Question

Using the method of integration, find the area of the region bounded by the lines $3x-2y+1=0,2x+3y-21=0$ and $x-5y+9=0$

Answer 1

Given lines are
$3x-2y+1=0$ ...(1)
$2x+3y-21=0$ ...(2)
$x-5y+9=0$ ...(3)

For intersection of (1) and (2)
Multiplying eq.(1) by 3 and eq.(2) by 2 and adding them, we get
$9x-6y+3+4x+6y-42=0$
$13x-39=0$
$x=3$
Putting it in eq.(1), we get
$9-2y+1=0$
$2y=10$
$y=5$
Intersection point of (1) and (2) is $(3,5)$

For intersection of (2) and (3)
Multiplying eq.(3) by 2 and subtracting it from eq.(2), we get
$2x+3y-21-2x+10y-18=0$
$13y-39=0$
$y=3$
Putting $y=3$ in eq,(2), we get
$2x+9-21=0$
$2x-12=0$
$x=6$
Intersection point of (2) and (3) is $(6,3)$.

For intersection of (1) and (3)
Multiplying (3) x 3 and subrating it from eq.(1), we get
$3x-2y+1-3x+15y-27=0$
$13y-26=0$
$y=2$
Putting y = 2 in (1), we get
$3x-4+1=0$
$x=1$
Intersection point of (1) and (3) is $(1,2)$

With the help of point of intersection we draw the graph of lines (1), (2) and (3)
shaded region is required region.
Therefore, Area of required region = ${\int }_1^3\frac{3x+1}{2}dx+{\int }_3^6\frac{-2x+21}{3}dx-{\int }_1^6\frac{x+9}{5}dx$

$=\frac{3}{2}{\int }_1^{3}xdx+\frac{1}{2}{\int }_1^{3}dx-\frac{2}{3}{\int }_3^{6}dx+7{\int }_3^{6}dx-\frac{1}{5}{\int }_1^{6}xdx-\frac{9}{5}{\int }_1^{6}dx$

$=\frac{3}{2}{\left[\frac{x^2}{3}\right]}_1^3+\frac{1}{2}{\left[x\right]}_1^3-\frac{2}{3}{\left[\frac{x^2}{2}\right]}_3^6+7{\left[x\right]}_3^6-\frac{1}{5}{\left[\frac{x^2}{2}\right]}_1^6-\frac{9}{5}{\left[x\right]}_1^6$

$=\frac{3}{4}(9-1)+\frac{1}{2}(3-1)-\frac{2}{6}(36-9)+7(6-3)-\frac{1}{10}(36-1)-\frac{9}{5}(6-1)$

$=6+1-9+21-\frac{7}{2}-9$

$=10-\frac{7}{2}=\frac{20-7}{2}=\frac{13}{2}$