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CBSE Class 10 Answered

Q18
Asked by sucharitasahoo1 | 26 Nov, 2017, 10:22: AM
answered-by-expert Expert Answer
begin mathsize 16px style using space mid space point space theorem let space left parenthesis straight x subscript 1 comma straight y subscript 1 right parenthesis comma space left parenthesis straight x subscript 2 comma straight y subscript 2 right parenthesis comma space left parenthesis straight x subscript 3 comma straight y subscript 3 right parenthesis space be space the space vertices space of space triangle let space left parenthesis 12 comma 5 right parenthesis space mid space point space of space side space having space end space points space left parenthesis straight x subscript 1 comma straight y subscript 1 right parenthesis space & space left parenthesis straight x subscript 2 comma straight y subscript 2 right parenthesis so space by space mid space point space theorem 12 equals fraction numerator straight x subscript 1 plus straight x subscript 2 over denominator 2 end fraction comma space space space 5 equals fraction numerator straight y subscript 1 plus straight y subscript 2 over denominator 2 end fraction similarly 8 equals fraction numerator straight x subscript 2 plus straight x subscript 3 over denominator 2 end fraction comma space space space 4 equals fraction numerator straight y subscript 2 plus straight y subscript 3 over denominator 2 end fraction 7 equals fraction numerator straight x subscript 3 plus straight x subscript 1 over denominator 2 end fraction comma space space space 8 equals fraction numerator straight y subscript 1 plus straight y subscript 3 over denominator 2 end fraction solve space these space simultaneously end style
Answered by Arun | 26 Nov, 2017, 04:20: PM

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CBSE 10 - Maths
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