What is the shortest distance from the origin to the line 2x+y-3=0?

We'll write the formula of the distance from the origin to
the given line:

D = sqrt [(x - 0)^2 + (y -
0)^2]

Now, we'll determine the expression of y with respect
to x:

2x + y - 3 = 0

We'll
keep y to the left moving all the rest to the right:

y =
-2x + 3

We'll re-write the formula of distance, replacing y
by the equivalent expression:

D = sqrt[x^2 + (3 -
2x)^2]

We'll expand the
square:

D = sqrt(x^2 + 9 - 12x +
4x^2)

D = sqrt(5x^2 - 12x +
9)

We'll differentiate both sides with respect to
x:

dD/dx = (5x^2 - 12x + 9)'/2sqrt(5x^2 - 12x +
9)

dD/dx = (10x - 12)/2sqrt(5x^2 - 12x +
9)

dD/dx = 2(5x - 6)/2sqrt(5x^2 - 12x +
9)

dD/dx = (5x - 6)/sqrt(5x^2 - 12x +
9)

We'll cancel dD/dx:

dD/dx =
0 <=> 5x - 6 = 0 => x = 6/5

Now, we'll
replace x = 6/5 into the formula of D:

D = sqrt(5x^2 - 12x
+ 9)

D = sqrt(180/25 - 72/5 +
9)

D =
sqrt[(180-360+225)/25]

D = sqrt
(45/25)

D =
(3sqrt5)/5

The shortest distance from the
origin to the line 2x + y - 3 = 0 is: D = (3sqrt5)/5
units.