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Vector Algebra

Question

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Using the method of integration, find the area of the triangular region whose vertices are (2, -2), (4, 3) and (1, 2).

Solution

Equation space of space line space AB colon negative space straight y plus 2 space equals fraction numerator 2 plus 3 over denominator 2 end fraction left parenthesis straight x minus 2 right parenthesis rightwards double arrow space 2 straight y space equals space 5 straight x minus 14 Equation space of space line space BC colon negative straight y minus 3 space equals 1 half left parenthesis straight x minus 4 right parenthesis rightwards double arrow space 3 straight y space equals space straight x plus 5 Equation space of space line space CA colon negative left parenthesis straight y minus 2 right parenthesis space equals negative 4 left parenthesis straight x minus 1 right parenthesis 4 straight x plus straight y equals 6 therefore space ar space left parenthesis increment ABC right parenthesis equals integral subscript negative 2 end subscript superscript 3 fraction numerator space 2 straight y space plus 14 over denominator 5 end fraction space dy minus integral subscript 2 superscript 3 3 straight y minus 5 dy space minus integral subscript negative 2 end subscript superscript 2 fraction numerator 6 minus straight y over denominator 4 space end fraction dy equals 75 over 5 minus 5 over 2 minus 24 over 4 equals fraction numerator 300 minus 120 minus 50 over denominator 20 end fraction space equals space 130 over 20 space equals 13 over 2 space sq. space units

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