Chapter 7 - Applications of Integration - 7.1 Exercises - Page 443: 55

$A = 14$

$$\eqalign{ & {\text{Let the points }}A\left( {2, - 3} \right),{\text{ }}B\left( {4,6} \right){\text{ and }}C\left( {6,1} \right) \cr & {\text{*Find the equation of the line between the points }}A{\text{ and }}B \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y + 3 = \frac{{6 + 3}}{{4 - 2}}\left( {x - 2} \right) \cr & y + 3 = \frac{9}{2}\left( {x - 2} \right) \cr & y = \frac{9}{2}x - 12 \cr & \cr & {\text{*Find the equation of the line between the points }}A{\text{ and }}C \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y + 3 = \frac{{1 + 3}}{{6 - 2}}\left( {x - 2} \right) \cr & y + 3 = \left( {x - 2} \right) \cr & y = x - 5 \cr & \cr & {\text{*Find the equation of the line between the points }}B{\text{ and }}C \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 6 = \frac{{1 - 6}}{{6 - 4}}\left( {x - 4} \right) \cr & y - 6 = - \frac{5}{2}\left( {x - 4} \right) \cr & y = - \frac{5}{2}x + 16 \cr & \cr & {\text{From the image shown below}}{\text{, we can define the area as}} \cr & A = \int_2^4 {\left( {\frac{9}{2}x - 12 - x + 5} \right)dx + \int_4^6 {\left( { - \frac{5}{2}x + 16 - x + 5} \right)dx} } \cr & A = \int_2^4 {\left( {\frac{7}{2}x - 7} \right)dx + \int_4^6 {\left( { - \frac{7}{2}x + 21} \right)dx} } \cr & {\text{Integrate and evaluate}} \cr & A = \left[ {\frac{7}{4}{x^2} - 7x} \right]_2^4 + \left[ { - \frac{7}{4}{x^2} + 21x} \right]_4^6 \cr & A = \left[ {\frac{7}{4}{{\left( 4 \right)}^2} - 7\left( 4 \right)} \right] - \left[ {\frac{7}{4}{{\left( 2 \right)}^2} - 7\left( 2 \right)} \right] + \left[ { - \frac{7}{4}{{\left( 6 \right)}^2} + 21\left( 6 \right)} \right] \cr & - \left[ { - \frac{7}{4}{{\left( 4 \right)}^2} + 21\left( 4 \right)} \right] \cr & {\text{Simplifying}} \cr & A = 0 + 7 + 63 - 56 \cr & A = 14 \cr} $$