# To divide a line segment AB in the ratio $5:7$, first a ray AX is drawn so that$\angle BAX$ is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is (A) 8                            (B) 10                          (C) 11                          (D) 12

Solution
Given: $\angle BAX$ is an acute angle.
The required ratio is $5:7$
Let m = 5, n = 7

Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Locate 12 points on AX at equal distances.   (Because here $m+n= 12$)
3. Join $A_{12}B$
4. Through the point $A_{5}$ draw a line parallel to  $A_{12}B$ intersecting AB at the point P.
Then $AP:PB= 5:7$
$\because A_{5}P\parallel A_{12}B$
$\therefore \frac{AA_{5}}{A_{5}A_{12}}= \frac{AP}{PB}$  (By Basic Proportionality theorem)

By construction $\frac{AA_{5}}{A_{5}A_{12}}= \frac{5}{7}$
$\therefore \frac{AP}{PB}= \frac{5}{7}$
Hence the number of points is 12.

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