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

 

 

 

 
 
 
 
 

Answers (1)

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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