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Vector Algebra

Question
CBSEENMA12032743

Using integration, find the area of the triangle ABC whose vertices are A (3, 0) B (4, 5) and C (5, 1).

Solution

The given vertices are A(3, 0), B(4, 5), C(5, 1).
The equation of AB is
                   straight y minus 0 space equals fraction numerator 5 minus 0 over denominator 4 minus 3 end fraction left parenthesis straight x minus 3 right parenthesis              
or              y = 5x - 5  ...(1)

The equation of BC is
                   straight y minus 5 space equals space fraction numerator 1 minus 5 over denominator 5 minus 4 end fraction left parenthesis straight x minus 4 right parenthesis
or              
or               straight y minus 5 space equals space minus 4 straight x space plus 16
or                  straight y space equals space minus 4 straight x plus 21                                      ...(2)
The equation of CA is
                   straight y minus 1 space equals fraction numerator 0 minus 1 over denominator 3 minus 5 end fraction left parenthesis straight x minus 5 right parenthesis space space space space or space space space space space straight y space minus 1 space equals space 1 half left parenthesis straight x minus 5 right parenthesis
or                  straight y minus 1 space equals space straight x over 2 minus 5 over 2          or     straight y space equals straight x over 2 minus 3 over 2               ...(3)
From B.  C draw BM. CN ⊥s on x-axis
Required area = Area of ∆AMB + area MNCB — area of ∆ANC
        equals space integral subscript 3 superscript 4 left parenthesis 5 straight x minus 15 right parenthesis space dx space plus space integral subscript 4 superscript 5 left parenthesis negative 4 straight x plus 21 right parenthesis space dx space minus space integral subscript 3 superscript 5 open parentheses straight x over 2 minus 3 over 2 close parentheses dx
equals space open square brackets fraction numerator 5 straight x squared over denominator 2 end fraction minus 15 straight x close square brackets subscript 3 superscript 4 plus open square brackets negative 2 straight x squared plus 21 straight x close square brackets subscript 4 superscript 5 space minus space open square brackets straight x squared over 4 minus 3 over 2 straight x close square brackets subscript 3 superscript 5
equals space left parenthesis 40 minus 60 right parenthesis space minus open parentheses 45 over 2 minus 45 close parentheses plus left parenthesis negative 50 plus 105 right parenthesis minus left parenthesis 32 plus 84 right parenthesis
                                                                         negative open parentheses 25 over 4 minus 15 over 2 close parentheses plus open parentheses 9 over 4 minus 9 over 2 close parentheses
equals negative 20 plus 45 over 2 plus 55 minus 52 plus 5 over 4 minus 9 over 4 equals negative 17 plus 45 over 2 minus 1 space equals space 9 over 2 space sq. space units.

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