#### Answer

The point on the line whose equation is \[2x+y-2=0\] that is closest to the origin is\[\left( \frac{4}{5},\frac{2}{5} \right)\].

#### Work Step by Step

Consider the equation of the line $2x+y-2=0$:
Solve for y,
$y=-2x+2$
Apply the distance formula for $y=-2x+2$ ,
$\begin{align}
& d=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
& =\sqrt{{{x}^{2}}+{{\left( -2x+2 \right)}^{2}}} \\
& =\sqrt{{{x}^{2}}+{{\left( -2x+2 \right)}^{2}}} \\
& =\sqrt{{{x}^{2}}+4-8x+4{{x}^{2}}}
\end{align}$
Simplify further,
$d=\sqrt{5{{x}^{2}}-8x+4}$
The minimum value of $5{{x}^{2}}-8x+4$ is the minimum value of the function $d=\sqrt{5{{x}^{2}}-8x+4}$ ,
Thus minimize $f\left( x \right)=5{{x}^{2}}-8x+4$.
Compare it with the standard form $f\left( x \right)=a{{x}^{2}}+bx+c$
The value of $a$ is $5$ , $b$ is $-8$ and $c$ is $4$.
Since, $a>0$ , the function has the minimum value at $x=\frac{-b}{2a}$.
$\begin{align}
& x=\frac{-b}{2a} \\
& =\frac{-\left( -8 \right)}{2\cdot 5} \\
& =\frac{8}{10} \\
& =\frac{4}{5}
\end{align}$
Substitute $\frac{4}{5}$ for x in $y=-2x+2$ .
$\begin{align}
& y=-2\left( \frac{4}{5} \right)+2 \\
& =\frac{2}{5}
\end{align}$
Hence, the point is $\left( \frac{4}{5},\frac{2}{5} \right)$.