Q18

Asked by sucharitasahoo1 | 26th Nov, 2017, 10:22: AM

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 | 26th Nov, 2017, 04:20: PM