Question -
Answer -
Given:
The right bisector of a line segment bisects the linesegment at 90°.
End-points of the line segment AB are given as A (3, 4) andB (–1, 2).
Let mid-point of AB be (x, y)
x = (3-1)/2= 2/2 = 1
y = (4+2)/2= 6/2 = 3
(x, y) = (1, 3)
Let the slope of line AB be m1
m1 = (2 – 4)/(-1 – 3)
= -2/(-4)
= 1/2
And let the slope of the line perpendicular to AB be m2
m2 = -1/(1/2)
= -2
The equation of the line passing through (1, 3) and having aslope of –2 is
(y – 3) = -2 (x – 1)
y – 3 = – 2x + 2
2x + y = 5
∴ The required equation of the line is 2x + y = 5