Question
Using intergation, find the area of the bounded by the triangle whose vertices are (-1, 2), (1, 5) and (3, 4).
✓
Answer
Equation of side AB,$\frac{\text{x}+1}{1+1}=\frac{\text{y}-2}{5-2}$
$\Rightarrow\frac{\text{x}+1}{2}=\frac{\text{y}-2}{3}$
$\Rightarrow3\text{x}+3=2\text{y}-4$
$\Rightarrow2\text{y}-3\text{x}=7$
$\text{y}=\frac{3\text{x}+7}{2}\ ...(\text{i})$
Equation of side BC,
$\frac{\text{x}-1}{3-1}=\frac{\text{y}-5}{4-5}$
$\Rightarrow\frac{\text{x}-1}{2}=\frac{\text{y}-5}{1}$
$\Rightarrow-\text{x}+1=2\text{y}-10$
$\Rightarrow2\text{y}=11-\text{x}$
$\text{y}=\frac{11-\text{x}}{2}\ ... (\text{ii})$
Equation of side AC,
$\frac{\text{x}+1}{3+1}=\frac{\text{y}-2}{4-2}$
$\Rightarrow\frac{\text{x}+1}{4}=\frac{\text{y}-2}{2}$
$\Rightarrow\frac{\text{x}+1}{2}=\frac{\text{y}-2}{1}$
$\Rightarrow\text{x}+1=2\text{y}-4$
$\Rightarrow2\text{y}=5+\text{x}$
$\text{y}=\frac{5+\text{x}}{2}\ ...(\text{iii})$
From eq.(i, (ii) and (iii),
$=\int\limits_{-1}^{1}\text{y}_{\text{AB}}\text{ dx} +\int\limits_{1}^{3}\text{y}_{\text{BC}}\text{ dx} +\int\limits_{-1}^{3}\text{y}_{\text{AC}}\text{ dx} $
$=\int\limits_{-1}^{1}\frac{3\text{x}+7}{2}\text{ dx} +\int\limits_{1}^{3}\frac{11-\text{x}}{2}\text{ dx} -\int\limits_{-1}^{3}\frac{5+\text{x}}{2}\text{ dx} $
$=\frac{1}{2}\Big[\frac{3\text{x}^{2}}{2}+7\Big]^{1}_{-1}+\frac{1}{2}\Big[11\text{x}-\frac{\text{x}^{2}}{2}\Big]^{3}_{1}-\frac{1}{2}\Big[5\text{x}+\frac{\text{x}^{2}}{2}\Big]^{3}_{-1}$
$=\frac{1}{2}\Big[\frac{3(1^{2}-1^{2})}{2}+7\text{x}\Big]+\frac{1}{2}\big[11(3-1)-\frac{3^{2}-1^{2}}{2}\Big]$
$=\frac{1}{2}[0+14]+\frac{1}{2}[22-4]-\frac{1}{2}[20+4]$
$=7+\frac{1}{2}\times18-\frac{1}{2}\times24$
$=7+9-12$
$=4\ \text{sq.}\ \text{units}$
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