Question

In $\Delta ABC,L,M,N$ are points on $BC,CA,AB$ respectively, dividing them in the ratio $1:2$, $2:3,3:5$. If the point $K$ divides $AB$ in the ratio $5:3$, then $\frac{|\overline{AL}+\overline{BM}+\overline{CN}|}{|\overline{CK}|}=$

• A. $\frac{5}{8}$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. $\frac{1}{15}$

Answer 1

Answer: (D)

We have
In $\Delta ABC,L,M,N$ are points on $BC,CA$ and $AB$ respectively, dividing in the ratio $1:2,2:3$ and $3:5$

Now, $L=\frac{2b+c}{3},M=\frac{2a+3c}{5}$
$N=\frac{3b+5a}{8},K=\frac{5b+3a}{8}$
$AL=\frac{2b+c}{3}-a=\frac{2b+c-3a}{3}$
$BM=\frac{2a+3c}{5}-b=\frac{2a+3c-5b}{5}$
$CN=\frac{3b+5a}{8}-c=\frac{3b+5a-8c}{8}$
$CK=\frac{5b+3a}{8}-c=\frac{5b+3a-8c}{8}$
$\because \left|\frac{AL+BM+CN}{CK}\right|=$
$\left|\frac{\frac{2b+c-3a}{3}+\frac{2a+3c-5b}{5}+\frac{3b+5a-8c}{8}}{\frac{5b+3a-8c}{8}}\right|$
$=\left|\frac{(5b+3a-8c)}{15(5b+3a-8c)}\right|=\frac{1}{15}$