Question

Alguém poderia me ajudar na 2,3,4 ??

Answer 1

Boa noite Bruna!

Solução!

Para resolver o exercício 2 temos que determinar um ponto,para qualquer uma das retas,pois como se vê temos retas paralelas.

$2)~~r:4x-3y-4=0\\\\\ s:4x+3y-2=0\\\\\\\\\\ s:4x+3y-2=0\\\\\ 3y=-4x+2\\\\ y= \dfrac{4x}{3}+ \frac{2}{3}~~Para~~x=0 \\\\\\ y= \dfrac{4(0)}{3}+ \dfrac{2}{3}\\\\\\ y=\dfrac{2}{3}\\\\\ Logo!\\\\\ P(0,\frac{2}{3})$


Vamos agora usar a formula da distancia entre um ponto e reta.

$d(P,r)= \dfrac{|axP+byP+c|}{ \sqrt{a^{2}+b^{2} } }\\\\\\\ d(P,r)= \dfrac{|4x-3y-4|}{ \sqrt{a^{2}+b^{2} } }\\\\\\\ d(P,r)= \dfrac{|4(0)-3( \frac{2}{3}) -4|}{ \sqrt{4^{2}+(-3)^{2} } }\\\\\\\ d(P,r)= \dfrac{|0-2 -4|}{ \sqrt{16+9 } }\\\\\\\ d(P,r)= \dfrac{|-6|}{ \sqrt{25 } }\\\\\\\ d(P,r)= \dfrac{|-6|}{ 5 }\\\\\\\ d(P,r)= \dfrac{6}{ 5 }\\\\\\\\\\ \boxed{Resposta: d(P,r)= \dfrac{6}{ 5 }~~ou~~1,2}$

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$3)\\\\\\\ Sendo~~ os ~~pontos!\\\\\\ A(4,-2)~~B(5,1)~~C(-2,3)\\\\\\\ Para ~~calcular~~ a~~ area ~~do~~ tri\~angulo~~ vamos~~ usar\\\\\ o ~~determinante~~ de ~~uma~~ matriz ~~que ~~e ~~dado~~ por.\\\\\\ S=|Det|= \left[\begin{array}{ccc}x1&y1&1\\x2&y2&1\\x3&y3&1\end{array}\right]$

$S= \dfrac{1}{2} \left[\begin{array}{ccccc}4&-2&1&4&-2\\5&1&1&5&1\\-2&3&1&-2&3\end{array}\right]\\\\\\\\\\ S= \dfrac{1}{2}. (4+4+15)-(-2+12-10)\\\\\\\\\ S= \dfrac{1}{2} (23)-(0)\\\\\\\ S= \dfrac{1}{2} (23)\\\\\\\ S= \dfrac{1}{2} (23)\\\\\\\ S= \dfrac{23}{2}~~ou~~11,5 \\\\\\\\\\\\ \boxed{Resposta:S= \dfrac{23}{2}~~ou~~11,5}$

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$4)\\\\\ Para ~~resolver~~ o ~~exercicio~~ 4~~ temos~~ algumas~~ condic\~oes!\\\\\\\\ d(C,r)\ \textless \ r~~\Rightarrow~~Secante\\\\\\\ d(C,r)=r~~\Rightarrow~~Tangente\\\\\\\ d(C,r)\ \textgreater \ r~~\Rightarrow~~Exterior$

Para determinar essas posições vamos calcular o raio e a distância!

 $A)\\\\\ r:x-y+3=0\\\\\\\ (x+2)^2+(y-3)^2=2\\\\\\\ C(-2,3)\\\\R= \sqrt{2}\\\\\ d(C,r)= \dfrac{|x+y+3|}{\sqrt{a^{2} +b^{2}}} \\\\\\\\\ d(C,r)= \dfrac{|1(-2)-1(3)+3|}{\sqrt{1^{2} +1^{2}}} \\\\\\ d(C,r)= \dfrac{|-2-3+3|}{\sqrt{2}} \\\\\\ d(C,r)= \dfrac{|-2+0|}{\sqrt{2}} \\\\\\ d(C,r)= \dfrac{|-2|}{\sqrt{2}} \\\\\\ d(C,r)= \dfrac{2}{\sqrt{2}}\times \dfrac{ \sqrt{2} }{ \sqrt{2} } \\\\\\ d(C,r)= \dfrac{2 \sqrt{2} }{\sqrt{4}} \\\\\\ d(C,r)= \dfrac{2\sqrt{2} }{2} \\\\\\ d(C,r)= \sqrt{2}$

$R= d(C,r)= \sqrt{2}~~ \Rightarrow~~Tangente$

$B)\\\\\\ r:x-y-2=0\\\\\ (x-4)^2+(y+2)^2=2\\\\\ C(4,2)\\\\\\\\ R= \sqrt{2}\\\\\ d(C,r)= \dfrac{|x-y-2|}{\sqrt{a^{2} +b^{2} }}\\\\\\\\ d(C,r)= \dfrac{|1(4)-1(-2)-2|}{\sqrt{1^{2} +(-1)^{2} }}\\\\\\\\\ d(C,r)= \dfrac{|4+2-2|}{\sqrt{1 +1 }}\\\\\\\ d(C,r)= \dfrac{|4|}{\sqrt{2 }}\\\\\\\\\ d(C,r)= \dfrac{|4|}{\sqrt{2 }} \times \dfrac{\sqrt{2}}{ \sqrt{2} } \\\\\\\\\ d(C,r)= \dfrac{4 \sqrt{2} }{\sqrt{4 }} \\\\\\\\\ d(C,r)= \dfrac{4 \sqrt{2} }{2} \\\\\\\\\ d(C,r)= 2 \sqrt{2}$

$C)\\\\\\ r:x-y-2=0\\\\\ (x-5)^2+(y+1)=8\\\\\ C(5,-1)\\\\\\\\ d(C,r)= \dfrac{|x-y-2|}{\sqrt{a^{2} +b^{2}}}\\\\\\ d(C,r)= \dfrac{|1(5)-(-1)-2|}{\sqrt{1^{2} +(-1)^{2}}}\\\\\\ d(C,r)= \dfrac{|5+1-2|}{\sqrt{1 +1}}\\\\\\ d(C,r)= \dfrac{|4|}{\sqrt{2}}\\\\\\ d(C,r)= \dfrac{4}{\sqrt{2}} \times \dfrac{ \sqrt{2} }{ \sqrt{2} } \\\\\\ d(C,r)= \dfrac{4 \sqrt{2} }{\sqrt{4}} \\\\\\\\\ d(C,r)= \dfrac{4 \sqrt{2} }{2} \\\\\\\\\ d(C,r)= 2 \sqrt{2}$

 $d(C,r)\ \textgreater \ r~~\Rightarrow Externa$

$C\\\\ r:x-y-2=0\\\\\ (x-5)^{2}+(y+1)^{2}=8\\\\\ C(5,-1)\\\\\\ R= \sqrt{8}\\\\\ R=2 \sqrt{2} \\\\\ d(C,r)= \dfrac{|x-y-2|}{ \sqrt{a^{2}+b^{2}}}\\\\\\\\\ d(C,r)= \dfrac{|5-(-1)-2|}{ \sqrt{1^{2}+(-1)^{2}}}\\\\\\\\\ d(C,r)= \dfrac{|5+1-2|}{ \sqrt{1+1}}\\\\\\\\\ d(C,r)= \dfrac{|5+1-2|}{ \sqrt{2}}\\\\\\\\\ d(C,r)= \dfrac{|6-2|}{ \sqrt{2}}\\\\\\\\\ d(C,r)= \dfrac{|4|}{ \sqrt{2}}\\\\\\\\\ d(C,r)= \dfrac{4}{ \sqrt{2}} \times \dfrac{ \sqrt{2} }{ \sqrt{2} }$

$d(C,r)= \dfrac{4 \sqrt{2} }{ \sqrt{4}}\\\\\\\\\ d(C,r)= \dfrac{4 \sqrt{2} }{ 2}\\\\\\\\\ d(C,r)= 2 \sqrt{2}\\\\\\\\\ d(C,r)=R~~ \Rightarrow~~Tangente$

$D)\\\\\ r:2x-y-3=0\\\\\ (x- \frac{3}{2})^{2}+(x+1)= \frac{25}{4}\\\\\ C( \frac{3}{2} ,-1)\\\\\ R= \sqrt{ \frac{25}{4} }\\\\\ R= \frac{5}{2}\\\\\ d(C,r)= \dfrac{|2x-y-3|}{ \sqrt{a^{2}+b^{2} } } \\\\\\\ d(C,r)= \dfrac{|2( \frac{3}{2}) -(-1)-3|}{ \sqrt{2^{2}+1^{2} } } \\\\\\\ d(C,r)= \dfrac{|3 +1-3|}{ \sqrt{4+1 } } \\\\\\\ d(C,r)= \dfrac{|1|}{ \sqrt{5 } } \\\\\\\ d(C,r)= \dfrac{1}{ \sqrt{5 } }\times \dfrac{ \sqrt{5} }{ \sqrt{5} }$

$d(C,r)= \dfrac{ \sqrt{5} }{ \sqrt{25 } }\\\\\\\ d(C,r)= \dfrac{ \sqrt{5} }{ 5 } }\\\\\\\ R\ \textgreater \ d(C,r)~~\Rightarrow ~~secantes$

Boa noite!

Bons estudos!