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Solução!
![P(3,4)\\\\\
r:12x-5y-6=0\\\\\\
d(P,r)= \dfrac{|axP+byP+c|}{ \sqrt{a^{2}+b^{2}}} \\\\\\\\
d(P,r)= \dfrac{|12(3)-5(4)-6|}{ \sqrt{(-12)^{2}+(-5)^{2}}} \\\\\\\\\
d(P,r)= \dfrac{|36-20-6|}{ \sqrt{144+25}} \\\\\\\\\
d(P,r)= \dfrac{|36-26|}{ \sqrt{169}} \\\\\\\\\
d(P,r)= \dfrac{|10|}{ \sqrt{169}} \\\\\\\\\
\boxed {d(P,r)= \frac{10}{13}}](https://tex.z-dn.net/?f=P%283%2C4%29%5C%5C%5C%5C%5C%0Ar%3A12x-5y-6%3D0%5C%5C%5C%5C%5C%5C%0Ad%28P%2Cr%29%3D+%5Cdfrac%7B%7CaxP%2BbyP%2Bc%7C%7D%7B+%5Csqrt%7Ba%5E%7B2%7D%2Bb%5E%7B2%7D%7D%7D+%5C%5C%5C%5C%5C%5C%5C%5C%0Ad%28P%2Cr%29%3D+%5Cdfrac%7B%7C12%283%29-5%284%29-6%7C%7D%7B+%5Csqrt%7B%28-12%29%5E%7B2%7D%2B%28-5%29%5E%7B2%7D%7D%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%0Ad%28P%2Cr%29%3D+%5Cdfrac%7B%7C36-20-6%7C%7D%7B+%5Csqrt%7B144%2B25%7D%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%0Ad%28P%2Cr%29%3D+%5Cdfrac%7B%7C36-26%7C%7D%7B+%5Csqrt%7B169%7D%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%0Ad%28P%2Cr%29%3D+%5Cdfrac%7B%7C10%7C%7D%7B+%5Csqrt%7B169%7D%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed+%7Bd%28P%2Cr%29%3D+%5Cfrac%7B10%7D%7B13%7D%7D+%0A%0A%0A "P(3,4)\\\\\
r:12x-5y-6=0\\\\\\
d(P,r)= \dfrac{|axP+byP+c|}{ \sqrt{a^{2}+b^{2}}} \\\\\\\\
d(P,r)= \dfrac{|12(3)-5(4)-6|}{ \sqrt{(-12)^{2}+(-5)^{2}}} \\\\\\\\\
d(P,r)= \dfrac{|36-20-6|}{ \sqrt{144+25}} \\\\\\\\\
d(P,r)= \dfrac{|36-26|}{ \sqrt{169}} \\\\\\\\\
d(P,r)= \dfrac{|10|}{ \sqrt{169}} \\\\\\\\\
\boxed {d(P,r)= \frac{10}{13}}")
Boa noite!
Bons estudos!
Solução!
Boa noite!
Bons estudos!
NETESSINHA:
obrigada JBK.valeu mesmo..
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