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Using integration, find the area of triangle ABC, whose vertices are A(2, 5), B(4, 7) and C(6, 2).

Answers (1)

R Ravindra Pindel

Given: A(2,5), B(4,7) and C(6,2)

be the vertices of a triangle

The equation of side AB

$(y - 5) = \frac{7-5}{4-2} (x -2)$

$(y - 5) =(x -2)$

$y = x + 3 \qquad - (1)$

The equation of side BC,

$(y - 7) = \frac{2-7}{6-4}(x - 4)$

$(y - 7) = \frac{-5}{2}(x - 4)$

$2y -14 = -5x + 20$

$2y = -5x + 34$

$y =\frac{1}{2}( -5x + 34) \qquad - (2)$

The equation of side AC,

$(y -5) = \frac{2-5}{6-2}(x -2)$

$(y -5) = \frac{-3}{4}(x -2)$

$4y - 20 = -3x + 6$

$4y = -3x + 26$

$y =\frac{1}{4}( -3x + 26) \qquad -(3)$

$\therefore$ Area of $\Delta ABC$

$= \int_2^4y_{AB}\ dx + \int_4^6y_{BC}\ dx - \int^6_2y_{AC}\ dx$

$= \int_2^4(x + 3)\ dx + \int_4^6\frac{-1}{2}(5x - 34)\ dx - \int^6_2\frac{-1}{4}(3x - 26)\ dx$

$\Rightarrow \left[\frac{x^2}{2} + 3x \right ]_2^4 -\frac{1}{2}\left[\frac{5x^2}{2} - 34x \right ]_4^6 + \frac{1}{4}\left[\frac{3x^2}{2} -26x\right ]_2^6$

$= \left[\left(\frac{16}{2} + 12\right ) - \left(\frac{4}{2} + 6 \right ) \right ] - \frac{1}{2}\left[]\left(\frac{180}{2} - 204 \right ) - \left(\frac{80}{2} - 136 \right ) \right ] + \frac{1}{4} \left[\left (\frac{108}{2} - 156 \right ) - \left (\frac{12}{2} - 52 \right )\right ]$

$\Rightarrow [(8 + 12) - (2+6)] - \frac{1}{2}[(90 - 204) - ( 40 - 136)] + \frac{1}{4}[(54 - 156) - (6-52)]$

$= 12 + \frac{1}{2}(18) - \frac{1}{4} (56) - 12 + 9 -14 = 7 \text{ sq units}$

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