1. **Problem Statement:** Find the ratio in which the line segment joining points $A(-2,-3)$ and $B(5,6)$ is divided by (i) the x-axis and (ii) the y-axis. Also, find the coordinates of the points of division in each case. 2. **Formula Used:** If a point $P(x,y)$ divides the line segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ratio $m:n$, then $$x=\frac{mx_2+nx_1}{m+n}, \quad y=\frac{my_2+ny_1}{m+n}$$ 3. **Important Rules:** - The x-axis has equation $y=0$. - The y-axis has equation $x=0$. - We use these to find the ratio by substituting $y=0$ or $x=0$ in the section formula and solving for $m:n$. --- **(i) Division by the x-axis ($y=0$):** 4. Let the ratio be $m:n$ and the point of division be $P(x,0)$. 5. Using the section formula for $y$ coordinate: $$0=\frac{m\times 6 + n\times (-3)}{m+n} = \frac{6m - 3n}{m+n}$$ 6. Multiply both sides by $m+n$: $$6m - 3n = 0 \implies 6m = 3n \implies 2m = n$$ 7. So, the ratio is $m:n = 1:2$. 8. Find $x$ coordinate of $P$: $$x=\frac{m\times 5 + n\times (-2)}{m+n} = \frac{1\times 5 + 2\times (-2)}{1+2} = \frac{5 - 4}{3} = \frac{1}{3}$$ 9. Coordinates of point of division on x-axis are $\left(\frac{1}{3}, 0\right)$. --- **(ii) Division by the y-axis ($x=0$):** 10. Let the ratio be $m:n$ and the point of division be $Q(0,y)$. 11. Using the section formula for $x$ coordinate: $$0=\frac{m\times 5 + n\times (-2)}{m+n} = \frac{5m - 2n}{m+n}$$ 12. Multiply both sides by $m+n$: $$5m - 2n = 0 \implies 5m = 2n \implies \frac{m}{n} = \frac{2}{5}$$ 13. So, the ratio is $m:n = 2:5$. 14. Find $y$ coordinate of $Q$: $$y=\frac{m\times 6 + n\times (-3)}{m+n} = \frac{2\times 6 + 5\times (-3)}{2+5} = \frac{12 - 15}{7} = \frac{-3}{7}$$ 15. Coordinates of point of division on y-axis are $\left(0, -\frac{3}{7}\right)$. --- **Final answers:** - Ratio on x-axis: $1:2$, point $\left(\frac{1}{3}, 0\right)$. - Ratio on y-axis: $2:5$, point $\left(0, -\frac{3}{7}\right)$.