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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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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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Let A(x1,y1), B(x2,y2) and C(x3,y3) be the vertices of the triangle ABC.

Consider AB

By midpoint formula, (x1+x2)/2 = 0

x1+x2Â = 0

x1Â = -x2Â ..(i)

By midpoint formula, (y1+y2)/2 = Â½

y1+y2Â = 1 â€¦(ii)

Consider AC

By midpoint formula, (x1+x3)/2 = Â½

x1+x3Â = 1 â€¦(iii)

By midpoint formula, (y1+y3)/2 = 0

y1+y3Â = 0

y1Â = -y3Â â€¦(iv)

Consider BC

By midpoint formula, (x2+x3)/2 = Â½

x2+x3Â = 1 â€¦(v)

By midpoint formula, (y2+y3)/2 = Â½

y2+y3Â = 1 â€¦(vi)

Substitute (i) in (iii)

Then (iii) becomes -x2+x3Â = 1

Equation (v) x2+x3Â = 1

Adding above two equations, we get

2x3Â = 2

x3Â = 2/2 = 1

Substitute x3Â = 1 in (iii), we get x1Â = 0

x2Â = 0 [From (i)]

So x1Â = 0, x2Â = 0, x3Â = 1

Substitute (iv) in (ii)

Then (ii) becomes -y3+y2Â = 1

Equation (vi) y2+y3Â = 1

Adding above two equations, we get

2y2Â = 2

y2Â = 2/2 = 1

Substitute y2Â = 1 in (i), we get y1Â = 0

y3Â = 0

So y1Â = 0, y2Â = 1, y3Â = 0

Hence the Co-ordinates of vertices are A(0,0), B(0,1) and C(1,0).

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