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Question

If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y =0.

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Solution

Let D1 be the distance between (x,y) and (2,1)
D2 be the distance between (x,y) and (1,2)
D1=(x2)2+(y1)2)
D1=(x24x+4+y22y+1)
D1=(x2+y24x2y+5)

D2=(x1)2+(y+2)2)
D2=(x22x+1+y2+4y+4)
D2=(x2+y22x+4y+5)

Here,
D1=D2
(x2+y24x2y+5)=(x2+y22x+4y+5)
4x2y+5=2x+4y+5
2x+6y=0
x+3y=0
Hence proved.


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