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(C) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$ $(1)$.Since the line passes through $(3, 4)$,we have $\frac{3}{a} + \frac{4}{b} =

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$4x + 3y = 24$ and $x + y = 7$

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(C) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$ $(1)$.Since the line passes through $(3, 4)$,we have $\frac{3}{a} + \frac{4}{b} = 1$ $(2)$.Given that the sum of intercepts is $a + b = 14$,so $b = 14 - a$.Substituting $b = 14 - a$ in $(2)$:$\frac{3}{a} + \frac{4}{14 - a} = 1$$3(14 - a) + 4a = a(14 - a)$$42 - 3a + 4a = 14a - a^2$$a^2 - 13a + 42 = 0$$(a - 7)(a - 6) = 0$So,$a = 6$ or $a = 7$.If $a = 6$,then $b = 14 - 6 = 8$. The equation is $\frac{x}{6} + \frac{y}{8} = 1$,which simplifies to $4x + 3y = 24$.If $a = 7$,then $b = 14 - 7 = 7$. The equation is $\frac{x}{7} + \frac{y}{7} = 1$,which simplifies to $x + y = 7$.

Find the equation of the straight line passing through the point $(3, 4)$ such that the sum of its intercepts on the axes is $14$.

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