Question
Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B(4, 5) and C(6, 3).
✓
Answer
The vertices of $\Delta\text{ ABC}$ are A(2, 0), B(4, 5), and C(6, 3).
Equation of line segment AB is
$\text{y}-0=\frac{5-0}{4-2}(\text{x}-2)$
$2\text{y}=5\text{x}-10$
$\text{y}=\frac52(\text{x}-2)...(1)$
Equation of line segment BC is
$\text{y}-5=\frac{3-5}{6-4}(\text{x}-4)$
$2\text{y}-10=-2\text{x}+8$
$2\text{y}=-2\text{x}+18$
$\text{y}=-\text{x}+9\dots(2)$
Equation of line segment CA is
$\text{y}-3=\frac{0-3}{2-6}(\text{x}-6)$
$-4\text{y}+12=-3\text{x}+18$
$4\text{y}=3\text{x}-6$
$\text{y}=\frac34(\text{x}-2)\dots(3)$
Area $(\Delta\text{ ABC})$ = Area (ABLA) + Area (BLMCB) - (ACMA)
$=\int\limits^4_2\frac52(\text{x}-2)\text{ dx}+\int\limits^6_4(-\text{x}+9)\text{dx}-\int\limits^6_2\frac34(\text{x}-2)\text{ dx}$
$=\frac52\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^4_2+\Big[\frac{-\text{x}^2}{2}+9\text{x}\Big]^6_2-\frac34\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^6_2$
$=\frac52[8-8-2+4]+[-18+54+8-36]$ $-\frac34[18-12-2+4]$
$=5+8-\frac34(8)$
$=13-6\\=7\text{ units}$
Equation of line segment AB is
$\text{y}-0=\frac{5-0}{4-2}(\text{x}-2)$
$2\text{y}=5\text{x}-10$
$\text{y}=\frac52(\text{x}-2)...(1)$
Equation of line segment BC is
$\text{y}-5=\frac{3-5}{6-4}(\text{x}-4)$
$2\text{y}-10=-2\text{x}+8$
$2\text{y}=-2\text{x}+18$
$\text{y}=-\text{x}+9\dots(2)$
Equation of line segment CA is
$\text{y}-3=\frac{0-3}{2-6}(\text{x}-6)$
$-4\text{y}+12=-3\text{x}+18$
$4\text{y}=3\text{x}-6$
$\text{y}=\frac34(\text{x}-2)\dots(3)$
Area $(\Delta\text{ ABC})$ = Area (ABLA) + Area (BLMCB) - (ACMA)
$=\int\limits^4_2\frac52(\text{x}-2)\text{ dx}+\int\limits^6_4(-\text{x}+9)\text{dx}-\int\limits^6_2\frac34(\text{x}-2)\text{ dx}$
$=\frac52\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^4_2+\Big[\frac{-\text{x}^2}{2}+9\text{x}\Big]^6_2-\frac34\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^6_2$
$=\frac52[8-8-2+4]+[-18+54+8-36]$ $-\frac34[18-12-2+4]$
$=5+8-\frac34(8)$
$=13-6\\=7\text{ units}$
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