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(B) Let the line intersect the $x$-axis at $B(a, 0)$ and the $y$-axis at $C(0, b)$.Since the point $A(4, 3)$ is nearer to the $x$-axis and trisects th

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$3x + 2y = 18$

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(B) Let the line intersect the $x$-axis at $B(a, 0)$ and the $y$-axis at $C(0, b)$.Since the point $A(4, 3)$ is nearer to the $x$-axis and trisects the line segment $BC$,it divides $BC$ in the ratio $2:1$ from $C$ to $B$.Using the section formula,the coordinates of $A$ are given by:$A = \left( \frac{1 	imes 0 + 2 	imes a}{1 + 2}, \frac{1 	imes b + 2 	imes 0}{1 + 2} ight) = \left( \frac{2a}{3}, \frac{b}{3} ight)$Given $A(4, 3)$,we have:$\frac{2a}{3} = 4$ $\Rightarrow 2a = 12$ $\Rightarrow a = 6$$\frac{b}{3} = 3 \Rightarrow b = 9$Thus,the intercepts are $a = 6$ and $b = 9$.The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.Substituting the values,we get $\frac{x}{6} + \frac{y}{9} = 1$.Multiplying by $18$,we get $3x + 2y = 18$.

If a line intercepted between the coordinate axes is trisected at a point $A(4, 3)$,which is nearer to the $x$-axis,then its equation is

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