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Question Number 209307 by efronzo1 last updated on 06/Jul/24
Commented by mr W last updated on 06/Jul/24
∣QR∣<(8/3)−1=(5/3) ((QR)/(PR))=(2/1) ⇒((PQ)/(QR))=((√5)/2) ∣PQ∣=((√5)/2)×QR<((√5)/2)×(5/3)=((5(√5))/6)<(√5) ⇒impossible that ∣PQ∣=(√5) ! ⇒question is wrong!
$$\mid{QR}\mid<\frac{\mathrm{8}}{\mathrm{3}}−\mathrm{1}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\frac{{QR}}{{PR}}=\frac{\mathrm{2}}{\mathrm{1}}\:\Rightarrow\frac{{PQ}}{{QR}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mid{PQ}\mid=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}×{QR}<\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}×\frac{\mathrm{5}}{\mathrm{3}}=\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{6}}<\sqrt{\mathrm{5}} \\ $$$$\Rightarrow{impossible}\:{that}\:\mid{PQ}\mid=\sqrt{\mathrm{5}}\:! \\ $$$$\Rightarrow{question}\:{is}\:{wrong}! \\ $$
Commented by mr W last updated on 06/Jul/24
Commented by efronzo1 last updated on 07/Jul/24
$$\:\cancel{ } \\ $$
Answered by A5T last updated on 07/Jul/24
For the corrected question: y=((2/3))^(x+1) +(8/3) P(p_1 ,p_2 ),Q(q_1 ,q_2 ) (√((q_1 −p_1 )^2 +(q_2 −p_2 )^2 ))=(√5) y−2x=k⇒p_2 −2p_1 =q_2 −2q_1 ⇒2(p_1 −q_1 )=p_2 −q_2 ...(i) ⇒(√((p_1 −q_1 )^2 +4(p_1 −q_1 )^2 ))=(√5) ⇒5(p_1 −q_1 )^2 =5⇒(p_1 −q_1 )=−1 p_2 =(2^(p_1 +3) /3^(p_1 +3) )+1; q_2 =(2^(q_1 +1) /3^(q_1 +1) )+(8/3) ⇒p_2 =((2^q_1 ×4)/(3^q_1 ×9))+1; q_2 =((2^q_1 ×2)/(3^q_1 ×3))+(8/3) p_2 =((2q_2 )/3)−(7/9)⇒p_2 −q_2 =((−q_2 )/3)−(7/9) ⇒−2+(7/9)=((−11)/9)=((−q_2 )/3)⇒q_2 =((11)/3)⇒p_2 =((15)/9) ⇒q_1 +1=0⇒q_1 =−1⇒p_1 =−2 ⇒k=p_2 −2p_1 =((15)/9)+4=((17)/3)
$${For}\:{the}\:{corrected}\:{question}:\:{y}=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}+\mathrm{1}} +\frac{\mathrm{8}}{\mathrm{3}} \\ $$$${P}\left({p}_{\mathrm{1}} ,{p}_{\mathrm{2}} \right),{Q}\left({q}_{\mathrm{1}} ,{q}_{\mathrm{2}} \right) \\ $$$$\sqrt{\left({q}_{\mathrm{1}} −{p}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({q}_{\mathrm{2}} −{p}_{\mathrm{2}} \right)^{\mathrm{2}} }=\sqrt{\mathrm{5}} \\ $$$${y}−\mathrm{2}{x}={k}\Rightarrow{p}_{\mathrm{2}} −\mathrm{2}{p}_{\mathrm{1}} ={q}_{\mathrm{2}} −\mathrm{2}{q}_{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}\left({p}_{\mathrm{1}} −{q}_{\mathrm{1}} \right)={p}_{\mathrm{2}} −{q}_{\mathrm{2}} …\left({i}\right) \\ $$$$\Rightarrow\sqrt{\left({p}_{\mathrm{1}} −{q}_{\mathrm{1}} \right)^{\mathrm{2}} +\mathrm{4}\left({p}_{\mathrm{1}} −{q}_{\mathrm{1}} \right)^{\mathrm{2}} }=\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{5}\left({p}_{\mathrm{1}} −{q}_{\mathrm{1}} \right)^{\mathrm{2}} =\mathrm{5}\Rightarrow\left({p}_{\mathrm{1}} −{q}_{\mathrm{1}} \right)=−\mathrm{1} \\ $$$${p}_{\mathrm{2}} =\frac{\mathrm{2}^{{p}_{\mathrm{1}} +\mathrm{3}} }{\mathrm{3}^{{p}_{\mathrm{1}} +\mathrm{3}} }+\mathrm{1};\:{q}_{\mathrm{2}} =\frac{\mathrm{2}^{{q}_{\mathrm{1}} +\mathrm{1}} }{\mathrm{3}^{{q}_{\mathrm{1}} +\mathrm{1}} }+\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\Rightarrow{p}_{\mathrm{2}} =\frac{\mathrm{2}^{{q}_{\mathrm{1}} } ×\mathrm{4}}{\mathrm{3}^{{q}_{\mathrm{1}} } ×\mathrm{9}}+\mathrm{1};\:{q}_{\mathrm{2}} =\frac{\mathrm{2}^{{q}_{\mathrm{1}} } ×\mathrm{2}}{\mathrm{3}^{{q}_{\mathrm{1}} } ×\mathrm{3}}+\frac{\mathrm{8}}{\mathrm{3}} \\ $$$${p}_{\mathrm{2}} =\frac{\mathrm{2}{q}_{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{7}}{\mathrm{9}}\Rightarrow{p}_{\mathrm{2}} −{q}_{\mathrm{2}} =\frac{−{q}_{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{7}}{\mathrm{9}} \\ $$$$\Rightarrow−\mathrm{2}+\frac{\mathrm{7}}{\mathrm{9}}=\frac{−\mathrm{11}}{\mathrm{9}}=\frac{−{q}_{\mathrm{2}} }{\mathrm{3}}\Rightarrow{q}_{\mathrm{2}} =\frac{\mathrm{11}}{\mathrm{3}}\Rightarrow{p}_{\mathrm{2}} =\frac{\mathrm{15}}{\mathrm{9}} \\ $$$$\Rightarrow{q}_{\mathrm{1}} +\mathrm{1}=\mathrm{0}\Rightarrow{q}_{\mathrm{1}} =−\mathrm{1}\Rightarrow{p}_{\mathrm{1}} =−\mathrm{2} \\ $$$$\Rightarrow{k}={p}_{\mathrm{2}} −\mathrm{2}{p}_{\mathrm{1}} =\frac{\mathrm{15}}{\mathrm{9}}+\mathrm{4}=\frac{\mathrm{17}}{\mathrm{3}} \\ $$
Answered by mr W last updated on 07/Jul/24
y_1 =((2/3))^(x_1 +3) +1=2x_1 +k ...(i) y_2 =((2/3))^(x_1 +1) +(8/3)=2x_2 +k ...(ii) PQ=(√5) ⇒x_2 −x_1 =1 ((2/3))^(x_2 +1) +(8/3)−((2/3))^(x_1 +3) −1=2(x_2 −x_1 ) [((2/3))^(x_2 −x_1 −2) −1]((2/3))^(x_1 +3) =(1/3) [((2/3))^(−1) −1]((2/3))^(x_1 +3) =(1/3) ((2/3))^(x_1 +3) =(2/3) ⇒x_1 +3=1 ⇒x_1 =−2 ((2/3))^(−2+3) +1=2(−2)+k ⇒k=((17)/3) ✓
$${y}_{\mathrm{1}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}_{\mathrm{1}} +\mathrm{3}} +\mathrm{1}=\mathrm{2}{x}_{\mathrm{1}} +{k}\:\:\:…\left({i}\right) \\ $$$${y}_{\mathrm{2}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}_{\mathrm{1}} +\mathrm{1}} +\frac{\mathrm{8}}{\mathrm{3}}=\mathrm{2}{x}_{\mathrm{2}} +{k}\:\:\:…\left({ii}\right) \\ $$$${PQ}=\sqrt{\mathrm{5}}\:\Rightarrow{x}_{\mathrm{2}} −{x}_{\mathrm{1}} =\mathrm{1} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}_{\mathrm{2}} +\mathrm{1}} +\frac{\mathrm{8}}{\mathrm{3}}−\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}_{\mathrm{1}} +\mathrm{3}} −\mathrm{1}=\mathrm{2}\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right) \\ $$$$\left[\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}_{\mathrm{2}} −{x}_{\mathrm{1}} −\mathrm{2}} −\mathrm{1}\right]\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}_{\mathrm{1}} +\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\left[\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{−\mathrm{1}} −\mathrm{1}\right]\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}_{\mathrm{1}} +\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}_{\mathrm{1}} +\mathrm{3}} =\frac{\mathrm{2}}{\mathrm{3}}\: \\ $$$$\Rightarrow{x}_{\mathrm{1}} +\mathrm{3}=\mathrm{1}\:\Rightarrow{x}_{\mathrm{1}} =−\mathrm{2} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{−\mathrm{2}+\mathrm{3}} +\mathrm{1}=\mathrm{2}\left(−\mathrm{2}\right)+{k}\:\Rightarrow{k}=\frac{\mathrm{17}}{\mathrm{3}}\:\:\checkmark \\ $$

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