**Finding the Vertices with Midpoints of the Triangle :**

Here we are going to see some example problems using the concept of midpoint.

Finding the vertices of the triangle from midpoints short cut

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})

**Question 1 :**

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

**Solution :**

Let (x_{1}, y_{1}) ==> (2, 4)

(x_{2}, y_{2}) ==> (-2, 3) and (x_{3}, y_{3}) ==> (5, 2)

By using the formula given above,

A (x_{1 }+ x_{3 }- x_{2, }y_{1 }+ y_{3 }- y_{2}) ==> [(2+5+2), (4+2-3)]

(9, 3)

B (x_{1 }+ x_{2 }- x_{3, }y_{1 }+ y_{2}_{ }- y_{3}) ==> [(2-2-5), (4+3-2)]

(-5, 5)

C (x_{2 }+ x_{3 }- x_{1, }y_{2 }+ y_{3}_{ }- y_{1}) ==> [(-2+5-2), (3+2-4)]

(1, 1)

Hence the required vertices are (9, 3) (-5, 5) and (1, 1).

**Question 2 :**

O(0,0) is the centre of a circle whose one chord is AB, where the points A and B are (8,6) and (10,0) respectively. OD is the perpendicular from the centre to the chord AB. Find the coordinates of the mid-point of OD.

**Solution :**

Midpoint of the chord AB = D

(x1 + x2)/2, (y1 + y2)/2

= (8 + 10)/2, (6 + 0)/2

= 18/2, 6/2

= (9, 3)

Now we we have to find the midpoint of OD, that is E

O(0, 0) D(9, 3)

= (0 + 9)/2, (0 + 3)/2

= (9/2, 3/2)

**Question 3 :**

The points A(−5, 4), B(−1, −2) and C(5, 2) are the vertices of an isosceles right-angled triangle where the right angle is at B. Find the coordinates of D so that ABCD is a square.

**Solution :**

Since it forms a square,

Midpoint of the diagonal AC and BD are equal.

**Midpoint of AC :**

A (-5, 4) C (5, 2)

= (-5 + 5)/2, (4 + 2)/2

= 0/2, 6/2

= (0, 3)

**Midpoint of BD :**

B (-1, -2) D (a, b)

= (-1 + a)/2, (-2 + b)/2

By equating the x and y coordinates, we get

(-1 + a)/2 = 0 -1 + a = 0 a = 1 |
(-2 + b)/2 = 3 -2 + b = 6 b = 6 + 2 b = 8 |

Hence the required vertex is (1, 8).

After having gone through the stuff given above, we hope that the students would have understood, "Finding the Vertices with Midpoints of the Triangle"

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