## point P(5,2) is equidistant from the point (b,10) and (0,b) .Find b​

Question

point P(5,2) is equidistant from the point (b,10) and (0,b) .Find b​

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4 weeks 2021-08-21T13:27:06+00:00 1 Answer 1 views 0

1. ## Given : –

Point P(5,2) is equidistant from the point (b,10) and (0,b) .

## Required to find : –

• Value of b ?

## Formulaused : –

The formula to find the distance between any two points is ;

Distance = √( x₂ – x₁ )² + ( y₂ – y₁ )²

## Solution : –

Point P(5,2) is equidistant from the point (b,10) and (0,b) .

We need to find the value of b ?

So,

Let the two points which are equidistant from the 2 point be : A & B

That is ;

A(b,10) & B(0,b)

Now,

From the word equidistant we can conclude that ;

The distance between these points is equal .

i.e.

AP = BP

So,

This implies ;

Let’s try find the distance between AP .

So,

The co – ordinates are :

P(5,2) & A(b,10)

Here,

5 = x₁ , 2 = y₁

b = x₂ , 10 = y₂

Using the formula ;

Distance = √( x₂ – x₁ )² + ( y₂ – y₁ )²

Substitute the values we get ;

AP = √( b – 5)² + ( 10 – 2 )²

AP = √( b² + 25 – 10b ) + ( 8 )²

AP = √b² + 25 – 10b + 64

AP = √b² – 10b + 89

Hence,

The distance between the point A , point P is

√b² – 10b + 89

Similarly,

Now,

Let’s find the distance between point B & point P

Here,

Point B(0,b)

Point P(5,2)

x₁ = 5 , y₁ = 2

x₃ = 0 , y₃ = b

This implies ;

The formula becomes as ;

Distance = √( x₃ – x₁ )² + ( y₃ – y₁ )²

Substituting the values we get :

PB = √( 0 – 5 )² + ( b – 2 )²

PB = √( – 5 )² + ( b² + 4 – 4b )

PB = √25 + b² + 4 – 4b

PB = √b² – 4b + 29

Hence,

Distance between the point P and point B is

√b² – 4b + 29 .

Since,

AP = BP

√b² – 10b + 89 = √b² – 4b + 29

square root get’s cancelled on both sides

b² – 10b + 89 = b² – 4b + 29

b² , b² get’s cancelled on both sides

10b + 89 = 4b + 29

10b + 4b = 29 89

6b = 60

Taking – ( minus ) common on both sides

( 6b ) = ( 60 )

Minus ( – ) get’s cancelled on both sides

6b = 60

b = 60/6

b = 10

Therefore,

Value of b = 10 units