Question

If the point A ( 0, 2 ) is equidistant from the points B ( 3, p ) and C( p, 5 ), find the value of p. Also, find the length of AB.

Answer 1

Note: Distance : √(x₂ - x₁)^2 + (y₂ - y₁)^2 (or) D^2 = (x₂ - x₁)^2 + (y₂ - y₁)^2.

Given that A(0,2) is equidistant from B(3,p) and C(p,5).

⇒ AB = BC

On Squaring both sides, we get

⇒ AB^2 = BC^2

⇒ (3 - 0)^2 + (p - 2)^2 = (p - 0)^2 + (5 - 2)^2

⇒ 9 + p^2 + 4 - 4p = p^2 + 9

⇒ 9 + p^2 + 4 - 4p - p^2 - 9 = 0

⇒ 4 - 4p = 0

⇒ 4p = 4

⇒ p = 1.

Now,

Length of AB = √(3 - 0)^2 + (1 - 2)^2

                      = √9 + 1

                      = √10.

Therefore, Length of AB = √10 units.

Hope it helps!

Answer 2

Given that the point A( 0 , 2 ) is equidistant from the points B ( 3, p ) and C( p, 5 ).

$\underleftrightarrow{\;\;B(3,p)\;\;\;\;\;\;\;\;A(0,2)\;\;\;\;\;\;\;\;C(p,5)\;\;}$

As point A ( 0, 2 ) is equidistant from the points B ( 3, p ) and C( p, 5 ), distance between A and B is equal to the distance between B and C.

     

   By Distance formula,

$\boxed{\sf{Distance = \sqrt{(x_{2} - x_{1} )^2 + ( y_{2} - y_{1} )^2}}}$

Where x₁ , x₂ , y₁ , y₂ are the co ordinates of the given points.

In the question, assuming A( 0 , 2 ) as ( x₂ , y₂ ) , B( 3 , p ) as ( x₁ , y₁ ) and C( p , 5 ) as ( x₃ , y₃ ).

 

$\Rightarrow \sqrt{ (x_{3} - x_{2} )^2 + ( y_{3} - y_{2})^2 } = \sqrt{ x_{1} - x_{2})^2 + ( y_{1} - y_{2} )^2 }$

$\Rightarrow \sqrt{(p-0)^2 + ( 5 - 2)^2 } = \sqrt{( 3 - 0)^2 + ( p - 2 )^2 }$

⇒  p^2 + 3^2 = 3^2 + ( p - 2 )^2

⇒ p^2  = ( p - 2 )^2

⇒ p^2 = p^2 + 4 - 4p

⇒  4p = 4

⇒ $\dfrac{4}{4}$ = - p

⇒ 1 = p  

Therefore the value of p is - 1 .

Again by distance formula,

Distance between A and B = $\sqrt{( 3 - 0 )^2 + ( 1 - 2 )^2}$

Distance between A and B = $\sqrt{9 + 1}$

Distance between A and B = √10 units

Therefore the value of p is  1 and distance between A and B is √10 units.