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To divide a line segment AB in the ratio...

To divide a line segment AB in the ratio 5:6, draw a ray AX such that `angleBAX` is an acute angle, the draw a ray BY parallel to AX and the points `A_(1),A_(2),A_(3),….." and " B_(1),B_(2),B_(3),…..` are located to equal distances on ray AX and BY, respectively. Then, the points joined are

A

`A_(5) " and " B_(6)`

B

`A_(6) " and " B_(5)`

C

`A_(4) " and " B_(5)`

D

`A_(5) " and " B_(4)`

Text Solution

Verified by Experts

The correct Answer is:
A
(a) Given , a line segment AB and we have to divide it in the ratio 5:6.

Steps of construction
1. Draw a AX making an acute `angleBAX`.
2. Draw a ray BY parallel to AX by making `angleABY` equal to `angleBAX`.
3. Now, locate the points `A_(1),A_(2),A_(3),A_(4) " and " A_(5)`(m=5) on AX and `B_(1),B_(2),B_(3),B_(4),B_(5) " and " B_(6)`s(n=6) such that all the points are at equal distance from each other.
4. Join `B_(6)A_(5)`. Let it intesect AB at a point C.
Then, AC:BC=5:6
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