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CBSE Class 12-science Answered

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Asked by akshitaasia | 20 Mar, 2018, 02:58: PM
answered-by-expert Expert Answer
begin mathsize 16px style Find space equation space of space all space the space three space lines space using space two space slope space point space form.
Let space straight A left parenthesis negative 1 comma space 2 right parenthesis comma space straight B left parenthesis 1 comma space 5 right parenthesis space and space straight C left parenthesis 3 comma space 4 right parenthesis
Equation space of space AB
fraction numerator straight y minus straight y subscript 1 over denominator straight x minus straight x subscript 1 end fraction equals fraction numerator straight y subscript 2 minus straight y subscript 1 over denominator straight x subscript 2 minus straight x subscript 1 end fraction
fraction numerator straight y minus 2 over denominator straight x minus open parentheses negative 1 close parentheses end fraction equals fraction numerator 5 minus 2 over denominator 1 minus open parentheses negative 1 close parentheses end fraction
fraction numerator straight y minus 2 over denominator straight x plus 1 end fraction equals 3 over 2
2 straight y minus 4 equals 3 straight x plus 3
3 straight x minus 2 straight y plus 7 equals 0
straight y equals fraction numerator 3 straight x plus 7 over denominator 2 end fraction......... left parenthesis straight i right parenthesis
Similarly space you space can space find space equations space of space BC space and space AC.
you space will space get
straight y equals fraction numerator 11 minus straight x over denominator 2 end fraction space is space equation space of space BC......... left parenthesis ii right parenthesis
straight y equals fraction numerator straight x plus 5 over denominator 2 end fraction space is space equation space of space AC......... left parenthesis iii right parenthesis
Required space area equals integral subscript negative 1 end subscript superscript 1 space open parentheses fraction numerator 3 straight x plus 7 over denominator 2 end fraction close parentheses dx plus integral subscript 1 superscript 3 space open parentheses fraction numerator 11 minus straight x over denominator 2 end fraction close parentheses dx minus integral subscript negative 1 end subscript superscript 3 space open parentheses fraction numerator straight x plus 5 over denominator 2 end fraction close parentheses dx
After space simplifying space we space get space 4 space square space unit space area.
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