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Using integration, find the area of the triangle whose vertices are (2, 3), (3, 5) and (4, 4).

Answers (1)

Given: The vertices $\Delta\text{ABC}$ A(2, 3), B(3, 5) and C(4, 4).

Equation of Line AB,

$(y-3) = \frac{5-3}{3-2}(x-2)$

$(y-3) = 2(x-2)$

$y = 2x -4 +3$

$\underline{y = 2x -1}$

Equation of Line BC,

$(y-5) = \frac{4-5}{4-3}(x-3)$

$(y-5) = -(x-3)$

$\underline{y = -x +8}$

Equation of Line AC

$(y-3) = \frac{4-3}{4-2}(x-2)$

$(y-3) = \frac{-1}{2}(x-2)$

$y = \frac{x-2}{2} + 3$

$\underline{y = \frac{x +4}{2}}$

$I = \int _2^3 (2x-1)dx + \int^4_3(-x + 8)dx - \int_2^4\frac{x+4}{2}dx$

Area of $\Delta\text{ABC}$

$I = \int _2^3 y_{AB}dx + \int^4_3y_{BC}dx - \int_2^4y_{AC}dx$

$\Rightarrow \left[x^2 \right ]_2^3 - \left[x \right ]_2^3 - \frac{1}{2}[x^2]_3^4 + 8\left[x \right ]_3^4 - \frac{1}{2}\left[\frac{x^2}{2} \right ]_2^4 + 2\left[x \right ]_2^4$

$\Rightarrow [9-4] - [3-2] - \frac{1}{2}[16-9] + 8[4-3] - \frac{1}{4}[16-4] + 2[4-2]$

$I \Rightarrow 5 -1 -\frac{7}{2} + 8 -\frac{12}{4} + 4$

$\Rightarrow 4 -\frac{7}{2} + 8 -3 + 4$

$I = 13-\frac{7}{2} \Rightarrow I = \frac{19}{2} \ \text{sq . units}$

Posted by

Ravindra Pindel

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