In what ratio is the line segment joining the points A(−2, −3) and B(3, 7) divided by the y-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

$P\left(\frac{3 k-2}{k+1}, \frac{7 k-3}{k+1}\right)$

But P lies on the y-axis; so, its abscissa is 0.

Therefore, $\frac{3 k-2}{k+1}=0$

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

Therefore, the required ratio is $\frac{2}{3}: 1$, which is same as $2: 3$.

Thus, the x-axis divides the line AB in the ratio 2:3 at the point P.

Applying $k=\frac{2}{3}$, we get the coordinates of point $P$ :

$P\left(0, \frac{7 k-3}{k+1}\right)$

$=P\left(0, \frac{7 \times \frac{2}{3}-3}{\frac{2}{3}+1}\right)$

$=P\left(0, \frac{\frac{\frac{14-9}{3}}{\frac{2+3}{3}}}{3}\right)$

$=P\left(0, \frac{5}{5}\right)$

$=P(0,1)$

Hence, the point of intersection of AB and the x-axis is P(0, 1).