If d(-1/5,5/2), e(7,3) and f(7/2,7/2) are mid points of sides of triangle ABC find the coordinate of vertices A,B,C.
Answers
coordinate of vertices A,B,C. A = ( -37/10 , 3) , B = (33/10 , 2) , C = (107/10 , 4)
Step-by-step explanation:
Let say the coordinate of vertices A,B,C.
are (Ax , Ay ) , (Bx , By) & (Cx , Cy) respectively
d - mid point of AB (-1/5 , 5/2)
(Ax + Bx)/2 = - 1/5 & (Ay + By)/2 = 5/2
Ax + Bx = -2/5 & Ay + By = 5
e - mid point of BC (7 , 3)
(Bx + Cx)/2 = 7 & (By + Cy)/2 = 3
Bx + Cx = 14 & By + Cy = 6
f - mid point of AC (7/2 , 7/2)
(Ax + Cx)/2 = 7/2 & (Ay + Cy)/2 = 7/2
Ax + Cx = 7 & Ay + Cy = 7
Cx - Ax = 14 + 2/5 = 72/5
Cx + Ax = 7
2Cx = 107/5
=> Cx = 107/10
2Ax = -37/5
Ax = - 37/10
Bx + Cx = 14
Bx = 14 - 107/10
Bx = 33/10
Cy - Ay = 1
Ay + Cy = 7
2Cy = 8
Cy = 4
Ay = 3
By = 2
A = ( -37/10 , 3)
B = (33/10 , 2)
C = (107/10 , 4)
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