# Knowing their linear relationship, find the expression of ^{\circ}C in

Russell Gillen 2021-12-18 Answered
Knowing their linear relationship, find the expression of $$\displaystyle^{\left\lbrace\circ\right\rbrace}{C}$$ in terms of $$\displaystyle^{\left\lbrace\circ\right\rbrace}{F}$$ using the point-slope form af analytic geomtry. With boiling of $$\displaystyle{100}^{{\circ}}{C}={212}^{{\circ}}{F}$$ and the freezing point of $$\displaystyle{0}^{{\circ}}{C}={32}^{{\circ}}{F}$$

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Karen Robbins
Step 1
If we have 2 points $$\displaystyle{\left({x}_{{{0}}},\ {y}_{{{0}}}\right)}$$ and $$\displaystyle{\left({x}_{{{1}}},\ {y}_{{{1}}}\right)}$$ then the point-slope form is $$\displaystyle{y}-{y}_{{{1}}}={m}{\left({x}-{x}_{{{1}}}\right)}$$, here, m is the slope given by:
$$\displaystyle{m}={\frac{{{y}_{{{1}}}-{y}_{{{0}}}}}{{{x}_{{{1}}}-{x}_{{{0}}}}}}$$
Hence, the point-slope form becomes $$\displaystyle{y}-{y}_{{{1}}}={\left({\frac{{{y}_{{{1}}}-{y}_{{{0}}}}}{{{x}_{{{1}}}-{x}_{{{0}}}}}}\right)}{\left({x}-{x}_{{{1}}}\right)}$$
Now, we have to find the relation between Fahrenheit and Celcius.
Let Fahrenheit be represented by F and Celcius by C.
Now, we are given that $$\displaystyle{100}^{{\circ}}{C}={212}^{{\circ}}{F}$$ and $$\displaystyle{0}^{{\circ}}{C}={32}^{{\circ}}{F}$$
This means that when the temperature is $$\displaystyle{100}^{{\circ}}{C}$$ then it is same as $$\displaystyle{212}^{{\circ}}{F}$$, and when the temperature is $$\displaystyle{0}^{{\circ}}{C}$$ it is same as $$\displaystyle{32}^{{\circ}}{F}$$
Therefore, let:
$$\displaystyle{C}_{{{0}}}={0}^{{\circ}}{C}$$ and $$\displaystyle{F}_{{{0}}}={32}^{{\circ}}{F}$$
$$\displaystyle{C}_{{{1}}}={100}^{{\circ}}{C}$$ and $$\displaystyle{F}_{{{1}}}={212}^{{\circ}}{F}$$
Step 2
Therefore, we receive two points
$$\displaystyle{\left({C}_{{{0}}},\ {F}_{{{0}}}\right)}={\left({0},\ {32}\right)}$$
$$\displaystyle{\left({C}_{{{1}}},\ {F}_{{{1}}}\right)}={\left({100},\ {212}\right)}$$
Now, we find the point-slope form:
$$\displaystyle{F}-{F}_{{{1}}}={\left({\frac{{{F}_{{{1}}}-{F}_{{{0}}}}}{{{C}_{{{1}}}-{C}_{{{0}}}}}}\right)}{\left({C}-{C}_{{{1}}}\right)}$$
$$\displaystyle\Rightarrow{F}-{212}={\left({\frac{{{212}-{32}}}{{{100}-{0}}}}\right)}{\left({C}-{100}\right)}$$
$$\displaystyle\Rightarrow{F}-{212}={\left({\frac{{{180}}}{{{100}}}}\right)}{\left({C}-{100}\right)}$$
$$\displaystyle\Rightarrow{F}-{212}={\left({\frac{{{9}}}{{{5}}}}\right)}{\left({C}-{100}\right)}$$
$$\displaystyle\Rightarrow{F}-{212}={\frac{{{9}}}{{{5}}}}{C}-{\left({\frac{{{9}}}{{{5}}}}\right)}{100}$$
$$\displaystyle\Rightarrow{F}-{212}={\frac{{{9}}}{{{5}}}}{C}-{9}\times{20}$$
$$\displaystyle\Rightarrow{F}-{212}={\frac{{{9}}}{{{5}}}}{C}-{180}$$
$$\displaystyle\Rightarrow{F}={\frac{{{9}}}{{{5}}}}{C}+{32}$$
$$\displaystyle\Rightarrow{F}-{32}={\frac{{{9}}}{{{5}}}}{C}$$
$$\displaystyle\Rightarrow{\frac{{{5}}}{{{9}}}}{\left({F}-{32}\right)}={C}$$
$$\displaystyle\Rightarrow{C}={\frac{{{5}}}{{{9}}}}{\left({F}-{32}\right)}$$
Hence, the relation between Fahrenheit and Celcius is a linear relation given by $$\displaystyle{C}={\frac{{{5}}}{{{9}}}}{\left({F}-{32}\right)}$$
Answer: The expression of Celcius in terms of Fahrenheit is $$\displaystyle{C}={\frac{{{5}}}{{{9}}}}{\left({F}-{32}\right)}$$
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Jenny Sheppard
Step 1
The relatijn between the celsium and Fanrenhcit scals
As the temperature risce on the celsius scale the equivalent temperature of fahrenhcit waild also be high
As the temperature falls on the celsius scale, the temp equal to fahrenhciet would also be low-
freezing point of water at $$\displaystyle^{\left\lbrace\circ\right\rbrace}{C}={0}^{{\circ}}{C}$$
and boiling point of untered $$\displaystyle^{\left\lbrace\circ\right\rbrace}{C}={100}^{{\circ}}{C}$$
freezing point of water at Fahrenhcit $$\displaystyle={32}^{{\circ}}{F}$$
Boiling point
According to thed
$$\displaystyle{\frac{{{C}-{L}\times{F}\times{P}}}{{{U}\times{F}\times{P}-{L}\times{F}\times{P}}}}={\frac{{{F}-{L}\times{F}\times{P}}}{{{U}\times{F}\times{P}-{L}\times{F}\times{P}}}}$$
$$\displaystyle{L}\times{F}\times{P}=\text{Lower Fixed point}$$
$$\displaystyle{U}\times{F}\times{P}=\text{Upper fixed point}$$
$$\displaystyle{\frac{{{C}-{O}}}{{{100}-{0}}}}={\frac{{{F}-{32}}}{{{212}-{32}}}}$$
$$\displaystyle{\frac{{{C}}}{{{100}}}}={\frac{{{F}-{32}}}{{{180}}}}$$
$$\displaystyle{\frac{{{C}}}{{{5}}}}={\frac{{{\left({F}-{32}\right)}}}{{{9}}}}$$
$$\displaystyle{C}={\frac{{{5}}}{{{9}}}}{\left({F}-{32}\right)}$$
$$\displaystyle{C}={\frac{{{5}}}{{{9}}}}{F}-{\frac{{{5}\times{32}}}{{{9}}}}$$
$$\displaystyle{c}={\frac{{{5}}}{{{9}}}}{F}-{\frac{{{160}}}{{{9}}}}$$
nick1337

Step 1
If C is a function, then we may represent the points as (F, C)
$$(C-C_{1})=m(F-F_{1})$$
(32,0)
(212.100)
$$(C-0)=\frac{100-0}{212-32}(F-32)$$
$$C=\frac{100}{180}(F-32)$$
$$C=\frac{5}{9}(F-32)$$

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