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Question Number 119875 by help last updated on 27/Oct/20

Commented by help last updated on 27/Oct/20

thx...same answer.no correct options

$${thx}...{same}\:{answer}.{no}\:{correct}\:{options} \\ $$

Answered by TANMAY PANACEA last updated on 27/Oct/20

3y+2x−24=0  dj=∣((3×3+2×1−24)/( (√(3^2 +2^2 ))))∣=∣((−13)/( (√(13))))∣=(√(13))   sorry i made mistake

$$\mathrm{3}{y}+\mathrm{2}{x}−\mathrm{24}=\mathrm{0} \\ $$$${dj}=\mid\frac{\mathrm{3}×\mathrm{3}+\mathrm{2}×\mathrm{1}−\mathrm{24}}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}\mid=\mid\frac{−\mathrm{13}}{\:\sqrt{\mathrm{13}}}\mid=\sqrt{\mathrm{13}}\:\:\:{sorry}\:{i}\:{made}\:{mistake} \\ $$

Answered by Dwaipayan Shikari last updated on 27/Oct/20

y=−(2/3)x+8  3y+2x−24=0  P≡(1,3)  Distance =((∣3.3+2.1−24∣)/( (√(3^2 +2^2 ))))=((13)/( (√(13))))=(√(13))

$${y}=−\frac{\mathrm{2}}{\mathrm{3}}{x}+\mathrm{8} \\ $$$$\mathrm{3}{y}+\mathrm{2}{x}−\mathrm{24}=\mathrm{0} \\ $$$${P}\equiv\left(\mathrm{1},\mathrm{3}\right) \\ $$$${Distance}\:=\frac{\mid\mathrm{3}.\mathrm{3}+\mathrm{2}.\mathrm{1}−\mathrm{24}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}=\frac{\mathrm{13}}{\:\sqrt{\mathrm{13}}}=\sqrt{\mathrm{13}} \\ $$

Answered by mr W last updated on 27/Oct/20

D=d^2 =(x−1)^2 +(y−3)^2 =(x−1)^2 +(−(2/3)x+8−3)^2   (dD/dx)=2(x−1)+2(−(2/3)x+5)(−(2/3))=0  9x−9+4x−30=0  ⇒x=3  d_(min) =(√((3−1)^2 +(−(2/3)×3+5)^2 ))=(√(13))

$${D}={d}^{\mathrm{2}} =\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} =\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{2}}{\mathrm{3}}{x}+\mathrm{8}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\frac{{dD}}{{dx}}=\mathrm{2}\left({x}−\mathrm{1}\right)+\mathrm{2}\left(−\frac{\mathrm{2}}{\mathrm{3}}{x}+\mathrm{5}\right)\left(−\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\mathrm{9}{x}−\mathrm{9}+\mathrm{4}{x}−\mathrm{30}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3} \\ $$$${d}_{{min}} =\sqrt{\left(\mathrm{3}−\mathrm{1}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{3}+\mathrm{5}\right)^{\mathrm{2}} }=\sqrt{\mathrm{13}} \\ $$

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