1. Find $m$

We have

$$m=\sum_{k=1}^{99} \frac{1}{\sqrt{k}+\sqrt{k+1}}.$$

Rationalize each term:

$$\frac{1}{\sqrt{k}+\sqrt{k+1}}=\frac{\sqrt{k+1}-\sqrt{k}}{(\sqrt{k+1}+\sqrt{k})(\sqrt{k+1}-\sqrt{k})}=\sqrt{k+1}-\sqrt{k}.$$

So,

$$m=\sum_{k=1}^{99}(\sqrt{k+1}-\sqrt{k}).$$

This is a telescoping series:

$$m=(\sqrt2-1)+(\sqrt3-\sqrt2)+\cdots +(10-\sqrt{99})=10-1=9.$$

Hence,

$$m=9.$$

2. Find $n$

We have

$$n=\sum_{k=1}^{99}\frac{1}{k(k+1)}.$$

Use partial fractions:

$$\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}.$$

Thus,

$$n=\sum_{k=1}^{99}\left(\frac{1}{k}-\frac{1}{k+1}\right).$$

Again this telescopes:

$$n=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac{1}{99}-\frac{1}{100}\right)=1-\frac{1}{100}=\frac{99}{100}.$$

Hence,

$$n=\frac{99}{100}.$$

3. Point $(m,n)$

So the point is

$$(m,n)=\left(9,\frac{99}{100}\right).$$

4. Check the options

• A: $11(x-1)-100y=0$

Substitute $x=9, y=\frac{99}{100}$:

$$11(9-1)-100\cdot\frac{99}{100}=88-99=-11\neq 0.$$

So A is false.

• B: $11x-100y=0$

$$11\cdot 9-100\cdot\frac{99}{100}=99-99=0.$$

So B is true.

• C: $11(x-1)-100(y-2)=0$

$$11(8)-100\left(\frac{99}{100}-2\right)=88-(-101)=189\neq 0.$$

So C is false.

• D: $11(x-2)-100(y-1)=0$

$$11(7)-100\left(\frac{99}{100}-1\right)=77-(-1)=78\neq 0.$$

So D is false.

Therefore, the correct option is

$$\boxed{\text{B}}.$$

5. Comparison with stored answer

Stored correct answer: B.

My derived answer also gives B, so they agree.