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

Ans:

We know that the distance between two points (X₁, Y₁) and (X₂, Y₂) is given by:

S = √ [(X₂ – X₁)² + (Y₂ – Y₁)² ]

Now distance between A(7,1) and P (X, Y):

AP = √ (X – 7)² + (Y – 1)² ]

Similarly, distance between P (X, Y) and B(3, 5):

PB = √ (3 – X)² + (5 – Y)² ]

Since P is equidistant from A & B,

∴ AP = PB

∴  (X – 7)² + (Y – 1)²  = (3 – X)² + (5 – Y)² 

∴ X² – 14 X + 49 + Y² – 2 Y + 1 = 9 – 6 X + X² + 25 – 10 Y + Y²

∴ – 14 X  + – 2 Y + 50 = – 6 X + – 10 Y + + 34

∴ – 14 X – 2 Y + 50 = – 6 X – 10 Y + 34

∴ – 14 X + 6 X – 2 Y + 10 Y = 34 – 50

∴ – 8 X + 8 Y = – 16

∴ X – Y = 2

Therefore, relation between X and Y is given by X – Y = 2.

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