(b) We know that $\frac{d\mathcal{Q}}{dt} = kA\frac{d\theta}{dx}$
In steady state flow of heat
$d\theta = \frac{d\mathcal{Q}}{dt}.\frac{1}{kA}dx$
$\Rightarrow$ $\theta_{H} - \theta = k'x$
$\Rightarrow$ $\theta = \theta_{H} - k'x$
Equation $\theta = \theta_{H} - k'x$ represents a straight line.