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Question
Using integration, find the area of the triangle whose vertices are (2, 3), (3, 5) and (4, 4).
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Solution 1
The Vertices of ΔABC are A (2, 3), B (3, 5), and C (4, 4)
Equation of line segment AB is
`("y" - 5) = (5-3)/(3-2) ("x" -3)`
=`"y" -5 = 2 ("x" -3)`
=`"y" = 2"x" -1`
Equation of line segment BC is
`("y" - 5) = (5-3)/(3-4) ("x" -3)`
=`"y" -5 = -1 ("x" -3)`
=`"y" = -"x" + 8`
Equation of line segment AC is
`("y" - 4) = (4-3)/(4-2) ("x" -4)`
=`"y" -4 = (1)/(2) ("x" -4)`
= `"y" = ("x")/(2) + 2`
∴ Area of ΔABC = `int_2^3 [(2"x" -1) - ("x"/2+2)] d"x" + int_3^4 [(-"x" + 8) - ("x"/2 + 2)] . d"x"`
= `int_2^3 ((3"x")/2 -3) . d"x" + int_3^4 ((-3"x")/2 + 6) . d"x"`
= `[(3"x"^2)/(4) - 3"x"]_2^3 + [ (-3"x"^2)/4 + 6"x"]_3^4`
= `(27/4 - 9) - (3 -6) + (-12 + 24) - (-27/4 + 18)`
= `(3)/(2) "sq. units"`.
Solution 2
The Vertices of ΔABC are A (2, 3), B (3, 5), and C (4, 4)
Equation of line segment AB is
`("y" - 5) = (5-3)/(3-2) ("x" -3)`
=`"y" -5 = 2 ("x" -3)`
=`"y" = 2"x" -1`
Equation of line segment BC is
`("y" - 5) = (5-3)/(3-4) ("x" -3)`
=`"y" -5 = -1 ("x" -3)`
=`"y" = -"x" + 8`
Equation of line segment AC is
`("y" - 4) = (4-3)/(4-2) ("x" -4)`
=`"y" -4 = (1)/(2) ("x" -4)`
= `"y" = ("x")/(2) + 2`
∴ Area of ΔABC = `int_2^3 [(2"x" -1) - ("x"/2+2)] d"x" + int_3^4 [(-"x" + 8) - ("x"/2 + 2)] . d"x"`
= `int_2^3 ((3"x")/2 -3) . d"x" + int_3^4 ((-3"x")/2 + 6) . d"x"`
= `[(3"x"^2)/(4) - 3"x"]_2^3 + [ (-3"x"^2)/4 + 6"x"]_3^4`
= `(27/4 - 9) - (3 -6) + (-12 + 24) - (-27/4 + 18)`
= `(3)/(2) "sq. units"`.
