The linear equation that converts Fahrenheit to Celsius is
F = (\(\frac{9}{5} \))C + 32
\(\Rightarrow \) 5F - 9C = 160....(i)
For plotting points, we get the following table :
C | 0 | -160/9 | -40 |
---|
F = (9/5)C + 32 | 32 | 0 | -40 |
---|
Points | A(0, 32) | B(-160/9, 0) | c(-40, -40) |
---|
Here, we plot the given points on graph paper and join all these points to form a line.
ii)

ii)If the temperature is \(30^{\circ}\)C i.e.,C = \(30^{\circ}\)C
Then we get,
F = (\(\frac{9}{5} \) )×(30) + 32
= 9 × 6 + 32
= 54 + 32 = 86
Therefore, the temperature in Fahrenheit is 86F
iii)If the temperature is 95F i.e.,F = 95
Then we get, from eq.(i)
5(95) - 9C = 160
\(\Rightarrow \) 9C = 475 - 160 = 315
\(\therefore \) C = \(35^{\circ}\)C.
Therefore, the temperature in Celsius is \(35^{\circ}\)C
iv)If the temperature is \(0^{\circ}\)C i.e.,C = \(0^{\circ}\)
Then we get,
F = (\(\frac{9}{5} \) )×(0) + 32 = 32
Therefore, the temperature in Fahrenheit is 32F.
If the temperature is 0F i.e.,F = 0
Then we get, from eq.(i)
5(0) - 9C = 160
\(\Rightarrow \) C = \(\frac{-160}{9} ^{\circ}\)C
Therefore, the temperature in celsius is \(\frac{-160}{9} ^{\circ}\)C.
V)For this, let us take C = F,
Thus from eq.(i) we get,
5F - 9F = 160
\(\Rightarrow \) -4F = 160
\(\Rightarrow \) F = -40
Therefore, C = F = \(-40^{\circ}\)