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how to find vertices of a triangle if the coordinates of the midpoints are given |
| Answer» Let the vertices of the triangle are A(x1, y1), B(x2, y2), C(x3, y3)D,E AND F are the mid points of sides AB, BC AND ACGiven, D(1,2),E(0,-1)and F(2,-1).Draw DE, DF,\xa0FE and BFAs D and F are mid points of AB and AC∴\xa0DF || BEE and F are mid points of BC and AC∴\xa0EF || BDHence, DBEF is a parallelogramWe know that, the diagonals of a parallelogram bisect each other.That means, both diagonals have same mid - point.Midpoint BF = Midpoint of DE{tex}\\Rightarrow \\left( \\frac { x _ { 2 } - 2 } { 2 } , \\frac { y _ { 2 } + 1 } { 2 } \\right) = \\left( \\frac { 1 + 0 } { 2 } , \\frac { 2 - 1 } { 2 } \\right){/tex}On comparing both sides, we get{tex}\\frac { x _ { 2 } - 2 } { 2 } = \\frac { 1 } { 2 } \\text { and } \\frac { y _ { 2 } + 1 } { 2 } = \\frac { 1 } { 2 }{/tex}{tex}\\Rightarrow{/tex}\xa0x2\xa0- 2 = 1, y2\xa0+ 1\xa0= 1{tex}\\therefore{/tex}\xa0x2\xa0= 3, y2\xa0= 0D is the midpoint of ABD =\xa0{tex}\\left( \\frac { x _ { 1 } + x _ { 2 } } { 2 } , \\frac { y _ { 1 } + y _ { 2 } } { 2 } \\right){/tex}(1, 2) =\xa0{tex}\\left( \\frac { x _ { 1 } + 3 } { 2 } , \\frac { y _ { 1 } + 0 } { 2 } \\right){/tex}{tex}\\Rightarrow{/tex}\xa0x1\xa0= -1 and y1\xa0= 4Now, F is the midpoint of ACF =\xa0{tex}\\left( \\frac { x _ { 1 } + x _ { 3 } } { 2 } , \\frac { y _ { 1 } + y _ { 3 } } { 2 } \\right){/tex}(2, -1) =\xa0{tex}\\left( \\frac { - 1 + x _ { 3 } } { 2 } , \\frac { 4 + y _ { 3 } } { 2 } \\right){/tex}{tex}\\Rightarrow{/tex}\xa0x3\xa0= 5 and y3\xa0= -6The vertices of the triangle are\xa0(1, 2), (3, 0) and (5, - 6) | |