Question

Find a relation between x and y such that the point (x, y) is equidistant from points (7, 1) and (3, 5).

Answer 1

Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).

So, AP = BP

Squaring on both sides, we get

⇒ (AP)² = (BP)²

Using, distance formula, 

Distance between (x₁​, y₁​) and (x₂​, y₂​) 

= \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)    .....(1)

Now,

⇒ (AP)² = (BP)²

⇒ (x − 7)² + (y − 1)² = (x − 3)² + (y − 5)²           [Using eq(1)]

⇒ x² + 49 − 14x + y² + 1 − 2y = x² + 9 − 6x + y² + 25 − 10y

⇒ −14x + 50 − 2y = −6x + 34 − 10y

⇒ −7x + 25 − y = −3x + 17 − 5y

⇒ −4x + 8 + 4y = 0

⇒ 4x − 4y = 8

⇒ 4(x − y) = 8

⇒ x − y = 2

Hence, this is the required relation between x and y.

Answer 2

Given points are A(7, 1) and B(3, 5). It is told that S(x, y) is equidistant from A and B.

So, we get SA = SB,

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

Now,

⇒ SA = SB

⇒ SA² = SB²

⇒ (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²

⇒ x² - 14x + 49 + y² - 2y + 1 = x² - 6x + 9 + y² - 10y + 25

By shifting all variables in R.H.S

x²-x²-6x+14x+y²-y²-10y+2y+9+25-49-1=0

⇒ 8x-8y+34-50=0

8x-8y = 50-34 = 16

⇒ 8x - 8y = 16

⇒ x - y = 2

∴ The relation between x and y is x - y = 2.