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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
aandb
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))
APis(y0)=m(x5)BQis(y0)=m(x+5)solvey=m(x5)and4x+3y=25tofindP4x+3m(x5)=254x+3mx15m=25x=25+15m4+3mandy=m(25+15m4+3m5)=m(25+15m2015m4+3m)=(5m4+3m)soP(25+15m4+3m,5m4+3m)nowsolvey=m(x+5)and4x+3y=25tofindQ4x+3m(x+5)=254x+3mx+15m=25x=2515m4+3m,y=m(2515m4+3m+5)Q(2515m4+3m,(2515m+20+15m4+3m)×m(2515m4+3m,45m4+3m)PQ={(25+15m4+3m)(2515m4+3m)}2+(45m5m4+3m)2=900m2(4+3m)2+1600m2(4+3m)2=550m4+3m=550m15m=20m=2035=4750m4+3m=550m=2015m65m=20m=413
Commented by peter frank last updated on 28/Apr/19
thank you
thankyou

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