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Vector Algebra

Question
CBSEENMA12032733

Using the method of integration find the area of the region bounded by lines:
2 x + y = 4, 3 x - 2 y = 6 and x - 3 y + 5 = 0.

Solution

The equations of the sides are
2 x + y - 4 = 0    ...(1)
3 x - 2 y - 6 = 0  ...(2)
x - 3 y + 5 = 0    ...(3)
Solving (1) and (2), we get
            fraction numerator straight x over denominator negative 6 minus 8 end fraction equals fraction numerator straight y over denominator negative 12 plus 12 end fraction equals fraction numerator 1 over denominator negative 4 minus 3 end fraction space space space space or space space fraction numerator straight x over denominator negative 14 end fraction equals straight y over 0 equals fraction numerator 1 over denominator negative 7 end fraction
therefore space space space space space space space straight x space equals space fraction numerator negative 14 over denominator negative 7 end fraction space equals space 2 comma space space space space straight y space equals space fraction numerator 0 over denominator negative 7 end fraction space equals space 0
therefore space space space space lines space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis space intersect space in space straight C left parenthesis 2 comma space 0 right parenthesis
Solving (2) and (3), we get,
                                  fraction numerator straight x over denominator negative 10 minus 18 end fraction equals fraction numerator straight y over denominator negative 6 minus 15 end fraction equals fraction numerator 1 over denominator negative 9 plus 2 end fraction space space space or space space fraction numerator straight x over denominator negative 28 end fraction equals fraction numerator straight y over denominator negative 21 end fraction equals fraction numerator 1 over denominator negative 7 end fraction
therefore space space space space space space space space space straight x equals fraction numerator negative 28 over denominator negative 7 end fraction space equals 4 comma space space space space straight y space equals space fraction numerator negative 21 over denominator negative 7 end fraction space equals space 3
therefore space space space space lines space left parenthesis 2 right parenthesis space and space left parenthesis 3 right parenthesis space intersect space in space straight A left parenthesis 4 comma space 3 right parenthesis
Solving (1) and (3), we get,
            fraction numerator straight x over denominator 5 minus 12 end fraction equals fraction numerator straight y over denominator negative 4 minus 10 end fraction equals space fraction numerator 1 over denominator negative 6 minus 1 end fraction  or  fraction numerator straight x over denominator negative 7 end fraction equals fraction numerator straight y over denominator negative 14 end fraction space equals space fraction numerator 1 over denominator negative 7 end fraction
therefore space space space space space space space space straight x space equals space fraction numerator negative 7 over denominator negative 7 end fraction space equals space 1 comma space space space space space straight y space equals space fraction numerator negative 14 over denominator negative 7 end fraction space equals space 2
therefore space space lines space left parenthesis 1 right parenthesis space and space left parenthesis 3 right parenthesis space intersect space in space straight B left parenthesis 1 comma space 2 right parenthesis
From B. draw BL ⊥ x-axis and from A. draw AM ⊥ x-axis.

Required area = Area of increment ABC = Area of region BLMA - area of increment BLC space - area of increment ACM
equals space integral subscript 1 superscript 4 open parentheses fraction numerator straight x plus 5 over denominator 3 end fraction close parentheses dx space minus space integral subscript 1 superscript 2 left parenthesis 4 minus 2 right parenthesis space dx space minus space integral subscript 2 superscript 4 fraction numerator 3 straight x minus 6 over denominator 2 end fraction dx
equals space 1 third integral subscript 1 superscript 4 left parenthesis straight x plus 5 right parenthesis space dx space minus space integral subscript 1 superscript 2 left parenthesis 4 minus 2 straight x right parenthesis space dx space minus 1 half integral subscript 2 superscript 4 left parenthesis 3 straight x minus 6 right parenthesis space dx
equals space 1 third open square brackets straight x squared over 2 plus 5 straight x close square brackets subscript 1 superscript 4 space minus space open square brackets 4 straight x minus straight x squared close square brackets subscript 1 superscript 2 space minus space 1 half open square brackets fraction numerator 3 straight x squared over denominator 2 end fraction minus 6 straight x close square brackets subscript 2 superscript 4
equals space 1 third open square brackets open parentheses 16 over 2 plus 20 close parentheses minus space open parentheses 1 half plus 5 close parentheses close square brackets space minus space left square bracket left parenthesis 8 minus 4 right parenthesis space minus space left parenthesis 4 minus 1 right parenthesis right square bracket minus 1 half left square bracket left parenthesis 24 minus 24 right parenthesis minus left parenthesis 6 minus 12 right parenthesis right square bracket
equals space 1 third open parentheses 28 minus 11 over 2 close parentheses minus left parenthesis 4 minus 3 right parenthesis minus 1 half left parenthesis 0 plus 6 right parenthesis space equals space 1 third cross times 45 over 2 minus 1 minus 3 space equals space 15 over 2 minus 4 space equals 7 over 2 space sq. space units

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