Plot the region of the points P (x,y) satisfying `|x|+|y| lt 1.`
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` (##CEN_CG_C02_S01_051_S01.png" width="80%"> Given that `|x|+|y| lt 1`. We have four cases. If x, `y gt 0`, then we have `x+y lt 1 " or " x+y-1 lt 0.` Points satisfying these inequalities lie in triangular region in first quadrant formed by axes and line x+y-1 = 0. Similarly, we have three trinangular regions in all other quadrants. Thus, points satisfying given inequality lie in a square as shown in the following figure.
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x+y lt 5
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If a x ^2+b x+c=0 has imaginary roots and a-b+c >0 then the set of points (x ,y) satisfying the equation |a(x^2+y/a)+(b+1)x+c|=|a x^2+b x+c|+|x+y| consists of the region in the x y-p l a n e which is on or above the bisector of I and III quadrant on or above the bisector of II and IV quadrant on or below the bisector of I and III quadrant on or below the bisector of II and IV quadrant
