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 \(\displaystyle{C}={\frac{{{5}}}{{{9}}}}{\left({F}-{32}\right)}\)

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 \(\displaystyle{C}={\frac{{{5}}}{{{9}}}}{\left({F}-{32}\right)}\)