**How to Find the Vertices of a Triangle If the Midpoints are Given :**

Here we are going to see, how to find the vertices of a triangle if the midpoints are given.

From the midpoints of the triangle, we may find the vertices in two ways.

(i) Using shortcut

(ii) Using midpoint formula

Now we are going to look into an example problem to show how to find the vertices of a triangle if the midpoints are given.

If (x_{1}, y_{1}) (x_{2}, y_{2}) and (x_{3}, y_{3}) are the mid-points of the sides of
a triangle, we may use the vertices of the triangle by using the formula given below.

A (x_{1 }+ x_{3 }- x_{2, }y_{1 }+ y_{3 }- y_{2})

B (x_{1 }+ x_{2 }- x_{3, }y_{1 }+ y_{2}_{ }- y_{3})

C (x_{2 }+ x_{3 }- x_{1, }y_{2 }+ y_{3}_{ }- y_{1})

**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 :**

x_{1} = 5, y_{1} = 1

x_{2} = 3, y_{2} = -5

x_{3} = -5, y_{3} = -1

Vertex A :

A (x_{1 }+ x_{3 }- x_{2, }y_{1 }+ y_{3 }- y_{2})

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

A (-3, 5)

Vertex B :

B (x_{1 }+ x_{2 }- x_{3, }y_{1 }+ y_{2}_{ }- y_{3})

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

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

B (13, -3)

Vertex C :

C (x_{2 }+ x_{3 }- x_{1, }y_{2 }+ y_{3}_{ }- y_{1})

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

C (3 - 10, -5 - 2)

C (-7, -7)

**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 the given given points be D(5, 1) E(3, -5) and F(-5, -1).

Midpoint of the side AC = D

Midpoint of the side AB = E

Midpoint of the side BC = F

Midpoint = (x_{1} + x_{2})/2 , (y_{1} + y_{2})/2

Midpoint of the side AC = (-5,- 1)

(x_{1} + x_{3})/2 , (y_{1} + y_{3})/2 = (-5,- 1)

By equating x and y coordinates, we get

x x |
y y |

Midpoint of the side AC = (5, 1)

(x_{1} + x_{2})/2 , (y_{1} + y_{2})/2 = (5, 1)

By equating x and y coordinates, we get

x x |
y y |

Midpoint of the side AC = F (3, -5)

(x_{2} + x_{3})/2 , (y_{2} + y_{3})/2 = (3, -5)

By equating x and y coordinates, we get

x x |
y y |

(1) + (3) + (5)

2 (x_{1} + x_{2} + x_{3}) = -10 + 10 + 6

2 (x_{1} + x_{2} + x_{3}) = 6

x_{1} + x_{2} + x_{3} = 3 ----(A)

(1) - (A)

(x_{1} + x_{3})_{ -} (x_{1} + x_{2} + x_{3}) = -10 - 3

-x_{2} = -13 ==> x_{2} = 13

(3) - (A)

(x_{1} + x_{2})_{ -} (x_{1} + x_{2} + x_{3}) = 10 - 3

-x_{3} = 7 ==> x_{3} = -7

(5) - (A)

(x_{2} + x_{3})_{ -} (x_{1} + x_{2} + x_{3}) = -6 - 3

-x_{1} = 3 ==> x_{1} = -3

(2) + (4) + (6)

2 (y_{1} + y_{2} + y_{3}) = -2 + 2 - 10

2 (y_{1} + y_{2} + y_{3}) = -10

y_{1} + y_{2} + y_{3} = -5 ----(B)

(2) - (B)

(y_{1} + y_{3})_{ -} (y_{1} + y_{2} + y_{3}) = -2 + 5

-y_{2} = 3 ==> y_{2} = -3

(4) - (B)

(y_{1} + y_{2})_{ -} (y_{1} + y_{2} + y_{3}) = 2 + 5

-y_{3} = 7 ==> y_{3} = -7

(6) - (C)

(y_{2} + y_{3})_{ -} (y_{1} + y_{2} + y_{3}) = -10 + 5

-y_{1} = -5 ==> y_{1} = 5

Hence the required vertices of triangle are A (-3, 5) B(13, -3) and C (-7, -7).

After having gone through the stuff given above, we hope that the students would have understood, "How to Find the Vertices of a Triangle If the Midpoints are Given"

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