the equation of perpendicular bisector of a line segment AB is x-y+5=0 if a=(1,-2) then AB is
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perpendicular bisector of AB is x-y+5=0 or y=x+5
its slope = 1
so slope of AB = -1
equation of line with slope -1 is y=-x+c
AB passes through (1,-2)
thus, -2=-1+c
⇒c = -1
Equation of AB : y=-x-1 or x+y+1=0
Intersection of both the lines is the mid point of AB.
x-y+5 = 0
x+y+1 = 0
solving, we the point of intersection as (-3,2)
let point B is (m,n)
(m+1)/2=-3
⇒ m = -7
(n-2)/2=2
⇒ n = 6
Point B is (-7,6)
its slope = 1
so slope of AB = -1
equation of line with slope -1 is y=-x+c
AB passes through (1,-2)
thus, -2=-1+c
⇒c = -1
Equation of AB : y=-x-1 or x+y+1=0
Intersection of both the lines is the mid point of AB.
x-y+5 = 0
x+y+1 = 0
solving, we the point of intersection as (-3,2)
let point B is (m,n)
(m+1)/2=-3
⇒ m = -7
(n-2)/2=2
⇒ n = 6
Point B is (-7,6)
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