Question

Let us find a point on x-axis which is equidistant from the two points (3,5) and (1,3)
Hints: the required point on the x-axis is (x, 0)
(x - 3)2 + (0 - 5)2 = (x - 1)2 + (0 - 3)​

Answer 1

Answer:

Let the point on the x−axis be P(a,0)

Let the other two point be A(−2,5) and B(2,−3)

So by distance formula we have,

Distance between two points =

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

Then,

PA=PB

PA

2

=PB

2

⇒(a+2)

2

+(0−5)

2

=(a−2)

2

+(0+3)

2

⇒a

2

+4+4a+25=a

2

+4−4a+9

⇒4a+29=−4a+13

⇒ 8a=−16⇒a=−2

∴ The required point isP(−2,0)

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Answer 2

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