Find the ratio in which the point

Let $\mathrm{P}\left(\frac{3}{4}, \frac{5}{12}\right)$ divide $\mathrm{AB}$ internally in the ratio $\mathrm{m}$ :n usina the section formula we get.

$\left(\frac{3}{4}, \frac{5}{12}\right)=\left(\frac{2 m-\frac{n}{2}}{m+n}, \frac{-5 m+\frac{3}{2} n}{m+n}\right)$

$[\because$ internal section formula, the coordinates of point $P$ divides the line segment joining the point $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ in the ratio $m_{1}: m_{2}$ internally is $\left.\left(\frac{m_{2} x_{1}+m_{1} x_{2}}{m_{1}+m_{2}}, \frac{m_{2} y_{1}+m_{1} y_{2}}{m_{1}+m_{2}}\right)\right]$

On equating, we get

$\frac{3}{4}=\frac{2 m-\frac{n}{2}}{m+n} \quad$ and $\quad \frac{5}{12}=\frac{-5 m+\frac{3}{2} n}{m+n}$

$\Rightarrow \quad \frac{3}{4}=\frac{4 m-n}{2(m+n)}$ and $\quad \frac{5}{12}=-\frac{10 m+3 n}{2(m+n)}$

$\Rightarrow \quad \frac{3}{2}=\frac{4 m-n}{m+n} \quad$ and $\quad \frac{5}{6}=\frac{-10 m+3 n}{m+n}$

$\Rightarrow \quad 3 m+3 r i=8 m-2 n$ and $5 m+5 n=-60 m+18 n$

$\Rightarrow \quad 5 n-5 m=0$ and $65 m-13 n=0$

$\Rightarrow \quad n=m$ and $13(5 m-n)=0$

$\Rightarrow \quad n=m$ and $5 m-n=0$

Since, $m=n$ does not satisty.

$\therefore \quad 5 m-n=0$

$\Rightarrow \quad 5 m=n$

$\therefore$ $\frac{m}{n}=\frac{1}{5}$

Hence, the required ratio is 1 : 5.