What is the perpendicular bisector equation of the line segment whose end points are at 7 3 and -6 1?

Answer 1

To solve this, four steps are needed:

1. Find the midpoint of the line segment (X, Y) which is a point on the perpendicular bisector
2. Find the slope m for the line segment: m = change_in_y/change_in_x
3. Find the slope m' of the perpendicular line; the slopes of the lines are related by mm' = -1 → m' = -1/m
4. Find the equation of the perpendicular bisector using the slope-point equation for a line passing through point (X, Y) with slope m': y - Y = m'(x - X)

Have a go before reading the solution below.

--------------------------------------------------------------------

The midpoint of (7, 3) and (-6, 1) is at ((7 + -6)/2, (3 + 1)/2) = (1/2, 2)

The slope of the line segment is: m = change_in_y/change_in_x = (1 - 3)/(-6 - 7) = -2/-13 = 2/13

The slope of the perpendicular bisector is m' = -1/m = -1/(2/13) = -13/2

The equation of the perpendicular bisector passing through point (X, Y) = (1/2, 2) with slope m' = -13/2 is given by:

y - Y = m'(x - Y)

→ y - 2 = -13/2(x - 1/2)

→ 4y - 8 = -26x + 13

→ 4y + 26x = 21