Verified answer

Solving 2x+4y = 1 for x in terms of y gives x = 0.5 - 2y. Then substitute and complete the square.

x^2 + y^2 = (0.5 - 2y)^2 + y^2

= 4y^2 - 2y + 0.25 + y^2

= 5y^2 - 2y + 0.25

= 5(y^2 - 0.4y + 0.05)

= 5[y^2 - 0.4y + (0.2)^2 + 0.05 - (0.2)^2]

= 5[(y - 0.2)^2 + 0.01]

>= 5(0.01)

= 1/20.

Lord bless you today!

(x² + y²) represents the square of the distance of the point (x,y) from origin .

If the point (x,y) lies on the line 2x + 4y =1, then it is required to show that the point , is at a distance 'd' from the origin , such that d² ≥ 1/20.

The distance ‘d’ from a point (m, n) to a line ax + by = c is given by :

d = ∣a*m + b*n -c ∣ / √(a² + b²) . In the given situation: a = 2, b= 4 and c = 1 and

at origin m = n = 0 , hence d = 1/√(2² + 4²) . Therefore d² = 1/20.

This is the shortest (perpendicular) distance of a point on the line 2x + 4y = 1 from origin.

Any other point on the line will be at a distance > d i.e., d² = (x² + y²) ≥ 1/20