In what ratio is the line segment joining A(2, −3) and B(5, 6) divided by the x-axis?

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are

$P=\left(\frac{5 k+2}{k+1}, \frac{6 k-3}{k+1}\right)$

But P lies on the x-axis; so, its ordinate is 0.

Therefore, $\frac{6 k-3}{k+1}=0$

$\Rightarrow 6 k-3=0 \Rightarrow 6 k=3 \Rightarrow k=\frac{3}{6} \Rightarrow k=\frac{1}{2}$

Therefore, the required ratio is $\frac{1}{2}: 1$, which is same as $1: 2$.

Thus, the x-axis divides the line AB in the ratio 1 : 2 at the point P.

Applying $k=\frac{1}{2}$, we get the coordinates of point:

$P\left(\frac{5 k+2}{k+1}, 0\right)$

$=P\left(\frac{5 \times \frac{1}{2}+2}{\frac{1}{2}+1}, 0\right)$

$=P\left(\frac{\frac{5+4}{2}}{\frac{1+2}{2}}, 0\right)$

$=P\left(\frac{9}{3}, 0\right)$

$=P(3,0)$

Hence, the point of intersection of AB and the x-axis is P(3, 0).