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Find the Equation of the Perpendicular Bisector of the Line Segment Joining Points (7, 1) and (3,5).

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Question

Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3,5).

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Solution

TO FIND The equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5)

Let P(xy) 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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Chapter 6: Co-ordinate Geometry - Exercise 6.2 [Page 17]

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R.D. Sharma Mathematics [English] Class 10
Chapter 6 Co-ordinate Geometry
Exercise 6.2 | Q 39 | Page 17
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