Question Number 119875 by help last updated on 27/Oct/20
Commented by help last updated on 27/Oct/20
$${thx}…{same}\:{answer}.{no}\:{correct}\:{options} \\ $$
Answered by TANMAY PANACEA last updated on 27/Oct/20
$$\mathrm{3}{y}+\mathrm{2}{x}−\mathrm{24}=\mathrm{0} \\ $$$${dj}=\mid\frac{\mathrm{3}×\mathrm{3}+\mathrm{2}×\mathrm{1}−\mathrm{24}}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}\mid=\mid\frac{−\mathrm{13}}{\:\sqrt{\mathrm{13}}}\mid=\sqrt{\mathrm{13}}\:\:\:{sorry}\:{i}\:{made}\:{mistake} \\ $$
Answered by Dwaipayan Shikari last updated on 27/Oct/20
$${y}=−\frac{\mathrm{2}}{\mathrm{3}}{x}+\mathrm{8} \\ $$$$\mathrm{3}{y}+\mathrm{2}{x}−\mathrm{24}=\mathrm{0} \\ $$$${P}\equiv\left(\mathrm{1},\mathrm{3}\right) \\ $$$${Distance}\:=\frac{\mid\mathrm{3}.\mathrm{3}+\mathrm{2}.\mathrm{1}−\mathrm{24}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}=\frac{\mathrm{13}}{\:\sqrt{\mathrm{13}}}=\sqrt{\mathrm{13}} \\ $$
Answered by mr W last updated on 27/Oct/20
$${D}={d}^{\mathrm{2}} =\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} =\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{2}}{\mathrm{3}}{x}+\mathrm{8}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\frac{{dD}}{{dx}}=\mathrm{2}\left({x}−\mathrm{1}\right)+\mathrm{2}\left(−\frac{\mathrm{2}}{\mathrm{3}}{x}+\mathrm{5}\right)\left(−\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\mathrm{9}{x}−\mathrm{9}+\mathrm{4}{x}−\mathrm{30}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3} \\ $$$${d}_{{min}} =\sqrt{\left(\mathrm{3}−\mathrm{1}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{3}+\mathrm{5}\right)^{\mathrm{2}} }=\sqrt{\mathrm{13}} \\ $$