PLease help me out! 

Asked by akshitaasia | 20th Mar, 2018, 02:58: PM

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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Answered by Sneha shidid | 21st Mar, 2018, 09:37: AM