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Question-58652




Question Number 58652 by peter frank last updated on 27/Apr/19
Commented by peter frank last updated on 27/Apr/19
a and b
$${a}\:{and}\:{b} \\ $$
Answered by tanmay last updated on 27/Apr/19
AP  is  (y−0)=m(x−5)  BQ is (y−0)=m(x+5)  solve y=m(x−5) and 4x+3y=25 to find P  4x+3m(x−5)=25  4x+3mx−15m=25  x=((25+15m)/(4+3m)) and y=m(((25+15m)/(4+3m))−5)                                         =m(((25+15m−20−15m)/(4+3m)))                                         =(((5m)/(4+3m)))  so P(((25+15m)/(4+3m)),((5m)/(4+3m)))  now solve y=m(x+5) and 4x+3y=25 to find Q    4x+3m(x+5)=25  4x+3mx+15m=25  x=((25−15m)/(4+3m)),y=m(((25−15m)/(4+3m))+5)  Q(((25−15m)/(4+3m)),(((25−15m+20+15m)/(4+3m)))×m  (((25−15m�)/(4+3m)),((45m)/(4+3m)))  PQ=(√({(_ ((25+15m)/(4+3m)))−_ (((25−15m_ )/(4+3m)))}^2 +(((45m−5m)/(4+3m)))^2 ))  =(√(((900m^2 )/((4+3m)^2 ))+((1600m^2 )/((4+3m)^2 ))))=5  ((50m)/(4+3m))=5  50m−15m=20   m=((20)/(35))=(4/7)  ((50m)/(4+3m))=−5  50m=−20−15m  65m=−20   m=((−4)/(13))
$${AP}\:\:{is}\:\:\left({y}−\mathrm{0}\right)={m}\left({x}−\mathrm{5}\right) \\ $$$${BQ}\:{is}\:\left({y}−\mathrm{0}\right)={m}\left({x}+\mathrm{5}\right) \\ $$$${solve}\:{y}={m}\left({x}−\mathrm{5}\right)\:{and}\:\mathrm{4}{x}+\mathrm{3}{y}=\mathrm{25}\:{to}\:{find}\:{P} \\ $$$$\mathrm{4}{x}+\mathrm{3}{m}\left({x}−\mathrm{5}\right)=\mathrm{25} \\ $$$$\mathrm{4}{x}+\mathrm{3}{mx}−\mathrm{15}{m}=\mathrm{25} \\ $$$${x}=\frac{\mathrm{25}+\mathrm{15}{m}}{\mathrm{4}+\mathrm{3}{m}}\:{and}\:{y}={m}\left(\frac{\mathrm{25}+\mathrm{15}{m}}{\mathrm{4}+\mathrm{3}{m}}−\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={m}\left(\frac{\mathrm{25}+\mathrm{15}{m}−\mathrm{20}−\mathrm{15}{m}}{\mathrm{4}+\mathrm{3}{m}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{5}{m}}{\mathrm{4}+\mathrm{3}{m}}\right) \\ $$$${so}\:{P}\left(\frac{\mathrm{25}+\mathrm{15}{m}}{\mathrm{4}+\mathrm{3}{m}},\frac{\mathrm{5}{m}}{\mathrm{4}+\mathrm{3}{m}}\right) \\ $$$${now}\:{solve}\:{y}={m}\left({x}+\mathrm{5}\right)\:{and}\:\mathrm{4}{x}+\mathrm{3}{y}=\mathrm{25}\:{to}\:{find}\:{Q} \\ $$$$\:\:\mathrm{4}{x}+\mathrm{3}{m}\left({x}+\mathrm{5}\right)=\mathrm{25} \\ $$$$\mathrm{4}{x}+\mathrm{3}{mx}+\mathrm{15}{m}=\mathrm{25} \\ $$$${x}=\frac{\mathrm{25}−\mathrm{15}{m}}{\mathrm{4}+\mathrm{3}{m}},{y}={m}\left(\frac{\mathrm{25}−\mathrm{15}{m}}{\mathrm{4}+\mathrm{3}{m}}+\mathrm{5}\right) \\ $$$${Q}\left(\frac{\mathrm{25}−\mathrm{15}{m}}{\mathrm{4}+\mathrm{3}{m}},\left(\frac{\mathrm{25}−\mathrm{15}{m}+\mathrm{20}+\mathrm{15}{m}}{\mathrm{4}+\mathrm{3}{m}}\right)×{m}\right. \\ $$$$\left(\frac{\mathrm{25}−\mathrm{15}{m}}{\mathrm{4}+\mathrm{3}{m}},\frac{\mathrm{45}{m}}{\mathrm{4}+\mathrm{3}{m}}\right) \\ $$$${PQ}=\sqrt{\left\{\left(_{} \frac{\mathrm{25}+\mathrm{15}{m}}{\mathrm{4}+\mathrm{3}{m}}\right)−_{} \left(\frac{\mathrm{25}−\mathrm{15}{m}_{} }{\mathrm{4}+\mathrm{3}{m}}\right)\right\}^{\mathrm{2}} +\left(\frac{\mathrm{45}{m}−\mathrm{5}{m}}{\mathrm{4}+\mathrm{3}{m}}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\frac{\mathrm{900}{m}^{\mathrm{2}} }{\left(\mathrm{4}+\mathrm{3}{m}\right)^{\mathrm{2}} }+\frac{\mathrm{1600}{m}^{\mathrm{2}} }{\left(\mathrm{4}+\mathrm{3}{m}\right)^{\mathrm{2}} }}=\mathrm{5} \\ $$$$\frac{\mathrm{50}{m}}{\mathrm{4}+\mathrm{3}{m}}=\mathrm{5} \\ $$$$\mathrm{50}{m}−\mathrm{15}{m}=\mathrm{20}\: \\ $$$${m}=\frac{\mathrm{20}}{\mathrm{35}}=\frac{\mathrm{4}}{\mathrm{7}} \\ $$$$\frac{\mathrm{50}{m}}{\mathrm{4}+\mathrm{3}{m}}=−\mathrm{5} \\ $$$$\mathrm{50}{m}=−\mathrm{20}−\mathrm{15}{m} \\ $$$$\mathrm{65}{m}=−\mathrm{20}\: \\ $$$${m}=\frac{−\mathrm{4}}{\mathrm{13}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by peter frank last updated on 28/Apr/19
thank you
$${thank}\:{you} \\ $$

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