Let us take X=x−2 and V=y+4 Now, let us draw ∣x∣+∣y∣<5 . When y=0.∣X∣<5 i.e.,−5<X<5 . When X=0,∣y∣<5 i.e.,−5<Y<5
Now, try to think of how the graph of | x - 2|+ | y + 4| <5 would look, 'x - 2' indicates a shift along the X-axis, 2 unit to the right and 'y + 4' indicates a shift along the y-axis, 4 units to the bottom
We can observe that shifting the origin has no impact on the area of the figure. This essentially means that area computed using |x-2| + |y+4| <5 has to be same as the area computed using | x| + | y| < 5. Required Area = 4×21​×5×5=50squnits