To divide a line segment BC internally in the ratio 3:5, we draw a ray BX such that ∠CBX is an acute angle. What will be the minimum number of points to be located at equal distances, on ray BX?

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answer is B.

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Detailed Solution

Given a line segment BC, to be divided internally in the ratio 3:5.
Dividing an interval AB internally in a given ratio produces a point between A and B.
∠CBX is an acute angle.
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We will add the ratio to get the total number of points that need to be located at equal distances.
So, the minimum number of points are 8.
Thus, the value of minimum number of points to be located at equal distances on ray BX is 8.
Therefore, option (2) is correct.

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