Question

Using the method of integration, find the area of triangular region whose vertices are $(2,-2),(4,3)$ and $(1,2)$.

Answer 1

Consider the coordinates of points $A(1,2),B(2,-2)$ and $C(4,3)$.

We mark the points on the axes and get the triangle $ABC$ as shown in the figure:


Equation of line $AB$

$y-2=\frac{-2-2}{2-1}(x-1)$
$\Rightarrow y-2=-4(x-1)$
$\Rightarrow y-2=-4x+4$
$\Rightarrow 4x=6-y\Rightarrow x=\frac{1}{4}(6-y)....\left(1\right)$

Equation of line $BC$:

$y+2=\frac{3+2}{4-2}(x-2)$
$\Rightarrow 2y+4=5x-10$
$\Rightarrow 2y+14=5x$
$\Rightarrow x=\frac{2y+14}{5}....\left(2\right)$

Equation of line $AC$:

$y-2=\frac{3-2}{4-1}(x-1)$
$\Rightarrow 3y-6=x-1$
$\Rightarrow x=3y-\mathrm{5....}\left(3\right)$

Now, area of $\Delta ABC$ =Area of trapezium $CDFB$ -area of trapezium $CDEA$ -area of trapezium $AEFB$

=${\int }_{-2}^3\frac{2y+14}{5}dy-{\int }_2^3(3y-5)dy-{\int }_{-2}^2\frac{6-y}{4}dy$

$=\frac{1}{5}{[\frac{2y^2}{2}+14y]}_{-2}^3-{[\frac{3y^2}{2}-5y]}_2^3-\frac{1}{4}{[6y-\frac{y^2}{2}]}_{-2}^2$

$=\frac{1}{5}[9+42-(4-28\left)\right]-[(\frac{27}{2}-15)-(6-10\left)\right]-\frac{1}{4}[12-2-(-12-2\left)\right]$

$=\frac{75}{5}-\frac{5}{2}-6=9-\frac{5}{2}=\frac{13}{2}$

$=6.5$ sq.units.