Write the ratio in which the line segment doining the points A (3, −6),

Let $\mathrm{P}(x, 0)$ be the point of intersection of $x$-axis with the line segment joining $\mathrm{A}(3,-6)$ and $\mathrm{B}(5,3)$ which divides the line segment $\mathrm{AB}$ in the ratio $\lambda: 1$.

Now according to the section formula if point a point $P$ divides a line segment joining $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ in the ratio $m$ : $n$ internally than,

$\mathrm{P}(x, y)=\left(\frac{n x_{1}+m x_{2}}{m+n}, \frac{n y_{1}+m y_{2}}{m+n}\right)$

Now we will use section formula as,

$(x, 0)=\left(\frac{5 \lambda+3}{\lambda+1}, \frac{3 \lambda-6}{\lambda+1}\right)$

Now equate the y component on both the sides,

$\frac{3 \lambda-6}{\lambda+1}=0$

On further simplification,

$\lambda=\frac{2}{1}$

So x-axis divides AB in the ratio 2:1.