Let’s A $(-2,-3)$ and B$(5,6)$ be the given points.

(i) Suppose that x-axis divides AB in the ratio k:$1$ at the point P

Now, the coordinates of the point of division are

$\left[ \frac{5k-2}{k+1},\frac{6k-3}{k+1} \right]$

As we know that, P lies in the x-axis, the y – coordinate is zero.

So,

$6k–3/k+1=0$

$6k–3=0$

$k=1/2$

So, the required ratio is $1:2$

By Using k in the coordinates of P

We have, P$(1/3,0)$

(ii) Suppose that y-axis divides AB in the ratio k:$1$ at point Q

Then, the coordinates of the point of division is given by

$\left[ \frac{5k-2}{k+1},\frac{6k-3}{k+1} \right]$

As we know, Q lies on the y-axis, the x – ordinate is zero.

Therefore

$5k–2/k+1=0$

$5k–2=0$

$k=2/5$

so, the required ratio is $2:5$

By Using k in the coordinates of Q

We get, Q$(0,-3/7)$