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Question Number 139639 by bemath last updated on 30/Apr/21

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{reflection}\mathrm{of}\mathrm{the}$$$$\mathrm{point}(2,2)\mathrm{in}\mathrm{the}\mathrm{line}\mathrm{x}+2\mathrm{y}=4?$$

Commented by bramlexs22 last updated on 30/Apr/21

$$\mathrm{Let}\mathrm{P}(2,2)\mathrm{\&}{\mathrm{P}}^{\prime }(\mathrm{a},\mathrm{b})\mathrm{is}\mathrm{the}\mathrm{reflectional}$$$$\mathrm{image}\mathrm{of}\mathrm{P}\mathrm{in}\mathrm{L}.$$$$\mathrm{let}\mathrm{Q}(\frac{\mathrm{a}+2}{2},\frac{\mathrm{b}+2}{2})\mathrm{is}\mathrm{the}\mathrm{midpoint}$$$${\mathrm{PP}}^{\prime }\mathrm{and}\mathrm{PQ}\mathrm{perpendicular}\mathrm{to}\mathrm{L}$$$$(\bullet ){\mathrm{m}}_{\mathrm{PQ}}=\frac{\mathrm{b}-2}{\mathrm{a}-2};{\mathrm{m}}_{\mathrm{L}}=-\frac{1}{2}$$$$\Rightarrow {\mathrm{m}}_{\mathrm{PQ}}\times {\mathrm{m}}_{\mathrm{L}}=-1$$$$\Rightarrow \frac{\mathrm{b}-2}{\mathrm{a}-2}=2\Rightarrow \mathrm{b}-2=2\mathrm{a}-4;2\mathrm{a}-\mathrm{b}=2$$$$(\bullet \bullet )\mathrm{line}\mathrm{L}\mathrm{passes}\mathrm{through}\mathrm{point}\mathrm{Q}$$$$\Rightarrow \frac{\mathrm{a}+2}{2}+2\left(\frac{\mathrm{b}+2}{2}\right)=4$$$$\Rightarrow \mathrm{a}+2+2\mathrm{b}+4=8;\mathrm{a}+2\mathrm{b}=2$$$$\mathrm{solving}\mathrm{for}\mathrm{a}\mathrm{\&}\mathrm{b}$$$$\Rightarrow \mathrm{a}+2\mathrm{b}=2...\left(\times 2\right)\Rightarrow 2\mathrm{a}+4\mathrm{b}=4$$$$2\mathrm{a}-\mathrm{b}=2$$$$\{\begin{array}{l}5\mathrm{b}=2;\mathrm{b}=\frac{2}{5}\\ \mathrm{a}=2-2\mathrm{b}=2-\frac{4}{5}=\frac{6}{5}\end{array}$$$$\mathrm{therefore}{\mathrm{P}}^{\prime }(\frac{6}{5},\frac{2}{5}).$$

Answered by mr W last updated on 30/Apr/21

$$thelinefrom(2,2)andperpendicular$$$$tolinex+2y=4is$$$$\frac{x-2}{1}=\frac{y-2}{2}=t$$$$orx=2+t,y=2+2t$$$$intersectionpointwiththeline:$$$$(2+t)+2(2+2t)=4$$$$\Rightarrow t=-\frac{2}{5}$$$$theimagepointisofdoubledistance,$$$$i.e.t=2\times (-\frac{2}{5})=-\frac{4}{5}$$$$\Rightarrow x=2-\frac{4}{5}=\frac{6}{5}$$$$\Rightarrow y=2-2\times \frac{4}{5}=\frac{2}{5}$$$$imageof(2,2)inlineis(\frac{6}{5},\frac{2}{5}).$$

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