HOW TO FIND THE VERTICES OF A TRIANGLE IF THE MIDPOINTS ARE GIVEN

Example 1 :

The mid-points of the sides of a triangle are (5, 1), (3, -5) and (-5, -1). Find the coordinates of the vertices of the triangle. 

Solution :

Let D, E and F be the mid points of the sides AB, BC and CA of ΔABC.

D(x₁, y₁) = (5, 1)

E(x₂, y₂) = (3, -5)

F(x₃, y₃) = (-5, -1)

Vertex A :

A(x₁ + x₃ - x_2, y₁ + y₃ - y₂)

A(5 - 5 - 3, 1 - 1 - (-5))

A(-3, 5)

Vertex B :

B(x₁ + x₂ - x_3, y₁ + y₂ - y₃)

B(5 + 3 - (-5), 1 - 5 - (-1))

B(8 + 5, 1 - 5 + 1)

B(13, -3)

Vertex C :

C(x₂ + x₃ - x_1, y₂ + y₃ - y₁)

C(3 - 5 - 5, -5 - 1 - 1)

C(3 - 10, -5 - 2)

C(-7, -7)

Example 2 :

The mid-points of the sides of a triangle are (5, 3), (4, 0) and (2, 2). Find the coordinates of the vertices of the triangle. 

Solution :

Let D, E and F be the mid points of the sides AB, BC and CA of ΔABC.

D(x₁, y₁) = (5, 3)

E(x₂, y₂) = (4, 0)

F(x₃, y₃) = (2, 2)

Vertex A :

A(x₁ + x₃ - x_2, y₁ + y₃ - y₂)

A(5 + 2 - 4, 3 + 2 - 0)

A(3, 5)

Vertex B :

B(x₁ + x₂ - x_3, y₁ + y₂ - y₃)

B(5 + 4 - 2, 3 + 0 - 2)

B(7, 1)

Vertex C :

C(x₂ + x₃ - x_1, y₂ + y₃ - y₁)

C(4 + 2 - 5, 0 + 2 - 3)

C(1, -1)

Example 3 :

If 𝐴(4, 2),𝐵(7, 6) and 𝐶(1, 4) are the vertices of a ∆ 𝐴𝐵𝐶 and 𝐴𝐷 is its median, prove that the median 𝐴𝐷 divides ∆ 𝐴𝐵𝐶 into two triangles of equal areas.

Solution :

Here D is the midpoint of the side BC.

Midpoint = (x₁ + x₂)/2, (y₁ + y₂)/2

= (7 + 1)/2, (6 + 4)/2

= 8/2, 10/2

= D(4, 5)

Now we are finding out the area created by the vertices ABD and ACD.

Distance between A and D = height of the triangle

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

A(4, 2) D(4, 5)

= √[(4 - 4)² + (5 - 2)²]

= √[0² + 3²]

= 3

Height of the triangle = 3 units

Since AD is the median, it will bisect the opposite side into two halves.

B(7, 6) D(4, 5)

Distance between BD

= √[(4 - 7)² + (5 - 6)²]

= √[3² + (-1)²]

= √10

Base of the triangle = √10 units

Area of the triangle = (1/2) x base x height

= (1/2) x 3 x  √10

= 1.5 √10 square units

Distance between CD

𝐶(1, 4) and D(4, 5)

= √[(4 - 1)² + (5 - 4)²]

= √[3² + 1²]

= √9 + 1

= √10

Base of the triangle = √10 units

Area of triangle ACD = (1/2) x CD x AD

= 1/2 x √10 x 3

= 1.5 √10 square units.

So, median is dividing the triangle into equal triangles consisting of same area.

Example 4 :

𝐴(4, −6),𝐵(3, −2) and 𝐶(5, 2) are the vertices of a ∆ 𝐴𝐵𝐶 and 𝐴𝐷 is its median. Prove that the median 𝐴𝐷 divides ∆𝐴𝐵𝐶 into two triangles of equal areas.

Solution :

Here D is the midpoint of the side BC.

Midpoint = (x₁ + x₂)/2, (y₁ + y₂)/2

= (3 + 5)/2, (-2 + 2)/2

= 8/2, 0/2

= D(4, 0)

Now we are finding out the area created by the vertices ABD and ACD.

Distance between A and D = height of the triangle

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

A(4, -6) D(4, 0)

= √[(4 - 4)² + (0 + 6)²]

= √[0² + 6²]

= 6

Height of the triangle = 6 units

Since AD is the median, it will bisect the opposite side into two halves.

B(3, -2) D(4, 0)

Distance between BD

= √[(4 - 3)² + (0 + 2)²]

= √[1² + 2²]

= √5

Base of the triangle = √5 units

Area of the triangle = (1/2) x base x height

= (1/2) x 6 x  √5

= 3 √5 square units

Distance between CD

𝐶(5, 2) and D(4, 0)

= √[(4 - 5)² + (0 - 2)²]

= √[1² + (-2)²]

= √1 + 4

= √5

Base of the triangle = √5 units

Area of triangle ACD = (1/2) x CD x AD

= 1/2 x 6 x √5

= 3 √5 square units.

So, median is dividing the triangle into equal triangles consisting of same area.

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