If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, - 3) and (3, 4). Find its vertices. PLEASE ANSWER FOR ME!!!!!! :( :(
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Let the coordinates of vertices A, B and C of the ∆ABC be (x1, y1), (x2, y2) and (x3, y3) respectively. Let P (3, 4), Q (4, 1) and R (2, 0) be the mid parts of sides BC, AC and AB respectively
Adding equations (1), (3) and (5)
2 (x1 + x2 + x3) = 18
⇒ x1 + x2 + x3 = 9 … (7)
Subtracting equation (1), (3) and (5) from equation (7), it is obtained
x1 = 9 – 6 = 3
x2 = 9 – 8 = 1
x3 = 9 – 4 = 5
Adding equations (2), (4) and (6)
2 (y1 + y2 + y3) = 10
⇒ y1 + y2 + y3 = 5 … (8)
Subtracting equation (2), (4) and (6) from equation (8), it is obtained
y1 = 5 – 8 = –3
y2 = 5 – 2 = 3
y3 = 5 – 0 = 5
Thus, the vertices of the triangles are (3, –3), (1, 3) and (5, 5).
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