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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 p(x, y), Q(2, 1), R(1, -2) be the given points
Here `x_1 = x`, `y_1 = y`
`x_2 = 2, y_2 = 1`
The distance between two points
p(x,y) and Q(2, 1) is given by
`PQ = sqrt((2- x)^2 + (1 - y)^2)`
Similarly
Now both these distance are given to be the same
PQ = PR
`sqrt((2- x)^2 + (1 - y)^2) = sqrt((1 - x)^2 + (-2 - y)^2)`
Squaring both the sides
`=> sqrt((2- x)^2 + (1 - y)^2) = sqrt((1 - x)^2 + (-2 - y))`
Squaring both the sides
`=> (2 - x)^2 + (1 - y)^2 = (1 - x)^2 + (-2 - y)^2`
`=> 4 + x^2 - 4x + 1 + y^2 - 2y = 1 + x^2- 2x + 4 + y^2 + 4y`
`=> 4 + x^2 - 4x + 1 + y^2 - 2y -1 - x^2 + 2x - 4 - y^2 - 4y = 0`
`=>-2x - 6y = 0`
`=> -2(x + 3y) = 0`
=> x + 3y = 0
Hence prove
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