Step 1: Find the ratio using similar triangles instead of the section formula.
Instead of assuming a ratio $k:1$ and solving an equation for $k$, we can read the ratio straight off the picture using similar triangles formed by dropping perpendiculars to the x-axis.
Step 2: Set up the picture.
Let $A(-6, 5)$ and $B(-4, -1)$ be the given points, and let $P$ be the point where segment $AB$ crosses the x-axis. Drop a perpendicular from $A$ down to the x-axis, meeting it at $A'(-6, 0)$, and drop a perpendicular from $B$ up to the x-axis, meeting it at $B'(-4, 0)$.
Step 3: Identify the pair of similar triangles.
Triangles $APA'$ and $BPB'$ are similar, because $AA'$ and $BB'$ are both perpendicular to the x-axis (so they are parallel to each other), and the angles at $P$ are vertically opposite. By AA similarity:
\[ \Delta APA' \sim \Delta BPB' \]
Step 4: Use the similarity to write the ratio $AP : PB$.
Corresponding sides of similar triangles are in the same ratio, so:
\[ \frac{AP}{PB} = \frac{AA'}{BB'} \]
Here $AA'$ is just the vertical distance from $A$ to the x-axis, which is $|5| = 5$, and $BB'$ is the vertical distance from $B$ to the x-axis, which is $|-1| = 1$. Since $A$ lies above the axis and $B$ lies below it, $P$ divides $AB$ internally, so:
\[ \frac{AP}{PB} = \frac{5}{1} \]
The x-axis divides the segment in the ratio $5:1$.
Step 5: Use the section formula only for the coordinates, now that the ratio is known.
With ratio $5:1$, the point dividing $A(-6,5)$ and $B(-4,-1)$ has x-coordinate:
\[ x = \frac{5(-4) + 1(-6)}{5 + 1} = \frac{-20 - 6}{6} = \frac{-26}{6} = -\frac{13}{3} \]
And as expected, the y-coordinate works out to 0:
\[ y = \frac{5(-1) + 1(5)}{6} = \frac{-5 + 5}{6} = 0 \]
Step 6: Final answer.
The x-axis divides the segment in the ratio $5:1$, and the point of intersection is $\left(-\frac{13}{3}, 0\right)$.
\[ \boxed{\text{Ratio } 5:1, \ P\left(-\dfrac{13}{3}, 0\right)} \]