Question

to divide a line segment AB in the ratio 5:7 first AX is drawn ,so that angle BAX is an acute angle and then at equal distance, point Are marked on the ray AX. find the minimum number of these points.

Answer 1

We know that,

no. of points n to be taken on the arm of the acute angl

n= A+B (where A:B is the ratio)

So, the lowest possible value of n is 2 (when A:B=1:1)

Dividing a line in 1:1 means drawing the bisector itself. Well, bisector can be drawn simply by drawing the perpendicular bisector of the line segment. But yeah, it can also be drawn by this method.

So the required number of minimum points is 2.

Hope this helps.

Thank you!!! :)