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Category: Coordinate Geometry

Question-226770

Question Number 226770 by gregori last updated on 14/Dec/25 Answered by TonyCWX last updated on 14/Dec/25 $$\mathrm{Stewart}'\mathrm{s}\:\mathrm{Theorem}: \\ $$$${b}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {y}=\left({x}+{y}\right)\left({z}^{\mathrm{2}} −{xy}\right) \\ $$$${b}^{\mathrm{2}} \left({x}+{y}\right)=\left({x}+{y}\right)\left({z}^{\mathrm{2}}…

Question-226338

Question Number 226338 by Spillover last updated on 25/Nov/25 Answered by Spillover last updated on 06/Dec/25 $${at}\:{point}\:{P}\left({a}\mathrm{cos}\:\theta+{b}\mathrm{sin}\:\theta\right){aa} \\ $$$$ \\ $$$${at}\:{point}\:{Q}\left({a}\mathrm{cos}\:\left(\theta+\frac{\pi}{\mathrm{2}}\right)+{i}\mathrm{sin}\left(\theta+\frac{\pi}{\mathrm{2}}\right)\right. \\ $$$${Q}=\left({a}\mathrm{sin}\:\theta,{b}\mathrm{cos}\:\theta\right. \\ $$$${from}\:{equation}\:{of}\:{normal}…

Question-225497

Question Number 225497 by gregori last updated on 31/Oct/25 Answered by Simurdiera last updated on 01/Nov/25 $$\begin{array}{|c|}{\mathrm{11}}\\\hline\end{array}\begin{cases}{{x}\:+\:{y}\:+\:\sqrt{{xy}}\:=\:\mathrm{28}\:\:\:\:\:\ldots\left(\mathrm{I}\right)}\\{{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{xy}\:=\:\mathrm{336}\:\:\:\:\:\ldots\left(\mathrm{II}\right)}\end{cases} \\ $$$$\mathrm{En}\:\left(\mathrm{II}\right),\:\mathrm{sumando}\:\mathrm{ceros} \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{xy}\:+\:{xy}\:−\:{xy}\:=\:\mathrm{336}…