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Line segment XY has endpoints X(5, 7) and Y-3, 3). Find the equation for the perpendicular bisector of line segment XY

Line segment XY has endpoints X(5, 7) and Y-3, 3). Find the equation for the perpendicular-example-1

1 Answer

4 votes

Answer:

Option (1)

Explanation:

Perpendicular bisector of the segment will pass through the midpoint of the segment joining two points (5, 7) and (-3, 3).

Midpoint of the segment will be,

(x, y) =
((x_1+x_2)/(2),(y_1+y_2)/(2))

=
((5-3)/(2),(7+3)/(2))

= (1, 5)

Slope of the line joining the given points
m_1 =
(y_2-y_1)/(x_2-x_1)


m_1=(7-3)/(5+3)

=
(1)/(2)

Let the slope of the line perpendicular to the segment joining the given points is
m_2.

By the property of perpendicular lines,


m_1* m_2=-1


(1)/(2)* m_2=-1


m_2=-2

Since, equation of a line passing through (x', y') and slope 'm' is,

y - y' = m(x - x')

Therefore, equation of the line passing through (1, 5) and slope (-2) will be,

y - 5 = (-2)(x - 1)

y = -2x + 2 + 5

y = -2x + 7

2x + y = 7

Therefore, Option (1) will be the answer.

User Ayoub Omari
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