Question

Determine a point which divides a line segment of length 7 cm internally in the ratio 3 : 5. Also, justify your construction.

Answer 1

Given: A line segment AB of length 7 cm. A point P divides AB internally in the ratio 3 : 5 (i.e. AP : PB = 3 : 5).

Step-wise calculation:

1. Numerical division (lengths):

Total parts = 3 + 5 = 8

`AP = 3/8 xx 7` cm 

= `21/8` cm 

= 2.625 cm

= `2 5/8` cm

`PB = 5/8 xx 7` cm 

= `35/8` cm 

= 4.375 cm

= `4 3/8` cm

2. Construction (ruler-and-compass / straightedge method):

Step 1: Draw AB = 7 cm.

Step 2: From A draw any ray AX making an acute angle with AB.

Step 3: On AX mark off 8 equal segments A₁, A₂, ..., A₈ (since 3 + 5 = 8).

Step 4: Join B to A₈.

Step 5: Through A₃ draw a line parallel to BA₈ use a set-square or construct a parallel with compass. Let this line meet AB at P.

Then P divides AB internally in the ratio 3 : 5. The standard justification uses the Basic Proportionality Theorem: because A₃P || A₈B, we get `(AP)/(PB) = (A A_3)/(A_3A_8) = 3/5`, so AP : PB = 3 : 5.

The required point P is located on AB at a distance `AP = 21/8` cm (2.625 cm or `2 5/8` cm) from A (so PB = `35/8` cm = 4.375 cm). The construction given above produces this point and is justified by the Basic Proportionality Theorem.