.1)
d 1 = ( 13 − 1 ) 2 + ( 5 − 0 ) 2 → d 1 = 144 + 25 → d 1 = 169 → d 1 = 13 d_{1}=\sqrt{(13-1)^2+(5-0)^2}\to d_{1}=\sqrt{144+25}\to d_{1}=\sqrt{169}\to\boxed{d_{1}=13} d 1 = ( 13 − 1 ) 2 + ( 5 − 0 ) 2 → d 1 = 144 + 25 → d 1 = 169 → d 1 = 13
d 2 = ( 13 − 13 ) 2 + ( 5 − 0 ) 2 → d 2 = 0 + 25 → d 2 = 25 → d 2 = 5 d_{2}=\sqrt{(13-13)^2+(5-0)^2}\to d_{2}=\sqrt{0+25}\to d_{2}=\sqrt{25}\to\boxed{d_{2}=5} d 2 = ( 13 − 13 ) 2 + ( 5 − 0 ) 2 → d 2 = 0 + 25 → d 2 = 25 → d 2 = 5
d 3 = ( 13 − 1 ) 2 + ( 0 − 0 ) 2 → d 3 = 144 + 0 → d 3 = 144 → d 3 = 12 d_{3}=\sqrt{(13-1)^2+(0-0)^2}\to d_{3}=\sqrt{144+0}\to d_{3}=\sqrt{144}\to\boxed{d_{3}=12} d 3 = ( 13 − 1 ) 2 + ( 0 − 0 ) 2 → d 3 = 144 + 0 → d 3 = 144 → d 3 = 12
a.2)
Pelo Teorema de Pitágoras: d 1 2 = d 2 2 + d 2 2 → 1 3 2 = 5 2 + 1 2 2 → 13 = 13 ✓ d_{1}^2=d_{2}^2+d_{2}^2\to 13^2=5^2+ 12^2\to \boxed{13=13}\quad\checkmark d 1 2 = d 2 2 + d 2 2 → 1 3 2 = 5 2 + 1 2 2 → 13 = 13 ✓
b.1)
d 1 = [ 9 − ( − 1 ) ] 2 + ( 4 − 1 ) 2 → d 1 = 100 + 9 → → d 1 = 109 d_{1}=\sqrt{[9-(-1)]^2+(4-1)^2}\to d_{1}=\sqrt{100+9}\to \to\boxed{d_{1}=\sqrt{109}} d 1 = [ 9 − ( − 1 ) ] 2 + ( 4 − 1 ) 2 → d 1 = 100 + 9 →→ d 1 = 109
d 2 = ( 9 − 9 ) 2 + ( 4 − 1 ) 2 → d 2 = 0 + 9 → d 2 = 9 → d 2 = 3 d_{2}=\sqrt{(9-9)^2+(4-1)^2}\to d_{2}=\sqrt{0+9}\to d_{2}=\sqrt{9}\to\boxed{d_{2}=3} d 2 = ( 9 − 9 ) 2 + ( 4 − 1 ) 2 → d 2 = 0 + 9 → d 2 = 9 → d 2 = 3
d 3 = [ 9 − ( − 1 ) ] 2 + ( 1 − 1 ) 2 → d 3 = 100 + 0 → d 3 = 100 → d 3 = 10 d_{3}=\sqrt{[9-(-1)]^2+(1-1)^2}\to d_{3}=\sqrt{100+0}\to d_{3}=\sqrt{100}\to\boxed{d_{3}=10} d 3 = [ 9 − ( − 1 ) ] 2 + ( 1 − 1 ) 2 → d 3 = 100 + 0 → d 3 = 100 → d 3 = 10
b.2)
Pelo Teorema de Pitágoras: d 1 2 = d 2 2 + d 2 2 → ( 109 ) 2 = 3 2 + 1 0 2 → 109 = 109 ✓ d_{1}^2=d_{2}^2+d_{2}^2\to \left(\sqrt{109}\right)^2=3^2+ 10^2\to \boxed{109=109}\quad\checkmark d 1 2 = d 2 2 + d 2 2 → ( 109 ) 2 = 3 2 + 1 0 2 → 109 = 109 ✓
0182
Mostre que os pontos a seguir definem os vértices de um triângulo retângulo:
a) ( 4 ; 0 ) (4;\,0) ( 4 ; 0 ) , ( 2 ; 1 ) (2;\,1) ( 2 ; 1 ) e ( − 1 ; − 5 ) (-1;\,-5) ( − 1 ; − 5 )
b) ( − 1 ; 3 ) (-1;\,3) ( − 1 ; 3 ) , ( 3 ; 5 ) (3;\,5) ( 3 ; 5 ) e ( 5 ; 1 ) (5;\,1) ( 5 ; 1 )
0182 - Demonstrações
Não podemos esquecer das condições de existência, válidas para quaisquer tipos de triângulos, isto é, para um triângulo qualquer, de lados "a a a ", "b b b " e "c c c ", devemos observar a veracidade dessas afirmações:
∣ b − c ∣ < a < b + c | b - c | < a < b + c ∣ b − c ∣ < a < b + c
∣ a − c ∣ < b < a + c | a - c | < b < a + c ∣ a − c ∣ < b < a + c
∣ a − b ∣ < c < a + b | a - b | < c < a + b ∣ a − b ∣ < c < a + b
Assim, vamos às resoluções:
a) ( 4 ; 0 ) (4;\,0) ( 4 ; 0 ) , ( 2 ; 1 ) (2;\,1) ( 2 ; 1 ) e ( − 1 ; − 5 ) (-1;\,-5) ( − 1 ; − 5 )
d 1 = ( 2 − 4 ) 2 + ( 1 − 0 ) 2 → d 1 = 4 + 1 → d 1 = 5 d_{1}=\sqrt{(2-4)^2+(1-0)^2}\to d_{1}=\sqrt{4+1}\to d_{1}=\sqrt{5} d 1 = ( 2 − 4 ) 2 + ( 1 − 0 ) 2 → d 1 = 4 + 1 → d 1 = 5
d 2 = ( − 1 − 2 ) 2 + ( − 5 − 1 ) 2 → d 2 = 9 + 36 → d 2 = 45 d_{2}=\sqrt{(-1-2)^2+(-5-1)^2}\to d_{2}=\sqrt{9+36}\to d_{2}=\sqrt{45} d 2 = ( − 1 − 2 ) 2 + ( − 5 − 1 ) 2 → d 2 = 9 + 36 → d 2 = 45
d 3 = ( − 1 − 4 ) 2 + ( − 5 − 0 ) 2 → d 3 = 25 + 25 → d 3 = 50 d_{3}=\sqrt{(-1-4)^2+(-5-0)^2}\to d_{3}=\sqrt{25+25}\to d_{3}=\sqrt{50} d 3 = ( − 1 − 4 ) 2 + ( − 5 − 0 ) 2 → d 3 = 25 + 25 → d 3 = 50
Esse triângulo é retângulo pois:
d 1 2 + d 2 2 = d 3 2 → ( 5 ) 2 + ( 45 ) 2 = ( 50 ) 2 → 50 = 50 ✓ d_{1}^2+d_{2}^2=d_{3}^2\to (\sqrt{5})^2+(\sqrt{45})^2=(\sqrt{50})^2\to 50=50\checkmark d 1 2 + d 2 2 = d 3 2 → ( 5 ) 2 + ( 45 ) 2 = ( 50 ) 2 → 50 = 50 ✓
b) ( − 1 ; 3 ) (-1;\,3) ( − 1 ; 3 ) , ( 3 ; 5 ) (3;\,5) ( 3 ; 5 ) e ( 5 ; 1 ) (5;\,1) ( 5 ; 1 )
d 1 = ( 3 + 1 ) 2 + ( 5 − 3 ) 2 → d 1 = 16 + 4 → d 1 = 20 d_{1}=\sqrt{(3+1)^2+(5-3)^2}\to d_{1}=\sqrt{16+4}\to d_{1}=\sqrt{20} d 1 = ( 3 + 1 ) 2 + ( 5 − 3 ) 2 → d 1 = 16 + 4 → d 1 = 20
d 2 = ( 5 − 3 ) 2 + ( 1 − 5 ) 2 → d 2 = 4 + 16 → d 2 = 20 d_{2}=\sqrt{(5-3)^2+(1-5)^2}\to d_{2}=\sqrt{4+16}\to d_{2}=\sqrt{20} d 2 = ( 5 − 3 ) 2 + ( 1 − 5 ) 2 → d 2 = 4 + 16 → d 2 = 20
d 3 = ( 5 + 1 ) 2 + ( 1 − 3 ) 2 → d 3 = 36 + 4 → d 3 = 40 d_{3}=\sqrt{(5+1)^2+(1-3)^2}\to d_{3}=\sqrt{36+4}\to d_{3}=\sqrt{40} d 3 = ( 5 + 1 ) 2 + ( 1 − 3 ) 2 → d 3 = 36 + 4 → d 3 = 40
Esse triângulo é retângulo pois:
d 1 2 + d 2 2 = d 3 3 → ( 20 ) 2 + ( 20 ) 2 = ( 40 ) 2 → 40 = 40 ✓ d_{1}^2+d_{2}^2=d_{3}^3\to(\sqrt{20})^2+(\sqrt{20})^2=(\sqrt{40})^2\to 40=40\checkmark d 1 2 + d 2 2 = d 3 3 → ( 20 ) 2 + ( 20 ) 2 = ( 40 ) 2 → 40 = 40 ✓
0181
Mostre que os pontos a seguir definem os vértices de um triângulo isósceles:
a) ( 1 ; − 3 ) (1;\,-3) ( 1 ; − 3 ) , ( 3 ; 2 ) (3;\,2) ( 3 ; 2 ) e ( − 2 ; 4 ) (-2;\,4) ( − 2 ; 4 )
b) ( 2 ; 3 ) (2;\,3) ( 2 ; 3 ) , ( 4 ; 9 ) (4;\,9) ( 4 ; 9 ) e ( − 2 ; 7 ) (-2;\,7) ( − 2 ; 7 )
0181 - Soluções
Não podemos esquecer das condições de existência, válidas para quaisquer tipos de triângulos, isto é, para um triângulo qualquer, de lados "a a a ", "b b b " e "c c c ", devemos observar a veracidade dessas afirmações:
∣ b − c ∣ < a < b + c | b - c | < a < b + c ∣ b − c ∣ < a < b + c
∣ a − c ∣ < b < a + c | a - c | < b < a + c ∣ a − c ∣ < b < a + c
∣ a − b ∣ < c < a + b | a - b | < c < a + b ∣ a − b ∣ < c < a + b
Assim, vamos às soluções:
a) ( 1 ; − 3 ) (1;\,-3) ( 1 ; − 3 ) , ( 3 ; 2 ) (3;\,2) ( 3 ; 2 ) e ( − 2 ; 4 ) (-2;\,4) ( − 2 ; 4 )
d 1 = ( 3 − 1 ) 2 + ( 2 − ( − 3 ) ) 2 → d 1 = 4 + 25 → d 1 = 29 d_{1}=\sqrt{(3-1)^2+(2-(-3))^2}\to d_{1}=\sqrt{4+25}\to d_{1}=\sqrt{29} d 1 = ( 3 − 1 ) 2 + ( 2 − ( − 3 ) ) 2 → d 1 = 4 + 25 → d 1 = 29
d 2 = ( − 2 − 3 ) 2 + ( 4 − 2 ) 2 → d 2 = 25 + 4 → d 2 = 29 d_{2}=\sqrt{(-2-3)^2+(4-2)^2}\to d_{2}=\sqrt{25+4}\to d_{2}=\sqrt{29} d 2 = ( − 2 − 3 ) 2 + ( 4 − 2 ) 2 → d 2 = 25 + 4 → d 2 = 29
d 3 = ( − 2 − 1 ) 2 + ( 4 − ( − 3 ) ) 2 → d 3 = 9 + 49 → d 3 = 58 d_{3}=\sqrt{(-2-1)^2+(4-(-3))^2}\to d_{3}=\sqrt{9+49}\to d_{3}=\sqrt{58} d 3 = ( − 2 − 1 ) 2 + ( 4 − ( − 3 ) ) 2 → d 3 = 9 + 49 → d 3 = 58
Esse triângulo é isósceles pois d 1 = d 2 d_{1}=d_{2}\,\, d 1 = d 2 e esse triângulo existe, pois:
∣ 29 − 58 ∣ < 29 < 29 + 58 → 2 , 2306 < 5 , 3852 < 13 , 001 ✓ | \sqrt{29} - \sqrt{58} | < \sqrt{29} < \sqrt{29} + \sqrt{58}\to 2,2306<5,3852<13,001\checkmark ∣ 29 − 58 ∣ < 29 < 29 + 58 → 2 , 2306 < 5 , 3852 < 13 , 001 ✓
∣ 29 − 58 ∣ < 29 < 29 + 58 → 2 , 2306 < 5 , 3852 < 13 , 001 ✓ | \sqrt{29} - \sqrt{58} | < \sqrt{29} < \sqrt{29} + \sqrt{58}\to 2,2306<5,3852<13,001\checkmark ∣ 29 − 58 ∣ < 29 < 29 + 58 → 2 , 2306 < 5 , 3852 < 13 , 001 ✓
∣ 29 − 29 ∣ < 58 < 29 + 29 → 0 < 7 , 6158 < 10 , 7703 ✓ | \sqrt{29} - \sqrt{29} | < \sqrt{58} < \sqrt{29} + \sqrt{29}\to 0<7,6158<10,7703\checkmark ∣ 29 − 29 ∣ < 58 < 29 + 29 → 0 < 7 , 6158 < 10 , 7703 ✓
b) ( 2 ; 3 ) (2;\,3) ( 2 ; 3 ) , ( 4 ; 9 ) (4;\,9) ( 4 ; 9 ) e ( − 2 ; 7 ) (-2;\,7) ( − 2 ; 7 )
d 1 = ( 4 − 2 ) 2 + ( 9 − 3 ) 2 → d 1 = 4 + 36 → d 1 = 40 d_{1}=\sqrt{(4-2)^2+(9-3)^2}\to d_{1}=\sqrt{4+36}\to d_{1}=\sqrt{40} d 1 = ( 4 − 2 ) 2 + ( 9 − 3 ) 2 → d 1 = 4 + 36 → d 1 = 40
d 2 = ( − 2 − 4 ) 2 + ( 7 − 9 ) 2 → d 2 = 36 + 4 → d 2 = 40 d_{2}=\sqrt{(-2-4)^2+(7-9)^2}\to d_{2}=\sqrt{36+4}\to d_{2}=\sqrt{40} d 2 = ( − 2 − 4 ) 2 + ( 7 − 9 ) 2 → d 2 = 36 + 4 → d 2 = 40
d 3 = ( − 2 − 2 ) 2 + ( 7 − 3 ) 2 → d 3 = 16 + 16 → d 3 = 32 d_{3}=\sqrt{(-2-2)^2+(7-3)^2}\to d_{3}=\sqrt{16+16}\to d_{3}=\sqrt{32} d 3 = ( − 2 − 2 ) 2 + ( 7 − 3 ) 2 → d 3 = 16 + 16 → d 3 = 32
Esse triângulo é isósceles pois d 1 = d 2 d_{1}=d_{2}\,\, d 1 = d 2 e esse triângulo existe, pois:
∣ 40 − 32 ∣ < 40 < 40 + 32 → 0 , 6677 < 6 , 3246 < 11 , 9815 ✓ | \sqrt{40} - \sqrt{32} | < \sqrt{40} < \sqrt{40} + \sqrt{32}\to 0,6677<6,3246<11,9815\checkmark ∣ 40 − 32 ∣ < 40 < 40 + 32 → 0 , 6677 < 6 , 3246 < 11 , 9815 ✓
∣ 40 − 32 ∣ < 40 < 40 + 32 → 0 , 6677 < 6 , 3246 < 11 , 9815 ✓ | \sqrt{40} - \sqrt{32} | < \sqrt{40} < \sqrt{40} + \sqrt{32}\to 0,6677<6,3246<11,9815\checkmark ∣ 40 − 32 ∣ < 40 < 40 + 32 → 0 , 6677 < 6 , 3246 < 11 , 9815 ✓
∣ 40 − 40 ∣ < 32 < 40 + 40 → 0 < 5 , 6569 < 12 , 6492 ✓ | \sqrt{40} - \sqrt{40} | < \sqrt{32} < \sqrt{40} + \sqrt{40}\to 0<5,6569<12,6492\checkmark ∣ 40 − 40 ∣ < 32 < 40 + 40 → 0 < 5 , 6569 < 12 , 6492 ✓
0180
Obtenha o ponto médio entre os pares de pontos a seguir:
a) ( 6 ; − 3 ) (6;\,-3) ( 6 ; − 3 ) e ( 6 ; 5 ) (6;\,5) ( 6 ; 5 )
b) ( 1 ; 4 ) (1;\,4) ( 1 ; 4 ) e ( 8 ; 4 ) (8;\,4) ( 8 ; 4 )
c) ( 1 ; 1 ) (1;\,1) ( 1 ; 1 ) e ( 9 ; 7 ) (9;\,7) ( 9 ; 7 )
d) ( 1 ; 12 ) (1;\,12) ( 1 ; 12 ) e ( 6 ; 0 ) (6;\,0) ( 6 ; 0 )
e) ( − 1 ; 2 ) (-1;\,2) ( − 1 ; 2 ) e ( 5 ; 4 ) (5;\,4) ( 5 ; 4 )
f) ( 2 ; 10 ) (2;\,10) ( 2 ; 10 ) e ( 10 ; 2 ) (10;\,2) ( 10 ; 2 )
g) ( − 16 , 8 ; 12 , 3 ) (-16,8;\,12,3) ( − 16 , 8 ; 12 , 3 ) e ( 5 , 6 ; 4 , 9 ) (5,6;\,4,9) ( 5 , 6 ; 4 , 9 )
h) ( 1 2 ; 1 ) \left(\frac{1}{2};\,1\right) ( 2 1 ; 1 ) e ( − 5 2 ; 4 3 ) \left(-\frac{5}{2};\,\frac{4}{3}\right) ( − 2 5 ; 3 4 )
0180 - Soluções
a) ( 6 ; − 3 ) (6;\,-3) ( 6 ; − 3 ) e ( 6 ; 5 ) (6;\,5) ( 6 ; 5 )
M ( x M ; y M ) = ( x 1 + x 2 2 ; y 1 + y 2 2 ) = ( 6 + 6 2 ; − 3 + 5 2 ) → M = ( 6 ; 1 ) ✓ M(x_{M};\,y_{M})=\left(\dfrac{x_{1}+x_{2}}{2};\,\dfrac{y_{1}+y_{2}}{2}\right)=\left(\dfrac{6+6}{2};\,\dfrac{-3+5}{2}\right)\to M=(6;\,1)\checkmark M ( x M ; y M ) = ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) = ( 2 6 + 6 ; 2 − 3 + 5 ) → M = ( 6 ; 1 ) ✓
b) ( 1 ; 4 ) (1;\,4) ( 1 ; 4 ) e ( 8 ; 4 ) (8;\,4) ( 8 ; 4 )
M ( x M ; y M ) = ( x 1 + x 2 2 ; y 1 + y 2 2 ) = ( 1 + 8 2 ; 4 + 4 2 ) → M = ( 9 2 ; 4 ) ✓ M(x_{M};\,y_{M})=\left(\dfrac{x_{1}+x_{2}}{2};\,\dfrac{y_{1}+y_{2}}{2}\right)=\left(\dfrac{1+8}{2};\,\dfrac{4+4}{2}\right)\to M=\left(\dfrac{9}{2};\,4\right)\checkmark M ( x M ; y M ) = ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) = ( 2 1 + 8 ; 2 4 + 4 ) → M = ( 2 9 ; 4 ) ✓
c) ( 1 ; 1 ) (1;\,1) ( 1 ; 1 ) e ( 9 ; 7 ) (9;\,7) ( 9 ; 7 )
M ( x M ; y M ) = ( x 1 + x 2 2 ; y 1 + y 2 2 ) = ( 1 + 9 2 ; 1 + 7 2 ) → M = ( 5 ; 4 ) ✓ M(x_{M};\,y_{M})=\left(\dfrac{x_{1}+x_{2}}{2};\,\dfrac{y_{1}+y_{2}}{2}\right)=\left(\dfrac{1+9}{2};\,\dfrac{1+7}{2}\right)\to M=\left(5;\,4\right)\checkmark M ( x M ; y M ) = ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) = ( 2 1 + 9 ; 2 1 + 7 ) → M = ( 5 ; 4 ) ✓
d) ( 1 ; 12 ) (1;\,12) ( 1 ; 12 ) e ( 6 ; 0 ) (6;\,0) ( 6 ; 0 )
M ( x M ; y M ) = ( x 1 + x 2 2 ; y 1 + y 2 2 ) = ( 1 + 6 2 ; 12 + 0 2 ) → M = ( 7 2 ; 6 ) ✓ M(x_{M};\,y_{M})=\left(\dfrac{x_{1}+x_{2}}{2};\,\dfrac{y_{1}+y_{2}}{2}\right)=\left(\dfrac{1+6}{2};\,\dfrac{12+0}{2}\right)\to M=\left(\dfrac{7}{2};\,6\right)\checkmark M ( x M ; y M ) = ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) = ( 2 1 + 6 ; 2 12 + 0 ) → M = ( 2 7 ; 6 ) ✓
e) ( − 1 ; 2 ) (-1;\,2) ( − 1 ; 2 ) e ( 5 ; 4 ) (5;\,4) ( 5 ; 4 )
M ( x M ; y M ) = ( x 1 + x 2 2 ; y 1 + y 2 2 ) = ( − 1 + 5 2 ; 2 + 4 2 ) → M = ( 2 ; 3 ) ✓ M(x_{M};\,y_{M})=\left(\dfrac{x_{1}+x_{2}}{2};\,\dfrac{y_{1}+y_{2}}{2}\right)=\left(\dfrac{-1+5}{2};\,\dfrac{2+4}{2}\right)\to M=\left(2;\,3\right)\checkmark M ( x M ; y M ) = ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) = ( 2 − 1 + 5 ; 2 2 + 4 ) → M = ( 2 ; 3 ) ✓
f) ( 2 ; 10 ) (2;\,10) ( 2 ; 10 ) e ( 10 ; 2 ) (10;\,2) ( 10 ; 2 )
M ( x M ; y M ) = ( x 1 + x 2 2 ; y 1 + y 2 2 ) = ( 2 + 10 2 ; 10 + 2 2 ) → M = ( 6 ; 6 ) ✓ M(x_{M};\,y_{M})=\left(\dfrac{x_{1}+x_{2}}{2};\,\dfrac{y_{1}+y_{2}}{2}\right)=\left(\dfrac{2+10}{2};\,\dfrac{10+2}{2}\right)\to M=\left(6;\,6\right)\checkmark M ( x M ; y M ) = ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) = ( 2 2 + 10 ; 2 10 + 2 ) → M = ( 6 ; 6 ) ✓
g) ( − 16 , 8 ; 12 , 3 ) (-16,8;\,12,3) ( − 16 , 8 ; 12 , 3 ) e ( 5 , 6 ; 4 , 9 ) (5,6;\,4,9) ( 5 , 6 ; 4 , 9 )
M ( x M ; y M ) = ( x 1 + x 2 2 ; y 1 + y 2 2 ) = ( − 16 , 8 + 5 , 6 2 ; 12 , 3 + 4 , 9 2 ) → M = ( − 5 , 6 ; 8 , 6 ) ✓ M(x_{M};\,y_{M})=\left(\dfrac{x_{1}+x_{2}}{2};\,\dfrac{y_{1}+y_{2}}{2}\right)=\left(\dfrac{-16,8+5,6}{2};\,\dfrac{12,3+4,9}{2}\right)\to M=\left(-5,6;\,8,6\right)\checkmark M ( x M ; y M ) = ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) = ( 2 − 16 , 8 + 5 , 6 ; 2 12 , 3 + 4 , 9 ) → M = ( − 5 , 6 ; 8 , 6 ) ✓
h) ( 1 2 ; 1 ) \left(\frac{1}{2};\,1\right) ( 2 1 ; 1 ) e ( − 5 2 ; 4 3 ) \left(-\frac{5}{2};\,\frac{4}{3}\right) ( − 2 5 ; 3 4 )
M ( x M ; y M ) = ( x 1 + x 2 2 ; y 1 + y 2 2 ) = ( 1 2 − 5 2 2 ; 1 + 4 3 2 ) → M = ( − 1 ; 7 6 ) ✓ M(x_{M};\,y_{M})=\left(\dfrac{x_{1}+x_{2}}{2};\,\dfrac{y_{1}+y_{2}}{2}\right)=\left(\dfrac{\frac{1}{2}-\frac{5}{2}}{2};\,\dfrac{1+\frac{4}{3}}{2}\right)\to M=\left(-1;\,\dfrac{7}{6}\right)\checkmark M ( x M ; y M ) = ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) = ( 2 2 1 − 2 5 ; 2 1 + 3 4 ) → M = ( − 1 ; 6 7 ) ✓
0179
Um avião voa da cidade "A", em uma linha reta até "B", que está a 120 quilômetros ao norte e 150 quilômetros a oeste de "A". Qual a distância percorrida pelo avião?
0179 - Solução
Se observarmos, a trajetória e as distâncias, formam um triângulo retângulo, onde as distâncias são os catetos e a "linha reta"(vamos chamar de 'd') é a hipotenusa. Assim, vamos utilizar o teorema de Pitágoras:
d 2 = 12 0 2 + 15 0 2 → d = 36.900 → d = 30 41 → d ≈ 192 , 09 km d^2=120^2+150^2\to d=\sqrt{36.900}\to d=30\sqrt{41}\to\boxed{d\approx 192,09\,\text{km}} d 2 = 12 0 2 + 15 0 2 → d = 36.900 → d = 30 41 → d ≈ 192 , 09 km
0178
Um segmento de reta tem o ponto ( x 1 ; y 1 ) (x_{1};\,y_{1}) ( x 1 ; y 1 ) e o ponto médio ( x m ; y m ) (x_{m};\,y_{m}) ( x m ; y m ) . Encontre o outro ponto ( x 2 ; y 2 ) (x_{2};\,y_{2}) ( x 2 ; y 2 ) em razão de x 1 x_{1} x 1 , y 1 y_{1} y 1 , x m x_{m} x m e y m y_{m} y m .
0178 - Solução
Da fórmula do ponto médio, temos:
x m = x 1 + x 2 2 → x 2 = 2 x m − x 1 x_{m}=\dfrac{x_{1}+x_{2}}{2}\to\boxed{x_{2}=2x_{m}-x_{1}} x m = 2 x 1 + x 2 → x 2 = 2 x m − x 1
e
y m = y 1 + y 2 2 → y 2 = 2 y m − y 1 y_{m}=\dfrac{y_{1}+y_{2}}{2}\to\boxed{y_{2}=2y_{m}-y_{1}} y m = 2 y 1 + y 2 → y 2 = 2 y m − y 1
Portanto:
( x 2 ; y 2 ) = ( 2 x m − x 1 ; 2 y m − y 1 ) \boxed{(x_{2};\,y_{2})=(2x_{m}-x_{1};\,\,2y_{m}-y_{1})} ( x 2 ; y 2 ) = ( 2 x m − x 1 ; 2 y m − y 1 )
0177
Nos mesmos termos do exercício anterior, encontre ( x 2 ; y 2 ) (x_{2};\,y_{2}) ( x 2 ; y 2 ) , para:
a) ( x 1 ; y 1 ) = ( 1 ; − 2 ) (x_{1};\,y_{1})=(1;\,-2) ( x 1 ; y 1 ) = ( 1 ; − 2 ) e ( x m ; y m ) = ( 4 ; − 1 ) (x_{m};\,y_{m})=(4;\,-1) ( x m ; y m ) = ( 4 ; − 1 )
b) ( x 1 ; y 1 ) = ( − 5 ; 11 ) (x_{1};\,y_{1})=(-5;\,11) ( x 1 ; y 1 ) = ( − 5 ; 11 ) e ( x m ; y m ) = ( 2 ; 4 ) (x_{m};\,y_{m})=(2;\,4) ( x m ; y m ) = ( 2 ; 4 )
0177 - Soluções
a) ( x 1 ; y 1 ) = ( 1 ; − 2 ) (x_{1};\,y_{1})=(1;\,-2) ( x 1 ; y 1 ) = ( 1 ; − 2 ) e ( x m ; y m ) = ( 4 ; − 1 ) (x_{m};\,y_{m})=(4;\,-1) ( x m ; y m ) = ( 4 ; − 1 )
( x 2 ; y 2 ) = ( 2 x m − x 1 ; 2 y m − y 1 ) → (x_{2};\,y_{2})=(2x_{m}-x_{1};\,\,2y_{m}-y_{1})\to ( x 2 ; y 2 ) = ( 2 x m − x 1 ; 2 y m − y 1 ) →
( x 2 ; y 2 ) = [ 2 ⋅ 4 − 1 ; 2 ⋅ ( − 1 ) − ( − 2 ) ] → (x_{2;\,y_{2}})=[2\cdot 4-1;\,\,2\cdot(-1)-(-2)]\to ( x 2 ; y 2 ) = [ 2 ⋅ 4 − 1 ; 2 ⋅ ( − 1 ) − ( − 2 )] →
( x 2 ; y 2 ) = ( 7 ; 0 ) \boxed{(x_{2};\,y_{2})=(7;\,0)} ( x 2 ; y 2 ) = ( 7 ; 0 )
b) ( x 1 ; y 1 ) = ( − 5 ; 11 ) (x_{1};\,y_{1})=(-5;\,11) ( x 1 ; y 1 ) = ( − 5 ; 11 ) e ( x m ; y m ) = ( 2 ; 4 ) (x_{m};\,y_{m})=(2;\,4) ( x m ; y m ) = ( 2 ; 4 )
( x 2 ; y 2 ) = ( 2 x m − x 1 ; y 2 = 2 y m − y 1 ) → (x_{2};\,y_{2})=(2x_{m}-x_{1};\,\,y_{2}=2y_{m}-y_{1})\to ( x 2 ; y 2 ) = ( 2 x m − x 1 ; y 2 = 2 y m − y 1 ) →
( x 2 ; y 2 ) = [ 2 ⋅ 2 − ( − 5 ) ; 2 ⋅ 4 − 11 ] ] → (x_{2};\,y_{2})=[2\cdot 2-(-5);\,\,2\cdot 4-11]]\to ( x 2 ; y 2 ) = [ 2 ⋅ 2 − ( − 5 ) ; 2 ⋅ 4 − 11 ]] →
( x 2 ; y 2 ) = ( 9 ; − 3 ) \boxed{(x_{2};\,y_{2})=(9;\,-3)} ( x 2 ; y 2 ) = ( 9 ; − 3 )
0176
Utilize a fórmula do ponto médio três vezes , a fim de encontrar três pontos que dividam o segmento de extremos ( x 1 ; y 1 ) (x_{1};\,y_{1}) ( x 1 ; y 1 ) e ( x 2 ; y 2 ) (x_{2};\,y_{2}) ( x 2 ; y 2 ) em quatro partes iguais.
0176 - Solução
→ \rightarrow → Primeiro ponto médio( P M 1 ) (P_{M1}) ( P M 1 ) , entre os pontos ( x 1 ; y 1 ) (x_{1};\,y_{1}) ( x 1 ; y 1 ) e ( x 2 ; y 2 ) (x_{2};\,y_{2}) ( x 2 ; y 2 ) :
P M 1 = ( x 1 + x 2 2 ; y 1 + y 2 2 ) P_{M1}=\left(\dfrac{x_{1}+x_{2}}{2};\,\,\dfrac{y_{1}+y_{2}}{2}\right) P M 1 = ( 2 x 1 + x 2 ; 2 y 1 + y 2 )
→ \rightarrow → Segunto ponto médio( P M 2 ) (P_{M2}) ( P M 2 ) , entre os pontos ( x 1 ; y 1 ) (x_{1};\,y_{1}) ( x 1 ; y 1 ) e ( x 1 + x 2 2 ; y 1 + y 2 2 ) \left(\dfrac{x_{1}+x_{2}}{2};\,\,\dfrac{y_{1}+y_{2}}{2}\right) ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) :
P M 2 = ( x 1 + x 1 + x 2 2 2 ; y 2 + y 1 + y 2 2 2 ) → P M 2 = ( 3 x 1 + x 2 4 ; 3 y 1 + y 2 4 ) P_{M2}=\left(\dfrac{x_{1}+\frac{x_{1}+x_{2}}{2}}{2};\,\,\dfrac{y_{2}+\frac{y_{1}+y_{2}}{2}}{2}\right)\to P_{M2}=\left(\dfrac{3x_{1}+x_{2}}{4};\,\,\dfrac{3y_{1}+y_{2}}{4}\right) P M 2 = ( 2 x 1 + 2 x 1 + x 2 ; 2 y 2 + 2 y 1 + y 2 ) → P M 2 = ( 4 3 x 1 + x 2 ; 4 3 y 1 + y 2 )
→ \rightarrow → Terceiro ponto médio( P M 3 ) (P_{M3}) ( P M 3 ) entre os pontos ( x 1 + x 2 2 ; y 1 + y 2 2 ) \left(\dfrac{x_{1}+x_{2}}{2};\,\,\dfrac{y_{1}+y_{2}}{2}\right) ( 2 x 1 + x 2 ; 2 y 1 + y 2 ) e ( x 2 ; y 2 ) (x_{2};\,y_{2}) ( x 2 ; y 2 ) :
P M 3 = ( x 1 + x 2 2 + x 2 2 ; y 1 + y 2 2 + y 2 2 ) → P M 3 = ( x 1 + 3 x 2 4 ; y 1 + 3 y 2 4 ) P_{M3}=\left(\dfrac{\frac{x_{1}+x_{2}}{2}+ x_{2}}{2};\,\,\dfrac{\frac{y_{1}+y_{2}}{2}+y_{2}}{2}\right)\to P_{M3}=\left(\dfrac{x_{1}+3x_{2}}{4};\,\,\dfrac{y_{1}+3y_{2}}{4}\right) P M 3 = ( 2 2 x 1 + x 2 + x 2 ; 2 2 y 1 + y 2 + y 2 ) → P M 3 = ( 4 x 1 + 3 x 2 ; 4 y 1 + 3 y 2 )