Question

A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.

Answer 1

We know that the distance between the two points (x₁​, y₁​) and (x₂​, y₂​) is

d = `sqrt((x_2​−x_1​)^2+(y_2​−y_1​)^2​)`

Let the given points be A = (a, 7) and B = (−3, a) and the third point given is P(2, −1).

We first find the distance between P(2, −1) and A =(a, 7) as follows:

PA = `sqrt((x_2​−x_1​)^2+(y_2​−y_1​)^2)`

​= `sqrt((a−2)^2+(7−(−1))^2)`

​= `sqrt((a−2)^2+(7+1)^2)`

​= `sqrt((a−2)^2+8^2)​`

= `sqrt((a−2)^2+64)`​ 

Similarly, the distance between P(2,−1) and B = (−3, a) is:

PB = `sqrt((x_2​−x_1​)^2+(y_2​−y_1​)^2)`

= `sqrt((−3−2)^2+(a−(−1))^2)`

​= `sqrt((−5)^2+(a+1)^2)`

​= `sqrt(25+(a+1)^2)​` 

Since the point P(2,−1) is equidistant from the points A(a, 7) and B = (−3, a), therefore, PA = PB that is:

`sqrt((a−2)^2+64)​=sqrt(25+(a+1)^2)​`

⇒ (a − 2)² + 64 = 25 + (a + 1)²

⇒ a² − 4a + 4 + 64 = 25 + a² + 2a + 1

⇒ a² − 4a + 68 =   a² + 2a + 26

⇒ −4a − 2a = 26 − 68

⇒ −6a = −42

⇒ a =`(−42)/(−6)` ​=7

Hence, a = 7.