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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).

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Toolbox:
• Suppose three lines intersect at three different points the enclosed area will be the area bounded by the three lines at their points of intersection.
• Equation of a line when two points $(x_1,y_1)$ and $(x_2,y_2)$ are given is$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
Given the vertices of the $\bigtriangleup ABC$ are A(2,0),B(4,5) and C(6,3).
By plotting these points on the graph ,we find the required area is the shaded portion ABC.
To find the equation of the lines,from the given points.(use the information in the toolbox.
segment AB is$\frac{y-0}{5-0}=\frac{x-2}{6-2}$
$\Rightarrow y=\frac{5}{2}(x-2)$-----(1)
similarly segment BC is $\frac{y-5}{3-5}=\frac{x-4}{6-4}$
$\Rightarrow y-5=\frac{-2}{2}(x-4)$
$\;\;\;\;\;\;\;\;\;\;y=-(x-4)+5$
$\;\;\;\;\;\;\;\;\;\;y=-x+9$-----(2)
segment CA is $\frac{y-3}{0-3}=\frac{x-6}{2-6}$
$\Rightarrow y-3=\frac{-3}{-4}(x-6)$
$\;\;\;\;\;\;\;\;\;\;y=\frac{3}{4}(x-6)+3$
$\;\;\;\;\;\;\;\;\;\;y=\frac{3}{4}(x-2)$------(3)
Hence the required area is the area of the triangle enclosed by these three lines.
A=(area enclosed by line AB and x-axis )+(area enclosed by line BC and x-axis )+(area enclosed by line AC and x-axis )
$A=\int_2^4\frac{5}{2}(x-2)dx+\int_0^1(-x+9)dx+\int_6^2\frac{3}{4}(x-2)dx$
on integrating we get,
$A=\frac{5}{2}\begin{bmatrix}\frac{x^2}{2}-2x\end{bmatrix}_2^4+\begin{bmatrix}\frac{-x^2}{2}+9x\end{bmatrix}_1^0-\frac{3}{4}\begin{bmatrix}\frac{x^2}{2}-2x\end{bmatrix}_2^6$
On applying limits we get,
$A=\frac{5}{2}\begin{bmatrix}8-8-2+4\end{bmatrix}+\begin{bmatrix}-18+54+8-36\end{bmatrix}-\frac{3}{4}\begin{bmatrix}18-12-2+4\end{bmatrix}$
$A=5+8-\frac{3}{4}(8)$
$\;\;\;=13-6=7sq.units$
Hence the required area is 7 sq. units.
answered Dec 21, 2013 by

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