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  • Find the ratio in which the point p(3 by 4 , 5 by 12) divides the line segment joining the points A(1 by 2, 3 by 2) and B (2, –5).

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).

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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

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infoexpert27

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