Question

Using the method of integration, find the area of the region bounded by the following lines $5x-2y-10=0$, $x+y-9=0$, and $2x-5y-4=0$

Answer 1

Given lines are
$5x-2y-10=0$ --- (1)
$x+y-9=0$--- (2)
$2x-5y-4=0$ ---- (3)
For intersecting point of (1) and (2)
(1) + (2)X (2) $\Rightarrow 5x-2y-10+2x+2y-18=0$
$\Rightarrow 7x-28=0\Rightarrow x=4$
Putting x = 4in (1), we get
$20-2y-10=0$
$y=5$
Intersecting point of (1) and (2) is (4, 5)
For intersecting point of (1) and (3)
$\left(1\right)\times 5-\left(3\right)\times 2\Rightarrow 25x-10y-50-4x+10y+8=0$
$\Rightarrow 21x-42=0$
$\Rightarrow x=2$
Putting x= 2 in (1), we get
$10x-2y-10=0$
$y=0$
Intersecting point of (1) and (3) is (2, 0)
For intersecting point of (2)and (3)
$2\times \left(2\right)\times \left(3\right)\Rightarrow 2x+2y-18-2x+5y+4=0$
$\Rightarrow 7y-14=0$
$\Rightarrow y=2$
Putting y = 2 in (2), we get
x +2 -9 = 0
x = 7
Intersecting point (2) and (3) is (7, 2)
Shaded region is required region
$\therefore$ Required area =${\int }_2^4\left(\frac{5x-10}{2}\right)dx+{\int }_4^7(-x+9)dx-{\int }_2^7\frac{2x-4}{5}dx$
$=\frac{5}{2}{\int }_2^{4}xdx-5{\int }_2^{4}dx-{\int }_4^{7}xdx+9{\int }_4^{7}dx-\frac{2}{5}{\int }_2^{7}xdx+\frac{4}{5}{\int }_2^{7}dx$
$=\frac{5}{2}{\left[\frac{x^2}{2}\right]}_2^4-5{\left[x\right]}_2^4-{\left[\frac{x^2}{2}\right]}_4^7+9{\left[x\right]}_4^7-\frac{2}{5}{\left[\frac{x^2}{2}\right]}_2^7+\frac{4}{5}{\left[x\right]}_2^7$
$=\frac{5}{4}(16-4)-5(4-2)-\frac{1}{2}(49-16)+9(7-4)-\frac{1}{5}(49-4)+\frac{4}{5}(7-2)$
$=15-10-\frac{33}{2}+27-9+4=27-\frac{33}{2}=\frac{54-33}{2}$
$=\frac{21}{2}$ sq.unit
Therefore the area is $\frac{21}{2}$ sq.units.