If a straight line passing through the point P(-3, 4) is such that its intercepted portion between the coordinate axes is bisected at P, then its equation is:

We have a straight line that cuts the coordinate axes at two points: let the x-intercept be $$A(a,0)$$ and the y-intercept be $$B(0,b)$$. In coordinate geometry, the equation of such a line in “intercept form” is first stated as

$$\frac{x}{a}+\frac{y}{b}=1.$$

The problem states that the fixed point $$P(-3,4)$$ lies on the line and, moreover, that the segment $$AB$$ is bisected at $$P$$. To translate “bisected at $$P$$”, we recall the mid-point formula: if $$M(x_m, y_m)$$ is the midpoint of $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$, then

$$x_m=\frac{x_1+x_2}{2},\qquad y_m=\frac{y_1+y_2}{2}.$$

Applying this to $$A(a,0)$$ and $$B(0,b)$$ with midpoint $$P(-3,4)$$, we write

$$\frac{a+0}{2}=-3,\qquad\frac{0+b}{2}=4.$$

Solving these two simple equations one by one:

First, $$\dfrac{a}{2}=-3 \;\Longrightarrow\; a=-6.$$

Second, $$\dfrac{b}{2}=4 \;\Longrightarrow\; b=8.$$

So the line meets the x-axis at $$(-6,0)$$ and the y-axis at $$(0,8)$$. We now substitute these values of $$a$$ and $$b$$ back into the intercept form of the line.

Starting with $$\dfrac{x}{a}+\dfrac{y}{b}=1$$ and inserting $$a=-6$$, $$b=8$$, we get

$$\frac{x}{-6}+\frac{y}{8}=1.$$

To clear the denominators, multiply both sides by the LCM $$24$$:

$$24\left(\frac{x}{-6}\right)+24\left(\frac{y}{8}\right)=24(1).$$

Simplifying term by term,

$$24\left(\frac{x}{-6}\right)=-4x,\qquad 24\left(\frac{y}{8}\right)=3y,\qquad 24(1)=24.$$

So the equation becomes

$$-4x+3y=24.$$

For a tidier appearance—and to match one of the given options—we multiply through by $$-1$$:

$$4x-3y=-24.$$

Finally, bringing all terms to the left-hand side gives the standard form

$$4x-3y+24=0.$$

Hence, the correct answer is Option B.