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  4. In Δ A B C, L , M , N are points on B C, C A, A B respectively, dividing them in the ratio 1: 2, 2: 3,3: 5. If the point K divides A B in the ratio 5: 3, then (| overlineA L+ overlineB M+ overlineC N|/| overlineC K|)=

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Q. In $\Delta A B C, L , M , N$ are points on $B C, C A, A B$ respectively, dividing them in the ratio $1: 2$, $2: 3,3: 5$. If the point $K$ divides $A B$ in the ratio $5: 3$, then $\frac{|\overline{A L}+\overline{B M}+\overline{C N}|}{|\overline{C K}|}=$

AP EAMCETAP EAMCET 2017

Solution:

We have
In $\Delta A B C, L, M, N$ are points on $B C, C A$ and $A B$ respectively, dividing in the ratio $1: 2,2: 3$ and $3: 5$
image
Now, $L=\frac{2 b + c }{3}, \, M=\frac{2 a +3 c }{5}$
$N=\frac{3 b +5 a }{8}, \, K=\frac{5 b +3 a }{8}$
$A L =\frac{2 b + c }{3}- a =\frac{2 b + c -3 a }{3}$
$B M =\frac{2 a +3 c }{5}- b =\frac{2 a +3 c -5 b }{5}$
$C N =\frac{3 b +5 a }{8}- c =\frac{3 b +5 a -8 c }{8}$
$C K =\frac{5 b +3 a }{8}- c =\frac{5 b +3 a -8 c }{8}$
$\because \left|\frac{ A L + B M + C N }{ C K }\right|=$
$\left|\frac{\frac{2 b + c -3 a }{3}+\frac{2 a +3 c -5 b }{5}+\frac{3 b +5 a -8 c }{8}}{\frac{5 b +3 a -8 c }{8}}\right|$
$=\left|\frac{(5 b +3 a -8 c )}{15(5 b +3 a -8 c )}\right|=\frac{1}{15}$