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

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### 1 Answer

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.**