The co-ordinates of the mid-points of sides BC, CA and AB of ∆

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 Multiple Choice QuestionsLong Answer Type

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86. The co-ordinates of the mid-points of sides BC, CA and AB of ∆ ABC are (2, 1, -1), (3, 2, 4) and (1, -2, 3) respectively. Find the co-ordinates of vertices A, B and C.


Let the co-ordinates of the vertices A, B and C respectively be space space space left parenthesis straight x subscript 1 comma space straight y subscript 1 comma space straight z subscript 1 right parenthesis comma space left parenthesis straight x subscript 2 comma space straight y subscript 2 comma space straight z subscript 2 right parenthesis space and space left parenthesis straight x subscript 3 comma space straight y subscript 3 comma space straight z subscript 3 right parenthesis.
                          (2, 1, -1) is mid-point of BC.
     ∴ space space open parentheses fraction numerator straight x subscript 2 plus straight x subscript 3 over denominator 2 end fraction comma space fraction numerator straight y subscript 2 plus straight y subscript 3 over denominator 2 end fraction comma space fraction numerator straight z subscript 2 plus straight z subscript 3 over denominator 2 end fraction close parentheses space left right arrow space left parenthesis 2 comma space 1 comma space minus 1 right parenthesis
rightwards double arrow                                     straight x subscript 2 plus straight x subscript 3 space equals space 4        ...(i)
                                          straight y subscript 2 plus straight y subscript 3 equals 2         ...(ii)
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#6 {main}</pre>     ....(iii)
                                                               
(3, 2, 4) is mid-point of CA.
∴ space space space space open parentheses fraction numerator straight x subscript 3 plus straight x subscript 1 over denominator 2 end fraction comma space fraction numerator straight y subscript 3 plus straight y subscript 1 over denominator 2 end fraction comma space fraction numerator straight z subscript 3 plus straight z subscript 1 over denominator 2 end fraction close parentheses space left right arrow space left parenthesis 3 comma space 2 comma space 4 right parenthesis
rightwards double arrow                               straight x subscript 3 plus straight x subscript 1 space equals space 6           ...(iv)
                                  space space space straight y subscript 3 plus straight y subscript 1 space equals space 4           ...(v)
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#6 {main}</pre>            ...(vi)
(1, -2, 3) is mid-point of AB.
∴ open parentheses fraction numerator straight x subscript 1 plus straight x subscript 2 over denominator 2 end fraction comma space fraction numerator straight y subscript 1 plus straight y subscript 2 over denominator 2 end fraction comma space fraction numerator straight z subscript 1 plus end subscript straight z subscript 2 over denominator 2 end fraction space close parentheses space left right arrow space left parenthesis 1 comma space minus 2 comma space 3 right parenthesis
rightwards double arrow                                    straight x subscript 1 plus straight x subscript 2 equals 2        ...(vii)
                                      space space space straight y subscript 1 plus straight y subscript 2 space equals space minus 4    ...(viii)
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#6 {main}</pre>         ...(ix)      
Adding (i), (iv) and (vii), we get 2 straight x subscript 1 plus 2 straight x subscript 2 plus 2 straight x subscript 3 equals 4 plus 6 plus 2 equals 12
rightwards double arrow                 straight x subscript 1 plus straight x subscript 2 plus straight x subscript 3 equals 6                      ...(x)
Using (i) in (x), we get     space space space straight x subscript 1 space equals space 2           
Using (iv) in (x), we get      straight x subscript 2 equals 0
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#6 {main}</pre>
Adding (ii), (v) and (viii), we get 2 straight y subscript 1 plus 2 straight y subscript 2 plus 2 straight y subscript 3 space equals space 2 plus 4 minus 4 equals 2
rightwards double arrow                       straight y subscript 1 plus straight y subscript 2 plus straight y subscript 3 equals 1                ...(xi)
Using (ii) in (xi), we get     space space straight y subscript 1 equals negative 1
Using (v) in (xi), we get      space straight y subscript 2 equals negative 3
Using (viii) in (xi), we get    space straight y subscript 3 equals 5
Adding (iii), (vi) and (ix), we get 2 straight z subscript 1 plus 2 straight z subscript 2 plus 2 straight z subscript 3 equals negative 2 plus 8 plus 6 equals 12
rightwards double arrow          straight z subscript 1 plus straight z subscript 2 plus straight z subscript 3 space equals space 6                          ...(xii)
Using (iii) in (xii), we get space space space space straight z subscript 1 equals 8
Using (vi) in (xii), we get     straight z subscript 2 equals negative 2
Using (ix) in (xii), we get     straight z subscript 3 equals 0
Hence, the three vertices A, B and C are (2, -1, 8), (0, -3, -2), (4, 5, 0) respectively.

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