Question

To divide a line segment $AB$ internally 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 ray $AX$ such that the minimum number of these points is

• A. $9$
• B. $10$
• C. $11$
• D. $12$

Answer 1

The correct option is D $12$

We know that to divide a line segment $AB$ in the ratio m:n we have to follow the following steps of construction:

1.Draw a line segment $AB$ of a given length by using a ruler.

2. Draw any Ray $AX$ making an acute angle with $AB$.

3. Along $AX$ mark off $(m+n)$ points $A1,A2,A3,\dots .Am,A(m+1),A(m+n)$ such that $AA1=AA2=A(m+n-1)A(m+n)$

4.Join $BA(m+n)$.

5. Through the point Am draw a line parallel to $A(m+n)$ by making an angle equal to $AA(m+n)$ which intersects the line segment $AB$ at point $P$.

The point $P$ so obtained is the required point which divides $AB$ internally in the ratio $m:n$.

Here, $m=5,n=7$

Hence,the minimum number of points on the ray $AX=5+7=12$.

Option D