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?

To divide a line segment BC internally in the ratio 3 : 5, we draw a r

1.	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?
Answer» ONG>ANSWER: the minimum number of points to be located at equal distances is 3

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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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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?