Hard Competition Geometry Problem

The shortest path that starts at the origin, touches the line y = 2x - 4, and then ends at (0, 1) is a straight line segment. The length of this line segment is sqrt(2^2 + (-4 - 1)^2) = sqrt(21).

 the length of the shortest path 

Hello BuilderBoy!

$y=2x-4\\\color{blue}P_1(0,0)\\ P(x,2x-4)\\ \color{blue}P_2(0,1)$

$L_{1,P}^2=x^2+y^2=x^2+(2x-4)^2\\ L_{P,2}^2=x^2+(1-y)^2=x^2+(1-2x+4)^2\\ L_{1,P}^2+L_{P,2}^2=x^2+(2x-4)^2+x^2+(5-2x)^2\\ $

$\frac{d(L_{1,P}^2+L_{P,2}^2)}{dx}=2x+2(2x-4)\cdot 2+2x+2(5-2x)\cdot (-2)=0\\ 4x+8x-16-20+8x=0\\ 20x=36\\ \color{blue}x_P=1.8\\ \color{blue}y_p=-0.4 \\ \color{blue}P(1.8,-0.4)$

You can calculate the distances between the points. The two-point equation is called: 

$​​L=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Have fun.

This is EZ if you know calculus.

Thanks Asinus and also Builderboi 

I have another method.

This is a question where the maths is super easy but understaning the concept may be a lot more challenging.

Rule: The shortest distance between 2 points is a line.

I have reflected the line AB  around  the line x=2 to get the line A'B'

Now to go from A to any point on the doted line and then back to B ist the same as going from A to that same point and then to B'

The shortest distance from A to B' is a straight line.  The length is     $\sqrt{4^2+1^1 }=\sqrt{17}$

Now you can find the actual point where the distance will be least using calculus but that is definitely not easy so I did it graphically.

I let the point giving the shortest distance be G

AG=EC   drew the circle

BG=AE   drew the circle.

Point G is where the circles and the line all cross each other.

(-1.6, 0.8)

Step-by-step explanation:

Given, equation of the line is y = 2x + 4 --- (1)

Closest point from origin will be the perpendicular distance from origin to the line.

We need to find out equation of the perpendicular from (0,0) on y = 2x + 4.

The equation is in slope intercept form i.e. y = mx + c

Slope, m = 2

Slope of the perpendicular = -(1/m) = -1/2

Equation of the perpendicular is found by (y - y1) = m (x - x1)

⇒ y - 0 = (-1/2) (x - 0)

⇒ y = (-1/2)x

⇒ 2y + x = 0 --- (2)

Solving (1) and (2), we get,

⇒ 5x = -8

⇒ x = -8/5

Substiute x = -8/5 in eq(2), we get

⇒ y = 4/5

x = -1.6 and y = 0.8

Therefore, the point on the line y = 2x + 4 closest to origin is (-1.6, 0.8)