A perpendicular bisector is a line that bisects (cuts in half) another line and it is at right angles to the line.
Let A and B be the endpoint of the line segment. To find the equation of the perpendicular bisector, we follow the steps given below.
(i) Find the midpoint of the line segment which has endpoints A and B.
(ii) Slope of the perpendicular line = -1/Slope of the line AB
(iii) To find equation, we use
y - y1 = m(x - x1)
Here (x1, y1) is the midpoint and m is the slope.
Write the equation of the line which is perpendicular bisector to each pair of points.
Problem 1 :
A(-4, -2) and (8, 4)
(i) Midpoint
(ii) Slope of AB
(iii) Perpendicular slope
(iv) Equation of the line
Solution :
A(-4, -2) and (8, 4)
(i) Midpoint = (x1 + x2)/2, (y1 + y2)/2
(x1, y1) ==> (-4, -2) and (x2, y2) ==> (8, 4)
= (-4 + 8)/2, (-2 + 4)/2
= 4/2, 2/2
= (2, 1)
(ii) Slope of AB = (y2 - y1)/(x2 - x1)
= (4 - (-2))/ (8 - (-4))
= (4 + 2) / (8 + 4)
= 6/12
= 1/2
(iii) Perpendicular slope = -1/(1/2) ==> -2
(iv) Equation of the perpendicular bisector :
(y - y1) = m(x - x1)
(y - 2) = -2(x - 1)
y - 2 = -2x + 2
y = -2x + 2 + 2
y = -2x + 4
Problem 2 :
A(-9, 11) and (-15,19)
(i) Midpoint
(ii) Slope of AB
(iii) Perpendicular slope
(iv) Equation of the line
Solution :
A(-9, 11) and (-15,19)
(i) Midpoint = (x1 + x2)/2, (y1 + y2)/2
(x1, y1) ==> (-9, 11) and (x2, y2) ==> (-15, 19)
= (-9 - 15)/2, (11 + 19)/2
= -24/2, 30/2
= (-12, 15)
(ii) Slope of AB = (y2 - y1)/(x2 - x1)
= (19 - 11)/ (-15 + 9)
= -8/6
= -4/3
(iii) Perpendicular slope = -1/(-4/3) ==> 3/4
(iv) Equation of the perpendicular bisector :
(y - y1) = m(x - x1)
(y - 15) = (3/4)(x - (-12))
4(y - 15) = 3(x + 12)
4y - 60 = 3x + 36
4y = 3x + 36 + 60
4y = 3x + 96
y = (3/4)x + 24
Problem 3 :
A(11, -5) and (1, -10)
(i) Midpoint
(ii) Slope of AB
(iii) Perpendicular slope
(iv) Equation of the line
Solution :
A(11, -5) and (1, -10)
(i) Midpoint = (x1 + x2)/2, (y1 + y2)/2
(x1, y1) ==> (11, -5) and (x2, y2) ==> (1, -10)
= (11+1)/2, (-5-10)/2
= 12/2, -15/2
= (6, -15/2)
(ii) Slope of AB = (y2 - y1)/(x2 - x1)
= (-10+5)/ (1-11)
= -5 / -10
= 1/2
(iii) Perpendicular slope = -1/(1/2) ==> -2
(iv) Equation of the perpendicular bisector :
(y - y1) = m(x - x1)
y + (15/2) = -2(x - 6)
(2y + 15) = -4(x - 6)
2y + 15 = -4x + 24
2y = -4x + 24 - 15
2y = -4x + 9
y = -2x + (9/2)
Problem 4 :
A(14, 18) and (-6, 10)
(i) Midpoint
(ii) Slope of AB
(iii) Perpendicular slope
(iv) Equation of the line
Solution :
A(14, -18) and (-6, 10)
(i) Midpoint = (x1 + x2)/2, (y1 + y2)/2
(x1, y1) ==> (14, -18) and (x2, y2) ==> (-6, 10)
= (14 - 6)/2, (-18 + 10)/2
= 8/2, -8/2
= (4, -4)
(ii) Slope of AB = (y2 - y1)/(x2 - x1)
= (10+18)/ (-6-14)
= 28 / -20
= -7/5
(iii) Perpendicular slope = -1/(-7/5) ==> 5/7
(iv) Equation of the perpendicular bisector :
(y - y1) = m(x - x1)
(y - (-4)) = (5/7)(x - 4)
7(y + 4) = 5(x - 4)
7y + 28 = 5x - 20
5x - 7y - 20 - 28 = 0
5x - 7y - 48 = 0
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM