Determine a point which divides a line segment of length 12 cm internally in the ratio $2 : 3$. Also, justify your construction.

Given:

A line segment of length 12 cm.

To do:

We have to determine a point which divides a line segment of length 12 cm internally in the ratio $2 : 3$.

Solution:

Steps of construction:

(i) Draw a line segment $AB = 12\ cm$.

(ii) Draw a ray $AX$ at $A$ making an acute angle with $AB$.

(iii) From $B$, draw another ray $BY$ parallel to $AX$.

(iv) Cut off two equal parts from $AX$ and three equal parts from $BY$.

(v) Join $A_2$ and $B_3$ which intersects $AB$ at $P$.

$P$ is the required point which divides $AB$ in the ratio of $2 : 3$ internally. 

Justification:

In $\triangle \mathrm{AA}_{2} \mathrm{P}$ and $\triangle \mathrm{BB}_{3} \mathrm{P}$,

$\angle A_{2} A P=\angle P B B_{3}$                   ($\angle A B Y=\angle B A X$)

$\angle \mathrm{APA}_{2}=\angle \mathrm{BPB}_{3}$         (Vertically opposite angles)

Therefore,  by AA similarity,

$\triangle \mathrm{AA}_{2} \mathrm{P} \sim \Delta \mathrm{BB}_{3} \mathrm{P}$

This implies,

$\frac{A A_{2}}{B B_{3}}=\frac{A P}{B P}$

$\frac{A P}{B P}=\frac{2}{3}$.

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Updated on: 10-Oct-2022

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