Question

Find the ratio in which the x-axis divides the line segment joining the points ($-$6, 5) and ($-$4, $-$1). Also, find the point of intersection.

Hint:

For any line segment divided by the x-axis, the ratio $k:1$ is always given by:
\[ k = -\frac{y_1}{y_2} \]
Substituting $y_1 = 5$ and $y_2 = -1$ gives:
\[ k = -\frac{5}{-1} = 5 \]
This is a robust shortcut for calculating division ratios!

Answer 1

Step 1: Find the equation of the line through the two given points.
For $A(-6,5)$ and $B(-4,-1)$, the slope is $m=\dfrac{-1-5}{-4-(-6)}=\dfrac{-6}{2}=-3$. Using point A: $y-5=-3(x+6)$, which simplifies to $y=-3x-13$.
Step 2: Set y = 0 to locate where this line meets the x-axis.
\[ 0=-3x-13 \implies x=-\dfrac{13}{3} \] So the line crosses the x-axis at $\left(-\dfrac{13}{3},0\right)$.
Step 3: Use the x-coordinates to recover the ratio k:1.
By the section formula, $-\dfrac{13}{3}=\dfrac{k(-4)+(-6)}{k+1}$. Cross-multiplying: $-13(k+1)=3(-4k-6) \implies -13k-13=-12k-18 \implies -k=-5 \implies k=5$.
\[ \boxed{5:1,\ \left(-\tfrac{13}{3},\,0\right)} \]