Question

A point lies on X-axis at a distance of 9 units from Y-axis. What are its coordinates? What
will be the coordinates of a point, if it lies on Y-axis at a distance of 9 units from X-axis in
negative direction? ​

Answer 1

Given P is equidistant from points A and B

PA=PB .....(1)

and Q is equidistant from points A and B

QA=QB .....(2)

In △PAQ and △PBQ

AP=BP from (1)

AQ=BQ from (2)

PQ=PQ (common)

So, △PAQ≅△PBQ (SSS congruence)

Hence ∠APQ=∠BPQ by CPCT

In △PAC and △PBC

AP=BP from (1)

∠APC=∠BPC from (3)

PC=PC (common)

△PAC≅△PBC (SAS congruence)

∴AC=BC by CPCT

and ∠ACP=∠BCP by CPCT ....(4)

Since, AB is a line segment,

∠ACP+∠BCP=180

∘

Thus, AC=BC and ∠ACP=∠BCP=90

∘

∴,PQ is perpendicular bisector of AB.

Hence proved.

Answer 2

Step-by-step explanation:

A point lies on X-axis at a distance of 9 units from Y-axis. What are its coordinates? What

will be the coordinates of a point, if it lies on Y-axis at a distance of 9 units from X-axis in

negative direction?