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**Hint:**The following diagram represents a graph between Celsius on the $y-axis$ and Fahrenheit on the $x-axis$. The curves $a,b,c,d$ are the curves on the graph out of which one shows the relation between Celsius and Fahrenheit. We will first write down the mathematical equation for the relation. Then make Celsius the subject of the equation and then compare it with the equation of a straight line. Then we find the slope using those two equations and see which slopes in the diagram match it.

**Formula Used:**

Relation between Celsius and Fahrenheit is given by equation (1) where $F,C$ are Fahrenheit and Celsius. Equation of straight line is given by equation (2).

$F = \dfrac{9}{5} \times C + 32......(1) $

$y = mx + c.......(2) $

**Complete step by step solution:**

We know that the relation between Celsius and Fahrenheit is $F = \dfrac{9}{5} \times C + 32$

We make Celsius the subject now,

$F - 32 = \dfrac{9}{5} \times C $

$C = \dfrac{5}{9} \times (F - 32) $

$C = (\dfrac{5}{9})F - \dfrac{{160}}{9}......\left( 3 \right) $

We make Celsius the subject because in the equation of straight line $y = mx + c$, $c,x,m,$ and $y$ represents the intercept of the straight line on $y - axis$ ,$x - axis$, slope of curve, and $y - axis$ respectively.

In the diagram $y - axis$ is Celsius, which is why we change the subject according to the equation of the straight line.

Now comparing equations (3) and (2) we can see that equation (3) is in the form of a straight-line equation. Where the slope is $\dfrac{5}{9}$ and the intercept of straight line on $y - axis$ is $\left( { - \dfrac{{160}}{9}} \right)$.

By this, we can say that the curve representing the relation between Celsius and Fahrenheit has a positive slope and a negative intercept on the y-axis. These conditions are satisfied by the curve $’a’$.

**Hence option (a) $a$ is the correct answer.**

**Note:**In this problem, we need to know in which quadrant a point falls in and should also know about the positive and negative slope. A positive slope expands on moving vertically upward whereas a negative slope declines on moving upward.

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