Question
Using integration, find the area of the region bounded by the triangle whose vertices are (2, –2), (4, 5) and (6, 2).
✓
Answer
yet A = (2 - 2)
B = (4, 5)
C = (6, 2)
We have to find the area of $\triangle\text{ABC}$
Equation of line AB $\text{y}-5=\Big(\frac{-2-5}{2-4}\Big)(\text{x}-4)$
$\text{y}-5=\frac72(\text{x}-4)$
$2\text{y}-10=7\text{x}-28$
$\text{7x}-2\text{y}=-10+28$
$\text{7x}-2\text{y}=18\dots(\text{i})$
Equation of line BC $\text{y}-2=\Big(\frac{5-2}{4-6}\Big)(\text{x}-6)$
$-2\text{y}+4=3\text{x}-18$
$4+18=3\text{x}+\text{2y}$
$\text{3x}+\text{2y}=22\dots{\text{(ii)}}$
Equation of line AC $\text{y}-2=\frac{-2-2}{2-6}(\text{x}-6)$
$\text{y}-2=\frac{-4}{-4}(\text{x}-6)$
$\text{y}-2=\text{x}-6$
$\text{x}-\text{y}=-2+6$
$\text{x}-\text{y}=4\dots(\text{iii})$
So the required are:
$\text{ar}(\triangle\text{ABC})=\int_\limits{2}^{2} (\text{y}+4)\text{dy}+\Big(\frac{-2}{3}\Big)\\\int_\limits{2}^{5}(\text{y}-11)\text{dy}-\int_\limits{-2}^{5}\frac27(\text{y}+9)\text{dy}$
$=\frac12\Big[(\text{y}+4)^2\Big]^2_{-2}-\frac13\Big[(\text{y}-11)^2\Big]^5_{2}-\frac17\Big[(\text{y}+9)^2\Big]^5_{-2}$
$=16+15-21=10.$
B = (4, 5)
C = (6, 2)
We have to find the area of $\triangle\text{ABC}$
Equation of line AB $\text{y}-5=\Big(\frac{-2-5}{2-4}\Big)(\text{x}-4)$
$\text{y}-5=\frac72(\text{x}-4)$
$2\text{y}-10=7\text{x}-28$
$\text{7x}-2\text{y}=-10+28$
$\text{7x}-2\text{y}=18\dots(\text{i})$
Equation of line BC $\text{y}-2=\Big(\frac{5-2}{4-6}\Big)(\text{x}-6)$
$-2\text{y}+4=3\text{x}-18$
$4+18=3\text{x}+\text{2y}$
$\text{3x}+\text{2y}=22\dots{\text{(ii)}}$
Equation of line AC $\text{y}-2=\frac{-2-2}{2-6}(\text{x}-6)$
$\text{y}-2=\frac{-4}{-4}(\text{x}-6)$
$\text{y}-2=\text{x}-6$
$\text{x}-\text{y}=-2+6$
$\text{x}-\text{y}=4\dots(\text{iii})$
So the required are:
$\text{ar}(\triangle\text{ABC})=\int_\limits{2}^{2} (\text{y}+4)\text{dy}+\Big(\frac{-2}{3}\Big)\\\int_\limits{2}^{5}(\text{y}-11)\text{dy}-\int_\limits{-2}^{5}\frac27(\text{y}+9)\text{dy}$
$=\frac12\Big[(\text{y}+4)^2\Big]^2_{-2}-\frac13\Big[(\text{y}-11)^2\Big]^5_{2}-\frac17\Big[(\text{y}+9)^2\Big]^5_{-2}$
$=16+15-21=10.$
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