M.L Aggarwal book Important Question of class 10 chapter Based on Section Formula for ICSE BOARD.

You have to find the coordinates of the vertices of the triangle the middle points of whose sides are given in this question so solve this question.

This is the Question Number 28, Exercise 11 of M.L Aggarwal.

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# Find the coordinates of the vertices of the triangle the middle points of whose sides are (0, 1/2 ) , (1/2 , 1/2) and ( 1/2 , 0).

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This answer was edited.Let A(x_{1},y_{1}), B(x_{2},y_{2}) and C(x_{3},y_{3}) be the vertices of the triangle ABC.Consider ABBy midpoint formula, (x_{1}+x_{2})/2 = 0x_{1}+x_{2}Â = 0x_{1}Â = -x_{2}Â ..(i)By midpoint formula, (y_{1}+y_{2})/2 = Â½y_{1}+y_{2}Â = 1 â€¦(ii)Consider ACBy midpoint formula, (x_{1}+x_{3})/2 = Â½x_{1}+x_{3}Â = 1 â€¦(iii)By midpoint formula, (y_{1}+y_{3})/2 = 0y_{1}+y_{3}Â = 0y_{1}Â = -y_{3}Â â€¦(iv)Consider BCBy midpoint formula, (x_{2}+x_{3})/2 = Â½x_{2}+x_{3}Â = 1 â€¦(v)By midpoint formula, (y_{2}+y_{3})/2 = Â½y_{2}+y_{3}Â = 1 â€¦(vi)Substitute (i) in (iii)Then (iii) becomes -x_{2}+x_{3}Â = 1Equation (v) x_{2}+x_{3}Â = 1Adding above two equations, we get2x_{3Â }= 2x_{3Â }= 2/2 = 1Substitute x_{3}Â = 1 in (iii), we get x_{1}Â = 0x_{2}Â = 0 [From (i)]So x_{1}Â = 0, x_{2}Â = 0, x_{3}Â = 1Substitute (iv) in (ii)Then (ii) becomes -y_{3}+y_{2}Â = 1Equation (vi) y_{2}+y_{3}Â = 1Adding above two equations, we get2y_{2}Â = 2y_{2}Â = 2/2 = 1Substitute y_{2}Â = 1 in (i), we get y_{1}Â = 0y_{3}Â = 0So y_{1}Â = 0, y_{2}Â = 1, y_{3}Â = 0Hence the Co-ordinates of vertices are A(0,0), B(0,1) and C(1,0).