**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 (X_{1}, Y_{1}) and (X_{2}, Y_{2}) is given by:

S = **√ **[(X_{2} – X_{1})^{2 } + (Y_{2} – Y_{1})^{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 }

∴ X^{2} – 14 X + 49 + Y^{2} – 2 Y + 1 = 9 – 6 X + X^{2} + 25 – 10 Y + Y^{2}

∴ – 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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