#### Find the ratio in which the point $P\left ( \frac{3}{4},\frac{5}{12} \right )$ divides the line segment joining the points $A\left ( \frac{1}{2},\frac{3}{2} \right )$ and  $B (2, -5)$.

Solution

$P\left ( X,Y \right )= \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Let the point $P\left ( \frac{3}{4},\frac{5}{12} \right )$  divides the line segment joining the points $A\left ( \frac{1}{2},\frac{3}{2} \right )$  and B(2, –5) in the ration k : 1.
(x1, y1) = (1/2, 3/2)                  (x2, y2) = (2, -5)
m1 = k, m2 = 1
Using section formula we have
$\left ( \frac{3}{4},\frac{5}{12} \right )= \left [ \frac{2k+1/2}{k+1} ,\frac{-5k+3/2}{k+1}\right ]$
$\left ( \frac{3}{4},\frac{5}{12} \right )= \left [ \frac{4k+1}{2k+1},\frac{-10k+3}{2k+2} \right ]$
$\frac{4k+1}{2k+1}= \frac{3}{4}$               $\frac{-10k+3}{2k+2}= \frac{5}{12}$

16k + 4 = 6k + 6                 –120k + 36 = 10k + 10
16k – 6k = 6 – 4                 –130k = –26
$k= \frac{2}{10}= \frac{1}{5}$
Therefore the required ratio is 1 : 5
i.e. 1 : 5