Question

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

Answer 1

Given : -

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

Required to find : -

• Value of b ?

Formula used : -

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