PRACTICE PROBLEMS ON PERPENDICULAR BISECTOR AND MEDIAN OF A TRIANGLE

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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)

Solution

Problem 2 :

A(-3, -5) and B(9, -2)

Solution

Problem 3 :

A(5, 10) and B(10, 7)

Solution

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.

equation-of-perpendicular-bisector-q1

Solution

Answer Key

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)

Solution

Problem 2 :

A (3, 8) and B (7, 14)

Solution

Problem 3 :

A (-5, 6) and B (1, 8)

Solution

Problem 4 :

A (-3, -6) and B (-1, 2)

Solution

Answer Key

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)

Solution

Problem 2 :

(3, 10), (4, 2), (10, 8)

Solution

Problem 3 :

(2, 6), (3, 1), (7, 5)

Solution

Problem 4 :

A triangle has vertices A(-3, 7), B(4, -5), and C(9, -3). Determine the equation of the median from B.

Solution

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.

Solution

Answer Key

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).

Solution

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.

Solution

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?

equation-of-perpendicular-line-q2.png

Solution

Problem 4 :

A = (-3, -2) B = (6, 4) The line C has equation 4y + 6x = 13. Show that C is perpendicular to AB.

Solution

Answer Key

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

Solution

Problem 2 :

A(-9, 11) and (-15,19)

(i) Midpoint

(ii) Slope of AB

(iii) Perpendicular slope

(iv) Equation of the line

Solution

Problem 3 :

A(11, -5) and (1, -10)

(i) Midpoint

(ii) Slope of AB

(iii) Perpendicular slope

(iv) Equation of the line

Solution

Problem 4 :

A(14, 18) and (-6, 10)

(i) Midpoint

(ii) Slope of AB

(iii) Perpendicular slope

(iv) Equation of the line

Solution

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?

Solution

Answer Key

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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