In what ratio does the X-axis divide the line segment

Let the required ratio be λ: 1. So, the coordinates of the point M of division A (- 4, – 6) and B(-1,7) are

$\left\{\frac{\lambda x_{2}+1 \cdot x_{1}}{\lambda+1}, \frac{\lambda y_{2}+1 \cdot y_{1}}{\lambda+1}\right\}$

Here, $x_{1}=-4, x_{2}=-1$ and $y_{1}=-6, y_{2}=7$

i.e., $\quad\left(\frac{\lambda(-1)+1(-4)}{\lambda+1}, \frac{\lambda(7)+1 \cdot(-6)}{\lambda+1}\right)=\left(\frac{-\lambda-4}{\lambda+1}, \frac{7 \lambda-6}{\lambda+1}\right)$

But according to the question, line segment joining $A(-4,-6)$ and $B(-1,7)$ is divided by the $x$-axis. So, $y$-coordinate must be zero.

$\therefore$ $\frac{7 \lambda-6}{\lambda+1} \Rightarrow 7 \lambda-6=0$

$\therefore$ $\lambda=\frac{6}{7}$

So, the required ratio is $6: 7$ and the point of division $M$ is $\left\{\begin{array}{ll}\frac{-\frac{6}{7}-4}{\frac{6}{7}+1}, & \frac{7}{\frac{6}{7}}+1\end{array}\right\}$

i.e.,  $\left(\frac{-34}{\frac{7}{13}}, \frac{6-6}{\frac{13}{7}}\right)$ i.e., $\left(\frac{-34}{13}, 0\right)$

Hence, the required point of division is $\left(\frac{-34}{13}, 0\right)$.