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Question Number 119007 by bramlexs22 last updated on 21/Oct/20

$$GivenpointA=(7,26)andB=(12,12),find$$$$allpointsPsuchthat\mid AP\mid =\mid BP\mid and\mathrm{\angle }APB=90°.$$

Answered by bemath last updated on 21/Oct/20

$$NotethatthetriangleABPisanisoscelesright$$$$trianglewith\mid AP\mid =\mid BP\mid .LetQbethemidpoint$$$$ofsegmentAB.ThenQ=(\frac{19}{2},19),where$$$$\mid QA\mid =\mid QB\mid =\mid QP\mid andQA\mathrm{\perp }QB$$$$Thus\overrightarrow{Q}\overrightarrow{A}=(\frac{5}{2},-7)and\overrightarrow{QP}=(7,\frac{5}{2})$$$$or(-7,-\frac{5}{2}).LetO=(0,0)betheorigin,itfollowsthat$$$$O\overrightarrow{P}=O\overrightarrow{Q}+Q\overrightarrow{P}=(\frac{19}{2},19)\pm (7,\frac{5}{2}).$$$$ConsequentlyP=(\frac{33}{2},\frac{43}{2})or(\frac{5}{2},\frac{33}{2})$$

Answered by 1549442205PVT last updated on 21/Oct/20

$$\overrightarrow{\mathrm{AB}}=(5,-14)$$$$\mathrm{Midpoint}\mathrm{of}\mathrm{AB}\mathrm{is}\mathrm{M}(\frac{19}{2},19),\mathrm{AP}=\mathrm{BP}$$$$\Rightarrow \mathrm{P}(\mathrm{x},\mathrm{y})\mathrm{lying}\mathrm{on}\mathrm{the}\mathrm{midline}\mathrm{d}\mathrm{of}\mathrm{AB},\mathrm{so}$$$$\overrightarrow{\mathrm{MP}}=(\frac{19}{2}-\mathrm{x},19-\mathrm{y})\mathrm{perpendicular}\mathrm{to}$$$$\overrightarrow{\mathrm{AB}}\iff \overrightarrow{\mathrm{AB}}.\overrightarrow{\mathrm{MP}}=0\iff 5(\frac{19}{2}-\mathrm{x})-14(19-\mathrm{y})=0$$$$\iff 5\mathrm{x}-14\mathrm{y}+218.5=0\Rightarrow \mathrm{P}(\mathrm{x},\frac{5\mathrm{x}+218.5}{14})$$$$\overrightarrow{\mathrm{AP}}=(\mathrm{x}-7,\frac{5\mathrm{x}-145.5}{14}),\overrightarrow{\mathrm{BP}}=(\mathrm{x}-12,\frac{5\mathrm{x}+50.5}{14})$$$$\widehat{\mathrm{APB}}=90°\iff \overrightarrow{\mathrm{AP}}\mathrm{\perp }\overrightarrow{\mathrm{BP}}\iff \overrightarrow{\mathrm{AP}}.\overrightarrow{\mathrm{BP}}=0$$$$\iff (\mathrm{x}-7)(\mathrm{x}-12)+\left(\frac{5\mathrm{x}-145.5}{14}\right)\left(\frac{5\mathrm{x}+50.5}{14}\right)=0$$$${\mathrm{x}}^2-19\mathrm{x}+84+\frac{{25\mathrm{x}}^2-475\mathrm{x}-7347.25}{196}=0$$$${221\mathrm{x}}^2-4199\mathrm{x}+9116.75=0$$$${\mathrm{x}}_1=16.5,{\mathrm{x}}_2=2.5.\mathrm{We}\mathrm{get}\mathrm{two}\mathrm{the}\mathrm{pointP}$$$${\mathrm{P}}_1(16.5,\frac{43}{2}),{\mathrm{P}}_2(2.5,\frac{33}{2})$$

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