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Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are $A(2,0), B(4,5)$ and $C(6,3)$
Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are $A(2,0), B(4,5)$ and $C(6,3)$
Applications of Integrals | CBSE | Class 12 | Maths | Miscellaneous | NCERT
Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are $A(2,0), B(4,5)$ and $C(6,3)$

Solution:
Vertices of $\triangle A B C$ are $A(2,0), B(4,5)$, and $C(6,3)$.


Eq. of line segment $A B$ is
$\begin{array}{l}
y-0=\frac{5-0}{4-2}(x-2) \\
2 y=5 x-10 \\
y=\frac{5}{2}(x-2)
\end{array}$
Eq. of line segment $B C$ is
$\begin{array}{l}
y-5=\frac{3-5}{6-4}(x-4) \\
2 y-10=-2 x+8 \\
2 y=-2 x+18 \\
y=-x+9
\end{array}$
Eq. of line segment CA is
$\begin{array}{l}
y-3=\frac{0-3}{2-6}(x-6) \\
-4 y+12=-3 x+18 \\
4 y=3 x-6 \\
y=\frac{3}{4}(x-2)
\end{array}$
$\begin{array}{l}
\text { Area }(\triangle \mathrm{ABC})=\text { Area (ABLA) }+\text { Area }(\mathrm{BLMCB})-\text { Area (ACMA) } \\
=\int_{2}^{+} \frac{5}{2}(x-2) d x+\int_{4}(-x+9) d x-\left[\frac{3}{4}(x-2) d x\right. \\
=\frac{5}{2}\left[\frac{x^{2}}{2}-2 x\right]_{2}^{4}+\left[\frac{-x^{2}}{2}+9 x\right]_{4}^{6}-\frac{3}{4}\left[\frac{x^{2}}{2}-2 x\right]_{2}^{6} \\
=\frac{5}{2}[8-8-2+4]+[-18+54+8-36]-\frac{3}{4}[18-12-2+4] \\
=5+8-\frac{3}{4}(8) \\
=13-6 \\
=7 \text {sq. units }
\end{array}$

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