Please explain that any line has only one mid-point .

Show that a line segment has only one mid-point.

Solution:

Let us suppose that a line segment PQ has two mid-points R and S.

Case I: When R is the mid-point, we have, PR = RQ.

We will use one of Euclid’s axioms, according to which, when equals are added to equals, then the wholes are equal.

On adding PR to both sides, we obtain

2PR = PR + RQ

Þ PQ = 2PR … (1)

Case II: When S is the mid-point, we have PS = SQ.

Again, we will use Euclid’s axiom that when equals are added to equals, the wholes are equal.

On adding PS to both sides, we obtain

2PS = PS + SQ

Þ PQ = 2PS … (2)

On comparing (1) and (2), we obtain

2PS = 2PR

PS = PR (Things which are double of the same things are equal to one another.)

This implies that the points S and R lie at the same place on line PQ. Thus, our assumption is wrong. Hence, a line segment can have only one mid-point.

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No This Answer is completely wrong i know the correct one 

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Show that a line segment has only one mid-point.

Solution:

Let us suppose that a line segment PQ has two mid-points R and S.

Case I: When R is the mid-point, we have, PR = RQ.

We will use one of Euclid’s axioms, according to which, when equals are added to equals, then the wholes are equal.

On adding PR to both sides, we obtain

2PR = PR + RQ

 PQ = 2PR … (1)

Case II: When S is the mid-point, we have PS = SQ.

Again, we will use Euclid’s axiom that when equals are added to equals, the wholes are equal.

On adding PS to both sides, we obtain

2PS = PS + SQ

 PQ = 2PS … (2)

On comparing (1) and (2), we obtain

2PS = 2PR

PS = PR (Things which are double of the same things are equal to one another.)

This implies that the points S and R lie at the same place on line PQ. Thus, our assumption is wrong. Hence, a line segment can have only one mid-point.

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