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(C) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively. The equation of the line is $\frac{x}{a} + \frac{y}{b} =

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$x+y=6$

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(C) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively. The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.Given that the sum of the intercepts is $a + b = 12$,so $b = 12 - a$.The area of the triangle formed by the line with the coordinate axes is $A = \frac{1}{2}ab$.Substituting $b = 12 - a$,we get $A = \frac{1}{2}a(12 - a) = 6a - \frac{1}{2}a^2$.To find the maximum area,we differentiate $A$ with respect to $a$ and set it to zero:$\frac{dA}{da} = 6 - a = 0 \Rightarrow a = 6$.Since $a = 6$,then $b = 12 - 6 = 6$.The equation of the line is $\frac{x}{6} + \frac{y}{6} = 1$,which simplifies to $x + y = 6$.

If a line is moving between the coordinate axes such that the sum of the intercepts made by it on the coordinate axes is always $12$,then the equation of that line which forms a triangle of maximum area with the coordinate axes is

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