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

## Answers ( )

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

Requiredtofind: –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 = y₁b = x₂ , 10 = y₂Using 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 + 89Hence,

The distance between the point A , point P is

√b² – 10b + 89Similarly,

Now,

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

Here,

Point B(0,b)

Point P(5,2)

x₁ = 5 , y₁ = 2x₃ = 0 , y₃ = bThis 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 + 29Hence,

Distance between the point P and point B is

√b² – 4b + 29 .Since,

AP = BP

➦√b² – 10b + 89 = √b² – 4b + 29square root get’s cancelled on both sides

➦b² – 10b + 89 = b² – 4b + 29b² , b² get’s cancelled on both sides

➦–10b+89=–4b+29➦–10b+4b=29–89➦–6b=–60Taking – ( minus ) common on both sides

➦–(6b)=–(60)Minus ( – ) get’s cancelled on both sides

➦6b=60➦b=60/6➦b=10Therefore,

Valueofb=10units