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:

Always assume the ratio as \(k:1\) instead of \(m:n\). It reduces the number of variables and makes the algebra much simpler.

Answer 1

Given:
The line segment joins the points \[ A(-6,\,5) \quad \text{and} \quad B(-4,\,-1) \] The x-axis divides this segment.
The point on x-axis will be of the form \((x,\,0)\).

Let the x-axis divide AB in the ratio \[ m:n \] where A corresponds to m and B corresponds to n.

Using the section formula for the y-coordinate:
For internal division, \[ 0 = \frac{m(-1) + n(5)}{m + n} \]

Step 1: Solve the equation
\[ m(-1) + 5n = 0 \] \[ -m + 5n = 0 \] \[ m = 5n \]

Thus, the ratio is:
\[ m:n = 5:1 \] So, the x-axis divides the line segment in the ratio 5 : 1.

Step 2: Find the point of intersection
Using section formula for x-coordinate: \[ x = \frac{m x_B + n x_A}{m + n} \] Substituting: \[ x = \frac{5(-4) + 1(-6)}{5 + 1} \] \[ = \frac{-20 - 6}{6} \] \[ = \frac{-26}{6} \] \[ = -\frac{13}{3} \]

Therefore, the point of intersection is:
\[ \left(-\frac{13}{3},\,0\right) \]

Final Answers:
Ratio = 5 : 1
Point of intersection = \[ \boxed{\left(-\frac{13}{3},\ 0\right)} \]