# To use the two-point formula to find the linear equation relating $C$ and $F$:

I've tried to solve a problem which I'm going to give below. What I don't understand is that which variable is dependent and which is independent among $C$ and $F$. I think we can relate $C$ and $F$ using two equations. In one of the equations $C$ can be the dependent variable and $F$ the independent and in the other equation $C$ can be independent and $F$ can be dependent. However, the problem is as follows:

Let $C$ denote Celsius temperature and $F$ Fahrenheit temperature. Thus, $C=0$ and $F=32$ at the freezing point of water, while $C=100$ and $F=212$ at the boiling point of water. Use the two point formula to find the linear equation relating $C$ and $F$.

My solution:

Let $C_1 = 0$ and $C_2 = 100$. Likewise, $F_1=32$ and $F_2=212$. If $P(C_1,F_1)$ and $Q(C_2,F_2)$ are the two points on the line $L$ then we have the following equation for $L$:

$F-F_1= \bigg( \frac{F_2-F_1}{C_2-C_1}\bigg)(C-C_1)$

OR

$F-32= \bigg( \frac{212-32}{100-0}\bigg)(C-0)$

OR

$F = \frac{180}{100}C + 32$

OR

$F = \frac{9}{5}C+32$

Do I need to find the other relation between $C$ and $F$?

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Your solution looks great to me. I'm not sure about what other relation you're referring to, but if you mean that you need to find $C$ in terms of $F$, then subtract 32 and multiply by $\frac{5}{9}$, giving you $C = \frac{5}{9}(F - 32)$. – Alex Wertheim May 12 '13 at 0:05
Yes, I meant the same thing and thanks. :) – Samama Fahim May 12 '13 at 0:15

Your steps in using the two point formula to find a relation between $F$ and $C$ is correct, and done well.

The resulting relation (equation) is your solution: $$F = \frac{9}{5}C+32$$

Note that given this equation, you can also represent $C$ as a function of $F$:

$$F = \frac{9}{5}C+32 \iff F - 32 = \frac 95 C \iff \frac 59(F - 32) = C$$

So $$F = \frac{9}{5}C+32 \quad \text{and} \quad C = \frac 59(F - 32)$$ represent equivalent equations, where the first expresses $F$ as a function of $C$, and the second expresses $C$ as a function of $F$.

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Nice graph, Samama! I just saw it, and see that you modified your earlier graphs, which I missed when you first added it. Nice work with the correction. Looks good to me! – amWhy May 12 '13 at 0:54
Thanks :). I made a correction in the earlier graph. The earlier graph was actually the graph of this equation: $F = \frac 59 C + 32$ which is incorrect. – Samama Fahim May 12 '13 at 1:49

Your computation and the resulting formula look fine, but there seems to be a problem with the graph. I think you plotted the intercept 32 on the vertical axis, assuming that the vertical direction represents $F$, but then I think you plotted a slope of $5/9$ rather than $9/5$, assuming that the vertical axis represents $C$ and the horizontal axis represents $F$. One thing I can say for sure, regardless of which axis represents which variable, is that the graph should go through the point $(-40,-40)$, which your graph doesn't. The reason is that $-40$ is the (one and only) temperature that comes out the same on the Fahrenheit and Celsius scales.

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Thanks for informing about the mistake. I've changed the screenshot. – Samama Fahim May 12 '13 at 0:50
Looks good now --- heading for $(-40,-40)$. – Andreas Blass May 12 '13 at 1:32