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Vector Algebra

Question
CBSEENMA12032737

Using the method of integration, find the area of the triangle ABC, co-ordinates of whose vertices are A (2, 0), B (4, 5), C (6, 3).

Solution

The equation of AB is
                  straight y minus 0 space space equals space fraction numerator 5 minus 0 over denominator 4 minus 2 end fraction left parenthesis straight x minus 2 right parenthesis
or               straight y space equals space 5 over 2 left parenthesis straight x minus 2 right parenthesis
The equation of BC is
               straight y minus 5 space equals fraction numerator 3 minus 5 over denominator 6 minus 4 end fraction left parenthesis straight x minus 4 right parenthesis
or             straight y minus 5 space equals space minus left parenthesis straight x minus 4 right parenthesis space space or space space space straight y space equals space minus straight x plus 9
The equation of CA is
                 straight y minus 3 space equals space fraction numerator 0 minus 3 over denominator 2 minus 6 end fraction left parenthesis straight x minus 6 right parenthesis space space or space straight y minus 3 space equals space 3 over 4 left parenthesis straight x minus 6 right parenthesis space or space space space straight y space equals space fraction numerator 3 straight x over denominator 4 end fraction minus 3 over 2

From C, draw CD perpendicular x-axis.
Required area = area of quad. ABCD - area of increment space ADC
                        equals space open square brackets integral subscript 2 superscript 4 5 over 2 left parenthesis straight x minus 2 right parenthesis space dx space plus space integral subscript 4 superscript 6 left parenthesis negative straight x plus 9 right parenthesis space dx close square brackets space minus space integral subscript 2 superscript 6 open parentheses fraction numerator 3 straight x over denominator 4 end fraction minus 3 over 2 close parentheses dx
equals space 5 over 2 open square brackets straight x squared over 2 minus 2 straight x close square brackets subscript 2 superscript 4 space plus open square brackets negative straight x squared over 2 plus 9 straight x close square brackets subscript 4 superscript 6 space minus space open square brackets fraction numerator 3 straight x squared over denominator 8 end fraction minus 3 over 2 straight x close square brackets subscript 2 superscript 6
equals space 5 over 2 open square brackets left parenthesis 8 minus 8 right parenthesis minus left parenthesis 2 minus 4 right parenthesis close square brackets plus open square brackets left parenthesis negative 18 plus 54 right parenthesis minus left parenthesis negative 8 plus 36 right parenthesis close square brackets minus open square brackets open parentheses 27 over 2 minus 9 close parentheses minus open parentheses 3 over 2 minus 3 close parentheses close square brackets
equals space 5 over 2 left parenthesis 0 plus 2 right parenthesis space plus space left parenthesis 36 minus 28 right parenthesis space minus open parentheses 9 over 2 plus 3 over 2 close parentheses
equals space 5 plus 8 minus 6 space equals space 7 space sq. space units.

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