Compute the length of the shortest path that starts at the origin, touches the line y = 2x - 4, and then ends at (0, 1).

I now know the answer but don't know how to solve it. Any help would be appreciated!

BuilderBoi Apr 29, 2023

#1**+2 **

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).

Guest Apr 29, 2023

#2**+1 **

Sorry, but that's incorrect. The correct answer is actually \(\sqrt{17}\), even though I don't know how to get it.

Thanks so much for your attempt, though!

BuilderBoi
Apr 30, 2023

#3**+2 **

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.**

asinus Apr 30, 2023

#4**+1 **

Then the distance from \((0, 0)\) to \((1.8, -0.4)\) is \(\sqrt{(1.8)^2 + (0.4)^2} = {\sqrt{85} \over 5} \) and the distance from \((0, 1) \) to \((1.8, -0.4)\) is \(\sqrt{(1.8)^2 + (1.4)^2} = {\sqrt{130} \over 5}\).

Adding the two gives us \({\sqrt {85 }+ \sqrt {130} \over 5} \approx 4.12426\) but the correct answer was \(\sqrt{17} \approx 4.12311\).

BuilderBoi
Apr 30, 2023

#6**+2 **

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.

Melody Apr 30, 2023

#7**+1 **

Thanks for the explanation Melody, but why did you reflect AB over x = 2 in particular?

BuilderBoi
Apr 30, 2023

#9**+1 **

That is a very good question and maybe one I will struggle with explaining.

You have the initial interval AB

I drew 2 lines that were perpendicular to AB and therefore parallel to each other.

I called them BC and AD but the exact position of C and D is not set yet

The line crossed AD at 2 units.

So the lines must go right 2 up one and left 2. I changed that to out 4 and up one.

So now I have the rectangle aa'b'b

going diagonally the closest way to get from A to B' is in a straight line.

Melody
Apr 30, 2023

#11**+1 **

(-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)

acyclics May 1, 2023