Use calculus to find length of y=-mx+b, then show that the answer agrees with the answer when using pythagorean theorem

Question

First question, sorry for poor formatting. This question is from my Calculus 2 class, and I am pretty sure I am supposed to be using arc length formula for the question.

Exact words:

1. Consider the line segment shown

   *shows simple right triangle with $y=-mx+b$ over hypotenuse, and the $90$ degree corner sitting at point $(0,0)$

   a. Use calculus to find its length (show your calculation of the integral involved)

   b. Now show that your answer to part a agrees with what you get when you simply use the Pythagorean Theorem (or distance formula)

So I think I got through the first part:

Taking the derivative of $y=-mx+b$ and putting it into the arc length formula I got $(b-a)\sqrt{1+m^2}$

The second part is where I am lost... I feel like I am missing something pretty easy but I don't understand how I would make the Pythagorean theorem or distance formula to agree with my answer. My answer to part a might just be wrong too but I am not sure how else to do it.

There are a lot more parts to the problem, if I can get past this hopefully it can show me how to approach the other problems.

Appreciate any help I can get, thanks!

Answer 1

Hint:when $x=0, y=b $ and when $y=0,x=\frac bm.$Now apply Pythagorean theorem to the right angled triangle constructed using the points $(0,0), (0,b)$ and $(b/m, 0)$.

But this will give you an answer different from the one you wrote above. Also you haven't mentioned anything about $a$.

Answer 2

The line $y=-mx+b$ looks like this

So if you have a right triangle with lengths $b$, and $b/m$ the hypotenuse should be (by Pythagorean theorem) length

$\sqrt{b^2+(b/m)^2} = b\sqrt{1+m^{-2}}$

But also we can compute the area of the arc length given by $\int_0^{\frac{b}{m}}\sqrt{1+f'\left(t\right)^2}dt$ where $f(x)=-mx+b$.

$\int_0^{b/m}\sqrt{1+f'\left(t\right)^2}dt$

$=\int_0^{b/m} \sqrt{1+m^2}dt= \frac{b}{m}\sqrt{1+m^2}$.

So we get the same result as when we employed Pythagorean theorem.