Question

Using integration find the area of region bounded by the triangle whose vertices are (-1, 0), (1, 3) and (3, 2).

Solution

Let A(-1, 0), B(1, 3), C(3, 2) be the vertices of the given triangle.
The equation of AB is
$straight y minus 0 space equals fraction numerator 3 minus 0 over denominator 1 plus 1 end fraction left parenthesis straight x plus 1 right parenthesis$
or    $straight y equals space 3 over 2 left parenthesis straight x plus 1 right parenthesis$ ...(1)

The equation of BC is
   $straight y minus 3 space equals space fraction numerator 2 minus 3 over denominator 3 minus 1 end fraction left parenthesis straight x minus 1 right parenthesis$
or $straight y minus 3 space equals negative 1 half left parenthesis straight x minus 1 right parenthesis space or space straight y minus 3 space equals space minus 1 half straight x plus 1 half$
or $straight y equals negative 1 half straight x plus 7 over 2$ ...(2)
The equation of CA is
   $straight y minus 2 space equals space fraction numerator 0 minus 2 over denominator negative 1 minus 3 end fraction left parenthesis straight x minus 3 right parenthesis space space or space space straight y space minus space 2 space equals space 1 half left parenthesis straight x minus 3 right parenthesis$
or    $straight y minus 2 space equals space 1 half straight x minus 3 over 2 space space space space or space space space space straight y space equals space 1 half straight x plus 1 half$ ...(3)
From B, draw BM ⊥ x-axis and from C, draw CN ⊥ x-axis.
Required area = Area of ∆ABC
= Area of $increment AMB$ + area of region BMNC - area of $increment ANC$
$equals space integral subscript negative 1 end subscript superscript 1 3 over 2 left parenthesis straight x plus 1 right parenthesis space dx plus integral subscript 1 superscript 3 open parentheses negative 1 half straight x plus 7 over 2 close parentheses dx space minus space integral subscript negative 1 end subscript superscript 3 open parentheses 1 half straight x plus 1 half close parentheses dx equals space 3 over 2 open square brackets straight x squared over 2 plus straight x close square brackets subscript negative 1 end subscript superscript 1 plus open square brackets negative straight x squared over 4 plus 7 over 2 straight x close square brackets subscript 1 superscript 3 space minus space open square brackets straight x squared over 4 plus straight x over 2 close square brackets subscript negative 1 end subscript superscript 3 equals space 3 over 2 open square brackets open parentheses 1 half plus 1 close parentheses minus open parentheses 1 half minus 1 close parentheses close square brackets plus open square brackets open parentheses negative 9 over 4 plus 21 over 2 close parentheses minus open parentheses negative 1 fourth plus 7 over 2 close parentheses close square brackets minus open square brackets open parentheses 9 over 4 plus 3 over 2 close parentheses minus open parentheses 1 fourth minus 1 half close parentheses close square brackets$
$equals space 3 over 2 open parentheses 3 over 2 plus 1 half close parentheses plus open parentheses 33 over 4 minus 13 over 4 close parentheses minus open parentheses 15 over 4 plus 1 fourth close parentheses equals space 9 over 4 plus 3 over 4 plus 33 over 4 minus 13 over 4 minus 15 over 4 minus 1 fourth equals 16 over 4 space equals space 4 space sq. space units.$

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Vector Algebra

Using integration find the area of region bounded by the triangle whose vertices are (-1, 0), (1, 3) and (3, 2).

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