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Vector Algebra

Question
CBSEENMA12035791

Using integration, find the area of region bounded by the triangle whose vertices are (–2, 1), (0, 4) and (2, 3).

Solution

the vertices of the triangle be A(–2, 1), B(0, 4) and C(2, 3).
Equation of line segment AB is
Equation space of space line space segment space AB space is
left parenthesis straight y minus 1 right parenthesis space equals space fraction numerator left parenthesis 4 minus 1 right parenthesis over denominator left parenthesis 0 plus 2 right parenthesis end fraction left parenthesis straight x plus 2 right parenthesis
rightwards double arrow space straight y space equals space fraction numerator 3 left parenthesis straight x plus 2 right parenthesis over denominator 2 end fraction space plus 1
rightwards double arrow space straight y space equals fraction numerator space 3 straight x space plus space 6 plus 2 over denominator 2 end fraction
rightwards double arrow space straight y space equals space fraction numerator 3 straight x plus 8 over denominator 2 end fraction
Equation space of space line space segment space BC space is
left parenthesis straight y minus 4 right parenthesis space equals space fraction numerator left parenthesis 3 minus 4 right parenthesis over denominator left parenthesis 2 minus 0 right parenthesis end fraction left parenthesis straight X minus 0 right parenthesis
rightwards double arrow space straight Y space equals space minus space 1 half space straight x space plus 4
Equation space of space line space segment space AC space is
left parenthesis straight y minus 3 right parenthesis space equals space fraction numerator left parenthesis 1 minus 3 right parenthesis over denominator left parenthesis negative 2 minus 2 right parenthesis end fraction left parenthesis straight X minus 2 right parenthesis
rightwards double arrow space straight Y space equals space space straight x over 2 space plus 2

AL, BO and CM are drawn perpendicular to x-axis.
It can be observed in the following figure that,
Area (ΔACB) = Area (ABOL) + Area (BOMCB) – Area (ALMCA)
Area space left parenthesis ABOL right parenthesis space equals space integral subscript negative 2 end subscript superscript 0 open parentheses fraction numerator 3 straight x space plus 8 over denominator 2 end fraction close parentheses space dx
space equals space integral subscript negative 2 end subscript superscript 0 space open parentheses fraction numerator 3 straight x over denominator 2 end fraction plus 4 close parentheses space dx
space equals fraction numerator 3 straight x squared over denominator 4 end fraction space plus right enclose 4 straight x end enclose subscript negative 2 end subscript superscript 0
space equals space left parenthesis 0 space plus space 0 right parenthesis minus open parentheses 12 over 4 minus 8 close parentheses
space equals space 5 space sq space units
Area space left parenthesis BOMCB right parenthesis space equals space integral subscript 0 superscript 2 open parentheses 1 half straight x space plus space 4 close parentheses dx
space equals space right enclose space open parentheses negative straight x squared over 4 plus 4 straight x squared close parentheses end enclose subscript 0 superscript 2
space equals space left parenthesis negative 1 plus space 8 right parenthesis minus left parenthesis 0 space plus space 0 right parenthesis
space equals space 7 space sq space units
Area space left parenthesis ALMCA right parenthesis space equals space integral subscript negative 2 end subscript superscript 2 space open parentheses straight x over 2 plus 2 close parentheses dx
space equals space space equals space right enclose space open parentheses straight x squared over 4 plus 2 straight x close parentheses end enclose subscript negative 2 end subscript superscript 2
space equals space left parenthesis 1 plus 4 right parenthesis minus left parenthesis 1 minus 4 right parenthesis
space equals space 8 space sq space units
space Required space area space equals space 4 space units

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