The perpendicular bisector of the line segment joining $$P(1, 4)$$ and $$Q(k, 3)$$ has y-intercept -4. Then a possible value of $$k$$ is

The line segment $$PQ$$ joins the points $$P(1,4)$$ and $$Q(k,3)$$.

1. Mid-point of $$PQ$$
$$M\Bigl(\frac{1+k}{2},\frac{4+3}{2}\Bigr)=\Bigl(\frac{1+k}{2},\frac{7}{2}\Bigr).$$

2. Slope of $$PQ$$
$$m_{PQ}=\frac{3-4}{k-1}=\frac{-1}{k-1}.$$

3. Slope of the perpendicular bisector
If two lines are perpendicular, the product of their slopes is $$-1$$. Hence the slope of the perpendicular bisector is the negative reciprocal:
$$m_{\perp}=-(1/m_{PQ})=k-1.$$

4. Equation of the perpendicular bisector
Using point-slope form with point $$M$$ and slope $$m_{\perp}$$,
$$y-\frac{7}{2}=(k-1)\Bigl(x-\frac{1+k}{2}\Bigr).$$

5. Imposing the given y-intercept
The y-intercept is obtained by setting $$x=0$$.
Substituting $$x=0$$ in the above equation:
$$y=\frac{7}{2}+(k-1)\Bigl(-\frac{1+k}{2}\Bigr)$$ This y-value must be $$-4$$:
$$\frac{7}{2}-(k-1)\frac{1+k}{2}=-4.$$

Multiply by 2 to clear the denominator:
$$7-(k-1)(k+1)=-8.$$

Simplify:
$$7+8=(k-1)(k+1)$$ $$15=k^{2}-1$$ $$k^{2}=16$$ $$k=\pm4.$$

6. Selecting from the options
Both $$k=4$$ and $$k=-4$$ satisfy the condition, but only $$k=-4$$ appears among the given options.

Hence a possible value of $$k$$ is
Option D which is: $$-4$$.