A straight line through the point $$A(3, 4)$$ is such that its intercept between the axes is bisected at $$A$$. Its equation is

Any straight line that cuts the coordinate axes at points $$P(a,0)$$ on the $$x$$-axis and $$Q(0,b)$$ on the $$y$$-axis can be written in intercept form as
$$\frac{x}{a}+\frac{y}{b}=1$$

The segment $$PQ$$ is called the intercept of the line between the axes. Its midpoint is obtained by averaging the coordinates of $$P$$ and $$Q$$:
midpoint $$M=\left(\frac{a+0}{2},\,\frac{0+b}{2}\right)=\left(\frac{a}{2},\,\frac{b}{2}\right)$$.

According to the question, this midpoint coincides with the given point $$A(3,4)$$. Therefore
$$\frac{a}{2}=3 \quad\text{and}\quad \frac{b}{2}=4$$

Solving these gives the actual intercepts:
$$a=6,\; b=8$$

Substitute $$a$$ and $$b$$ back into the intercept form:
$$\frac{x}{6}+\frac{y}{8}=1$$

Eliminate denominators by multiplying throughout by the least common multiple $$24$$:
$$24\left(\frac{x}{6}+\frac{y}{8}\right)=24(1)$$
$$4x+3y=24$$

Thus the required equation of the line is $$4x+3y=24$$.

Option C which is: $$4x + 3y = 24$$