Question

Alguém poderia me ajudar?

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Answer 1

Boa tarde Bruna!

Para resolver um calculo da distancia de um ponto a uma reta basta fazer a substituição na formula acima.Antes precisamos saber que é a e b.

$Formula!\\\\\ d= \dfrac{a.xp+b.yp+c}{ \sqrt{ a^{2}+b^{2} }} }$

$A)\\\\\ P(3,1)~~~~~r:3x-4y+5=0\\\\\ Veja,~~vou~~ substituir ~~a~~equc\~ao~~na~~formula,passo~~a~~passo..\\\\\\\ d= \dfrac{|3x-4y+5|}{ \sqrt{ a^{2}+b^{2} }}}\\\\\ a=3\\\\ b=4\\\\\ c=5\\\\\\\\\\ P(3,1)\\\\\ x=3\\\\ y=1\\\\\ d= \dfrac{|3x-4y+5|}{ \sqrt{ a^{2}+b^{2} }}}\\\\\\\\\\ d= \dfrac{|3(3)-4(1)+5|}{ \sqrt{ 3^{2}+(-4)^{2} }}}\\\\\\\\\ d= \dfrac{|9-4+5|}{ \sqrt{ 9+16 }}}\\\\\\\\\ d= \dfrac{|10|}{ \sqrt{ 25 }}}$

$d= \dfrac{10}{ 5}}\\\\\\\ d=2$

Todas seguem o mesmo raciocinio!
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$B)\\\\\ P(2,-2)~~~~r:3x-2y+1=0\\\\\\ a=3\\\ b=-2\\\\\\ d= \dfrac{|3x-2y+1|}{ \sqrt{ a^{2}+b^{2} }}\\\\\\\\ d= \dfrac{|3(2)-2(-2)+1|}{ \sqrt{ 3^{2}+(-2)^{2} }}\\\\\\\ d= \dfrac{|6+4+1|}{ \sqrt{ 9+4 }}\\\\\\\ d= \dfrac{|11|}{ \sqrt{ 13 }}\\\\\\\ d= \dfrac{11}{ \sqrt{ 13 }}\times \dfrac{ \sqrt{13} }{ \sqrt{13} } \\\\\\\ d= \dfrac{11 \sqrt{13} }{ \sqrt{ 169}} \\\\\\\ d= \dfrac{11 \sqrt{13} }{ 13}$

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$C)\\\\\\ P(0,0)~~r:5x+2y-7=0\\\\\\ a=5\\\\\ b=2\\\\\\\ d= \dfrac{|5x+2y-7|}{ \sqrt{ a^{2}+b^{2} }}}\\\\\\\\\\ d= \dfrac{|5(0)+2(0)-7|}{ \sqrt{ 5^{2}+2^{2} }}}\\\\\\\\\\ d= \dfrac{|0+0-7|}{ \sqrt{ 25+4 }}}\\\\\\\\\\ d= \dfrac{|-7|}{ \sqrt{ 29 }}}$

Observe que deu-7 um valor negativo,como ele esta em modulo o valor é 7,pois não existe um numero negativo modular.Em outras palavras não existe distância negativa.

$d= \dfrac{7}{ \sqrt{ 29 }}} \times \dfrac{ \sqrt{29} }{\sqrt{29}}\\\\\\\\\ d= \dfrac{7 \sqrt{29} }{ \sqrt{841 }}} \\\\\\\\\ d= \dfrac{7 \sqrt{29} }{ 29}}$

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$D)\\\\\\ P(2,3)~~r:2x+y-7=0\\\\\\ a=2\\\\ b=1\\\\\\\ d= \dfrac{|2x+y-7|}{ \sqrt{ a^{2}+b^{2} }}}\\\\\\\\\ d= \dfrac{|2(2)+1(3)+5|}{ \sqrt{ 2^{2}+1^{2} }}}\\\\\\\\\\ d= \dfrac{|4+3+5|}{ \sqrt{ 4+1}}}\\\\\\\\\\ d= \dfrac{|12|}{ \sqrt{ 5}}}\\\\\\\\\\ d= \dfrac{12}{ \sqrt{ 5}}}\times \dfrac{ \sqrt{5}}{ \sqrt{5} } \\\\\\\\\\ d= \dfrac{12 \sqrt{5} }{ \sqrt{ 25}}}$

$d= \dfrac{12 \sqrt{5} }{ 5}}$

Boa tarde!
Bons estudos!