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(A) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively.Then the coordinates of the points where the line intersec

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$2x + 3y = 12$

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(A) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively.Then the coordinates of the points where the line intersects the axes are $A(a, 0)$ and $B(0, b)$.The midpoint of the segment $AB$ is given by $(\frac{a+0}{2}, \frac{0+b}{2}) = (\frac{a}{2}, \frac{b}{2})$.Given that the midpoint is $(3, 2)$,we have:$\frac{a}{2} = 3 \Rightarrow a = 6$$\frac{b}{2} = 2 \Rightarrow b = 4$The intercept form of the equation of a line is $\frac{x}{a} + \frac{y}{b} = 1$.Substituting $a=6$ and $b=4$,we get $\frac{x}{6} + \frac{y}{4} = 1$.Multiplying by $12$,we get $2x + 3y = 12$.

If the coordinates of the midpoint of the portion of a line intercepted between the coordinate axes are $(3, 2)$,then the equation of the line is:

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