Đáp án:
g) \(d\left( {A;\left( \Delta \right)} \right) = 0\)
Giải thích các bước giải:
\(\begin{array}{l}
a)d\left( {A;\left( d \right)} \right) = \dfrac{{\left| {2.4 + 5 - 3} \right|}}{{\sqrt {{2^2} + 1} }} = \dfrac{{10}}{{\sqrt 5 }} = 2\sqrt 5 \\
b)d\left( {A;\left( \Delta \right)} \right) = \dfrac{{\left| {3 + 5 - 7} \right|}}{{\sqrt {1 + 1} }} = \dfrac{1}{{\sqrt 2 }}\\
c)d\left( {A;\left( d \right)} \right) = \dfrac{{\left| {4.4 - 5 - 5} \right|}}{{\sqrt {{4^2} + 1} }}\\
= \dfrac{6}{{\sqrt {17} }}\\
d)d\left( {A;\left( d \right)} \right) = \dfrac{{\left| { - 2.2 + 5.\left( { - 3} \right) - 4} \right|}}{{\sqrt {{{\left( { - 2} \right)}^2} + {5^2}} }}\\
= \dfrac{{23}}{{\sqrt {29} }}\\
e)vtcp:{\overrightarrow u _\Delta } = \left( {1; - 4} \right)\\
\to vtpt:{\overrightarrow n _\Delta } = \left( {4;1} \right)\\
PTTQ:\left( \Delta \right):4\left( {x - 2} \right) + y - 3 = 0\\
\to 4x + y - 11 = 0\\
\to d\left( {A;\left( \Delta \right)} \right) = \dfrac{{\left| {4.3 - 2 - 11} \right|}}{{\sqrt {{4^2} + 1} }} = \dfrac{1}{{\sqrt {17} }}\\
f)d\left( {A;\left( d \right)} \right) = \dfrac{{\left| {4 - 4.\left( { - 3} \right) + 5} \right|}}{{\sqrt {1 + {{\left( { - 4} \right)}^2}} }} = \dfrac{{21}}{{\sqrt {17} }}\\
g)vtcp:{\overrightarrow u _\Delta } = \left( {1; - 2} \right)\\
\to vtpt:{\overrightarrow n _\Delta } = \left( {2;1} \right)\\
PTTQ:\left( \Delta \right):2\left( {x - 2} \right) + y - 3 = 0\\
\to 2x + y - 7 = 0\\
\to d\left( {A;\left( \Delta \right)} \right) = \dfrac{{\left| {2.1 + 5 - 7} \right|}}{{\sqrt {{2^2} + 1} }} = 0
\end{array}\)
