# If the Coordinates of the Mid-points of the Sides of a Triangle Are (1, 1), (2, —3) and (3, 4), Find the Vertices of the Triangle. - Mathematics

If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, —3) and (3, 4), find the vertices of the triangle.

#### Solution

The coordinates of the midpoint (x_m,y_m) between two points (x_1,y_1) and (x_2, y_2) is given by,

(x_m,y_m) = (((x_1 + x_2)/2)"," ((y_1 + y_2)/2))

Let the three vertices of the triangle be A(x_A,y_A),B(x_B, y_B) and C(x_C, y_C)

The three midpoints are given. Let these points be M_(AB) (1,1), M_(BC) (2, -3)and M_(CA) (3,4)

Let us now equate these points using the earlier mentioned formula,

(1,1) = (((x_A + x_B)/2)","((y_A + y_B)/2))

Equating the individual components we get,

x_A + x_B = 2

y_A + y_B = 2

Using the midpoint of another side we have,

(2,-3) = (((x_B + x_C)/2)","((y_B + y_C)/2))

Equating the individual components we get,

x_A + x_C= 6

y_A + y_C= 8

Adding up all the three equations which have variable ‘x’ alone we have,

x_A + x_B + x_B + x_C + x_A + x_C =  2 + 4 + 6

2(x_A + x_B + x_C) = 12

x_A + x_B +x_C = 6

Substituting x_B + x_C = 4 in the above equation we have,

x_A + x_B +  x_C = 6

x_A + 4 = 6

x_A = 2

Therefore,

x_A + x_C= 6

x_C = 6 -  2

x_C = 4

And

x_A + x_B  2

x_b = 2 - 2

x_B = 0

Adding up all the three equations which have variable ‘y’ alone we have,

y_A + y_B + y_B + y_C + y_A + y_C= 2 - 6 + 8

2(y_A + y_B + y_C) = 4

y_A + y_B + y_C = 2

Substituting y_B + y_C = -6 in the above equation we have,

y_A + y_B + y_C = 2

y_A - 6 = 2

y_A = 8

Therefore,

y_A + y_C = 8

y_C = 8 - 8

y_C = 0

And

y_A + y_B = 2

y_B = 2 - 8

y_B = -6

Therefore the co-ordinates of the three vertices of the triangle are A(2,8), B(0, -6), C(4, 0)

Is there an error in this question or solution?
Chapter 6: Co-Ordinate Geometry - Exercise 6.3 [Page 30]

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.3 | Q 43 | Page 30
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