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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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