Answer:
Given lines are \[y=\frac{5}{2}x-5\] ?(i) \[x+y=9\] ?(ii) and \[y\frac{3}{4}x-\frac{3}{2}\] ?(iii) For finding the points of intersection, we solve in pairs. On solving Eqs. (i) and (ii), we get Coordinates of \[C=(4,\,\,5)\] On solving Eqs. (ii) and (iii), we get Coordinates of \[B=(6,\,\,3)\] On solving Eqs. (i) and (iii), we get Coordinates of \[A=(2,\,\,0)\] 
\[\therefore \]Required area = Area of \[\Delta \,ANC\] \[+\]Area of quadrilatiral NCBD \[-\] Area of\[\Delta \,ABM\] \[=\int\limits_{2}^{4}{(line\,\,AC)\,\,dx+\int\limits_{4}^{6}{(line\,\,BC)\,\,dx}-\int\limits_{2}^{6}{(line\,\,AB)\,\,dx}}\] \[=\int\limits_{2}^{4}{\,\,\left( \frac{5}{2}x-5 \right)\,\,dx+\int\limits_{4}^{6}{\,\,(9-x)\,\,dx-\int\limits_{2}^{6}{\,\,\left( \frac{3}{4}x-\frac{3}{2} \right)\,\,dx}}}\]\[=\left[ \frac{5}{2}\cdot \frac{{{x}^{2}}}{2}-5x \right]_{2}^{4}+\left[ 9x-\frac{{{x}^{2}}}{2} \right]_{4}^{6}-\left[ \frac{3}{4}\cdot \frac{{{x}^{2}}}{2}-\frac{3}{2}x \right]_{2}^{6}\] \[=\frac{5}{4}[{{x}^{2}}]_{2}^{4}-5\,[x]_{2}^{4}+9\,[x]_{4}^{6}-\frac{1}{2}[{{x}^{2}}]_{4}^{6}-\frac{3}{8}[{{x}^{2}}]_{2}^{6}+\frac{3}{2}[x]_{2}^{6}\]\[=\frac{5}{4}(16-4)-5\,(4-2)+9\,(6-4)-\frac{1}{2}[36-16]\] \[-\frac{3}{8}(36-4)+\frac{3}{2}(6-2)\] \[=\frac{5}{4}(12)-5\,(2)+9\,(2)-\frac{1}{2}(20)-\frac{3}{8}(32)+\frac{3}{2}(4)\] \[=15-10+18-10-12+6\] \[=\,\,|39-32|\] \[=7\,\,sq\,\,units\]
You need to login to perform this action.
You will be redirected in
3 sec
