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Question
Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3,5).
Solution
TO FIND The equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5)
Let P(x, y) be any point on the perpendicular bisector of AB. Then,
PA=PB
`=> sqrt((x - 7)^2 + (y - 1)^2) = sqrt((x - 3)^2 + (y - 5)^2)`
`=> (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2`
`=> x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25`
`=> -14x + 6x + 10y - 2y + 49 + 1 - 9 - 25 = 0`
=-8x + 8y + 16 = 0
=> x - y = 2
Hence the equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5) is x - y = 2
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