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Write an equation for the perpendicular bisector of the line segment joining
Problem 1 :
A(1, 4) and B(6, -6)
Problem 2 :
A(-3, -5) and B(9, -2)
Problem 3 :
A(5, 10) and B(10, 7)
Problem 4 :
For each of the following segments :
(i) Find the coordinates of the midpoint of the segment
(ii) Find the gradient of the segment
(iii) Find the equation of the perpendicular bisector of the segment.

1) equation of perpendicular bisector is 2x - 4y = 11
2) 8x + 2y = 17
3) 10x - 6y = 24
4) i) Midpoint is (3, 5)
ii) slope = 1/5
iii) 5x + y - 20 = 0
Problem 1 :
A (-3, 4) and B (5, 6)
Problem 2 :
A (3, 8) and B (7, 14)
Problem 3 :
A (-5, 6) and B (1, 8)
Problem 4 :
A (-3, -6) and B (-1, 2)
1) y = -4x + 9
2) 2x + 3y = 43
3) y = -3x + 1
4) x + 4y = -10
The ordered pairs represent vertices of a triangle. Write an equation of the line containing the median that joins the first vertex to the side opposite it.
Problem 1 :
(8, 4), (0, 0), (10, 0)
Problem 2 :
(3, 10), (4, 2), (10, 8)
Problem 3 :
(2, 6), (3, 1), (7, 5)
Problem 4 :
A triangle has vertices A(-3, 7), B(4, -5), and C(9, -3). Determine the equation of the median from B.
Problem 5 :
A triangle has vertices Q(6, -4), R(5, 2), and S(1, 4). Determine the equation of the perpendicular bisector of QR.
1) 4x - 3y - 20 = 0
2) 5x + 4y - 55 = 0
3) x + y - 8 = 0
4) equation of median is 7x + y = 23.
5) equation of perpendicular bisector is 10x + 9y - 46 = 0.
Problem 1 :
Find the equation of the perpendicular bisector of P(9, 5) and Q(-1, 3).
Problem 2 :
Given that the triangle QRS have vertices (-3, 1), (-4, -2) and (7, 0).Find the equation of the perpendicular bisector of RS.
Problem 3 :
Pictured to the right is the straight line S1 and the points R and T.
(a) Find the equation of S1.
(b) S2 is perpendicular to S1 and passes through R. Find the equation of S2.
(c) Does S2 pass through T?

Problem 4 :
A = (-3, -2) B = (6, 4) The line C has equation 4y + 6x = 13. Show that C is perpendicular to AB.
1) y = -5x + 24
2) y = (-11/2)x + (29/4)
3) a) y = (-5/2) x + 4
b) 2x - 5y + 24 = 0
c) Since the point T does not satisfy the equation, T does not lie on the line S2.
4) Since the product of these two slopes is equal to -1. These two lines are perpendicular.
5) the value of k is 0.
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
Problem 2 :
A(-9, 11) and (-15,19)
(i) Midpoint
(ii) Slope of AB
(iii) Perpendicular slope
(iv) Equation of the line
Problem 3 :
A(11, -5) and (1, -10)
(i) Midpoint
(ii) Slope of AB
(iii) Perpendicular slope
(iv) Equation of the line
Problem 4 :
A(14, 18) and (-6, 10)
(i) Midpoint
(ii) Slope of AB
(iii) Perpendicular slope
(iv) Equation of the line
Problem 5 :
A triangle has vertices K(-2, -3), L(5, -7) and M(6, 1).
(a) Find the equation of the perpendicular bisector of the line KM.
(b) Show that the point L lies on this line.
(c) What kind of triangle is triangle KLM?
1) i) Midpoint (2, 1)
ii) Slope = 1/2
iii) Perpendicular slope = -1/(1/2) ==> -2
iv) Equation of perpendicular bisector :
y = -2x + 4
2) i) Midpoint (-12, 15)
ii) Slope = -4/3
iii) Perpendicular slope = 3/4
iv) Equation of perpendicular bisector :
y = (3/4)x + 24
3) i) Midpoint (6, -15/2)
ii) Slope = 1/2
iii) Perpendicular slope = -1/(1/2) ==> -2
iv) Equation of perpendicular bisector :
y = -2x + (9/2)
4) i) Midpoint (4, -4)
ii) Slope = -7/5
iii) Perpendicular slope = -1/(-7/5) ==> 5/7
iv) Equation of perpendicular bisector :
5x - 7y - 48 = 0
5) i) Midpoint of KM = (2, -1)
b) Since the point L satisfies the equation, we decide it is on the line.
c) The triangle KLM must be right triangle since it has altitude
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May 21, 24 08:51 PM
May 21, 24 08:51 AM
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