Using integration, find the area of the triangle $P Q R$ , whose vertices are at $P (2, 5), Q (4, 7)$ and $R (6, 2)$ .

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Solution

$e q^{n} o f l i n e P Q :$

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

$y_{P Q} = x - 3$

$e q^{n} o f l i n e Q R :$

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

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

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

$2 y = - 5 x + 34$

$y_{Q R} = \frac{- 5 x}{2} + 17$

$e q^{n} o f l i n e P R :$

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

$y - 5 = - x + 2$

$y_{P R} = - x + 3$

$N o w A r e a = \int_{2}^{4} y_{P S} d x + \int_{4}^{6} y_{Q X} d n - \int_{2}^{6} y_{P R} d x$

$\int_{2}^{4} (x + 3) d x + \int_{4}^{6} (\frac{- 5 x}{2} + 17) d n - \int_{2}^{6} (- x + 3) d x$

${(\frac{x^{2}}{2} + 3 x)}_{2}^{4} + {(\frac{- 5 x^{2}}{4} + 17 x)}_{4}^{6} - {(\frac{- x}{2} + 3 x)}_{2}^{6}$

$(8 + 12 - 2 - 6) + (- 45 + 102 + 20 - 68) - (- 18 + 18 + 2 - 6)$

$12 + 9 + 4$

$25 s q u n i t s .$

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