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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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प्रश्न

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

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उत्तर

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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अध्याय 6: Co-ordinate Geometry - Exercise 6.2 [पृष्ठ १७]

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आर.डी. शर्मा Mathematics [English] Class 10
अध्याय 6 Co-ordinate Geometry
Exercise 6.2 | Q 39 | पृष्ठ १७

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