 Good morning students, I am Purva and today we will discuss the following question 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 let's begin with the solution now now the coordinates of the vertices of the triangle ABC are a 2, 0, b 4, 5 and c 6, 3 now suppose we have a line AB whose coordinates are x1 y1 and x2 y2 then the equation of the line AB is given by y minus y1 upon y2 minus y1 is equal to x minus x1 upon x2 minus x1 so this is the equation of the line AB now using this formula in this question we get the equation of line AB is given by y minus y1 and here y1 is 0 so we get y minus 0 upon y2 minus y1 and here y2 is 5 so we get 5 minus y1 that is 0 this is equal to x minus x1 and here x1 is 2 so we get x minus 2 upon x2 minus x1 here x2 is 4 so we get 4 minus 2 this implies y minus 0 is y so we get y upon 5 minus 0 is equal to 5 so we get y upon 5 is equal to x minus 2 upon 4 minus 2 is equal to 2 so we get 2 and solving this equation we get this implies 5x minus 2y is equal to 10 now the equation of line AC is given by y minus y1 here y1 is equal to 0 so we have y minus 0 upon y2 minus y1 and here y2 is equal to 3 so we have 3 minus 0 and this is equal to x minus x1 x minus here x1 is equal to 2 so we have x minus 2 upon x2 minus x1 here x2 is equal to 6 so we have 6 minus 2 and this implies y minus 0 is equal to y upon 3 minus 0 is equal to 3 so we get y upon 3 is equal to x minus 2 upon 6 minus 2 is equal to 4 so we get 4 and simplifying this we get this implies 3x minus 4y is equal to 6. Now we find the equation of the line BC so we have the equation of the line BC is y minus y1 that is y minus 3 upon y2 minus y1 that is 5 minus 3 and this is equal to x minus x1 that is x minus 6 upon x2 minus x1 that is 4 minus 6 and this implies y minus 3 upon 5 minus 3 is equal to 2 so we have 2 and this is equal to x minus 6 upon 4 minus 6 is equal to minus 2 and solving this we get equation of the line BC is x plus y is equal to 9 now we will plot the points 2 comma 0 4 comma 5 and 6 comma 3 on the graph so here we have plotted the points A B and C on the graph and we have joined the lines A B B C and C A to get the triangle ABC we have to find the area of this shaded region now we have area of triangle ABC is equal to area of triangle ABD plus area of trapezium BDEC minus area of triangle ACE so we have area of triangle ABC is equal to area of triangle ABD plus area of trapezium BDEC minus area of triangle ACE therefore area of triangle ABC that is the required area is equal to area of triangle ABD and here we have limiters from 2 to 4 and equation of line AB is given by 5x minus 2 y is equal to 10 that is y is equal to 5x minus 10 upon 2 so we have integral limiters from 2 to 4 5x minus 10 upon 2 dx plus area of trapezium BDEC now here limiters from 4 to 6 and equation of line BC is given by x plus y is equal to 9 so we have y is equal to 9 minus x therefore we get plus integral limiters from 4 to 6 9 minus x dx minus area of triangle ACE now here limiters from 2 to 6 and equation of line AC is given by 3x minus 4y is equal to 6 therefore we have y is equal to 3x minus 6 upon 4 therefore we get minus integral limiters from 2 to 6 3x minus 6 upon 4 dx we can write this as this is equal to 1 upon 2 integral limiters from 2 to 4 5x minus 10 dx plus integral limiters from 4 to 6 9 minus x dx minus 1 upon 4 integral limiters from 2 to 6 3x minus 6 dx this is equal to 1 upon 2 now integrating 5x minus 10 we get 5x square upon 2 minus 10x and limiters from 2 to 4 plus integrating 9 minus x we get 9x minus x square upon 2 and limiters from 4 to 6 minus 1 upon 4 now integrating 3x minus 6 we get 3x square upon 2 minus 6x and limiters from 2 to 6 this is equal to 1 upon 2 into now putting the limits we get 40 minus 40 minus 10 minus 20 now putting the upper limit 4 in place of x we get 40 minus 40 minus putting lower limit 2 in place of x we get 10 minus 20 plus again putting the limits we get 54 minus 18 minus 36 minus 8 now putting the upper limit 6 in place of x we get 54 minus 18 minus putting lower limit 4 in place of x we get 36 minus 8 minus 1 upon 4 now again putting the limits we get 54 minus 36 minus 6 minus 12 putting upper limit 6 in place of x we get 54 minus 36 minus putting lower limit 2 in place of x we get 6 minus 12 this is equal to 1 upon 2 into here 40 and 40 will cancel out we get minus 10 minus into minus will become plus so plus 20 plus 54 minus 18 is equal to 36 minus 36 minus 8 is equal to 28 minus 1 upon 4 into 54 minus 36 is equal to 18 minus 6 minus 12 is equal to minus 6 this is equal to 1 upon 2 into minus 10 plus 20 gives 10 plus 36 minus 28 gives 8 minus 1 upon 4 into 18 minus into minus will become plus so we get 18 plus 6 is equal to 24 and this is equal to 5 plus 8 minus 6 and this is equal to 7 so we get the required area is equal to 7 so we get our answer as 7 hope you have understood the solution bye and take care