Determining the mass of a substance using calibration curve produced by HPLC I hope everyone is doing well and safe! Yesterday, I was fortunate enough to be taught about how to read a chromatography calibration curve through this stackexchange! Special thanks to those who explained to me patiently and meticulously. Anyways, I've tried to solve some questions regarding this field but ended up no-where with this one. Could someone run me through the procedure of how to solve this question? Huge thanks in advance!!! &lt;3 &lt;3

Info: A $\pu{1.0mL}$ aliquot of a drink was diluted to $\pu{50.0mL}$ with deionized water. A sample of the drink was run through the HPLC to produce the calibration curve below. In this experiment, the mass of caffeine in the drink was determined by HPLC. The peak area obtained for this diluted sample was $2800$ arbitrary units.


Question: What is the mass of caffeine in grams, in a $\pu{330mL}$ can of this drink?

Answer = $\pu{825mg}$
P.S.: Could someone explain to me the significance of the peak area ($2800$ arbitrary units) that is obtained? And thank you for the constant support and help. Much love to this community!!!
Warning: This is also a homework question. However, this is more like mathematics (geometry) than chemistry, I decided to help you. Nonetheless, please keep in mind that we won't help you in all future perspectives, if you do not show your effort to solve the problem, regardless more mathematics involved.
As  M. Farooq pointed out that this problem can be solved using the answers to your previous question, specially my old-fashioned approach. Here is how you could do it. Analyze your graph thoroughly to find two different points of graph where it perfectly crossed the cross-lines:

For me, these two points are: $(35,2000)$ and $(60,3500)$. Therefore the tangent of the slope ($\alpha$) of line is:
$$\tan \alpha = \frac{3500-2000}{60-35} = \frac{1500}{25}=60$$
Now, take another arbitrary point in the graph. Say it is $(x,y)$. Then:
$$\tan \alpha = \frac{y-2000}{x-35} = 60 \ \Rightarrow \ y = 60x - 100 \tag1$$
The equation $(1)$ is the straight-line equation of the graph.
When $y = 2800$, $x= \frac{2800+100}{60} = \pu{48.33 mg/L}$. This is for the diluted sample. I think it is fair to leave you some work to do. So you can back calculate to get your answer. When I did, it came close to $\approx \pu{798 mg/can}$.
Note: The error ($3.27\%$) is due to the finding of the slope. Also there are few flaws in the question. The calibration curve is made with HPLC traces of known standard caffeine solutions (not using diluted samples of the drink). The peak area of $2800$ arbitrary units was obtained by the injection of diluted sample of the drink ($1$ to $\pu{50 mL}$ dilution).