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Question Number 153479 by naka3546 last updated on 07/Sep/21
Find  area  of  region  that  satisfy       ∣x−2∣ + ∣y+3∣ < 3
$${Find}\:\:{area}\:\:{of}\:\:{region}\:\:{that}\:\:{satisfy}\:\: \\ $$$$\:\:\:\mid{x}−\mathrm{2}\mid\:+\:\mid{y}+\mathrm{3}\mid\:<\:\mathrm{3} \\ $$
Answered by aleks041103 last updated on 07/Sep/21
Translation doesn′t change the region′s  area. Therefore we can find the area  of the region ∣x∣+∣y∣<3 instead.  The boundary is  ∣x∣+∣y∣=3  1st quadrant: y=3−x  2nd quadrant: y=x−3  3rd quadrant: y=−3−x  4th quadrant: y=3+x    Then the area is simply  A=4(((3.3)/2))=18
$${Translation}\:{doesn}'{t}\:{change}\:{the}\:{region}'{s} \\ $$$${area}.\:{Therefore}\:{we}\:{can}\:{find}\:{the}\:{area} \\ $$$${of}\:{the}\:{region}\:\mid{x}\mid+\mid{y}\mid<\mathrm{3}\:{instead}. \\ $$$${The}\:{boundary}\:{is} \\ $$$$\mid{x}\mid+\mid{y}\mid=\mathrm{3} \\ $$$$\mathrm{1}{st}\:{quadrant}:\:{y}=\mathrm{3}−{x} \\ $$$$\mathrm{2}{nd}\:{quadrant}:\:{y}={x}−\mathrm{3} \\ $$$$\mathrm{3}{rd}\:{quadrant}:\:{y}=−\mathrm{3}−{x} \\ $$$$\mathrm{4}{th}\:{quadrant}:\:{y}=\mathrm{3}+{x} \\ $$$$ \\ $$$${Then}\:{the}\:{area}\:{is}\:{simply} \\ $$$${A}=\mathrm{4}\left(\frac{\mathrm{3}.\mathrm{3}}{\mathrm{2}}\right)=\mathrm{18} \\ $$
Commented by talminator2856791 last updated on 08/Sep/21
 how did you make this animation?
$$\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{make}\:\mathrm{this}\:\mathrm{animation}?\: \\ $$

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