Question

AB is a line segment and line l is its perpendicular bisector. If a point P lies on l, show that P is equidistant from A and B.
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Answer 1

Answer:

Question:-

AB is a line segment and line L is its perpendicular bisector. If a point P lies on L,show that P is equidistant from A and B

▪Solution:-

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since , AB is a line segment,

L is drawn perpendicular to AB,

A point 'p' lies on line L.

To prove : P is equidistant from A and B.

In ∆AOP and ∆BOP ,

OP = OP ( common side )

∠ POA = ∠ POB

AO = OB

\therefore∴ ∆AOP\cong≅ ∆BOP (By S.A.S.)

=> AP = BP

Hence,"P" is equidistant from A and B .

Step-by-step explanation:

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Answer 2

Line l⊥AB and passes through C which is the midpoint of AB

To show that PA=PB

In △PCA and △PCB

AC=BC since C is the mid-point of AB

∠PCA=∠PCB=90∘(given)

PC=PC(common)

So,△PCA≅△PCB by SAS rule.

and so PA=PB as they are corresponding sides of congruent triangles.