the area bounded by the lines (2x-3)+(3y+4)=5 is
Hello Aspirant
This problem is not difficult, but tedious. Here is what needs to be done.
Consider four cases
2x-3>0, 3y+4>0 then (2x-3)+(3y+4) =5 or y=-(2/3)x+2
2x-3>0, 3y+4<0 then (2x-3)-(3y+4) =5 or y= (2/3)x +4
2x-3<0, 3y+4>0 then -(2x-3)+(3y+4) =5 or y =(2/3)x-2/3
2x-3<0, 3y-2<0 then -(2x-3)-(3y+4) =6 or y= -(2/3)x -7/3
Draw the four straight lines and look at the rectangle. It is a parallelogram because the first and the third line have the same slope, so do the second and the fourth line. We also know the coordinates of all four vertices. We can easily determine the area between the top two segments and the x-axis (sum of two trapezoids) and the area between bottom two segments and the x-axis (sum of two other trapezoids) and then deduct the two areas from each other.
Hope it helps
Thank you & Good Luck!!!
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