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Category: Coordinate Geometry

Question-209307

Question Number 209307 by efronzo1 last updated on 06/Jul/24 Commented by mr W last updated on 06/Jul/24 $$\mid{QR}\mid<\frac{\mathrm{8}}{\mathrm{3}}−\mathrm{1}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\frac{{QR}}{{PR}}=\frac{\mathrm{2}}{\mathrm{1}}\:\Rightarrow\frac{{PQ}}{{QR}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mid{PQ}\mid=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}×{QR}<\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}×\frac{\mathrm{5}}{\mathrm{3}}=\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{6}}<\sqrt{\mathrm{5}} \\ $$$$\Rightarrow{impossible}\:{that}\:\mid{PQ}\mid=\sqrt{\mathrm{5}}\:! \\…

Question-208327

Question Number 208327 by efronzo1 last updated on 12/Jun/24 Answered by A5T last updated on 12/Jun/24 $${DE}=\mathrm{5}{x},{DF}=\mathrm{12}{x} \\ $$$${Let}\:{the}\:{perpendicular}\:{from}\:{A}\:{to}\:{BC}\:{meet}\:{it}\:{at}\:{H}; \\ $$$${AH}×{BC}=\mathrm{3}×\mathrm{4}=\mathrm{12}\Rightarrow{AH}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\Rightarrow{BH}=\sqrt{\mathrm{9}−\frac{\mathrm{144}}{\mathrm{25}}}=\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\frac{{AD}}{{AB}}=\frac{{DE}}{{BC}}={x}\Rightarrow{AD}=\mathrm{3}{x}\Rightarrow{AE}=\mathrm{4}{x}…

Question-207879

Question Number 207879 by efronzo1 last updated on 29/May/24 Answered by mr W last updated on 29/May/24 $${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{12}=\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{3}\geqslant\mathrm{3} \\ $$$${f}\left({a}\right)=\mathrm{65539}={a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{12} \\ $$$$\Rightarrow{a}^{\mathrm{2}}…