Question Number 121418 by deleteduser12 last updated on 08/Nov/20
$${a}\:{changeable}\:{straight}\:{line}\:{going}\:{the}\:{point}\:\left({a},{b}\right) \\ $$$${creates}\:{a}\:{triangle}\:{with}\:{axes}.\:{what}\:{is}\: \\ $$$${the}\:{equation}\:{of}\:{locus}\:{of}\:{centroid}\:{of} \\ $$$${the}\:{triangle} \\ $$
Answered by TANMAY PANACEA last updated on 08/Nov/20
$${let}\:{the}\:{eqn}\:{be} \\ $$$$\frac{{x}}{{p}}+\frac{{y}}{{q}}=\mathrm{1}\rightarrow{st}\:{line}\:{meet}\:{x}\:{axis}\:{at}\left({p},\mathrm{0}\right)\:\:{yaxis} \\ $$$${at}\:\left(\mathrm{0},{q}\right) \\ $$$${centroid}=\left(\frac{{p}+\mathrm{0}+\mathrm{0}}{\mathrm{3}},\frac{\mathrm{0}+{q}+\mathrm{0}}{\mathrm{3}}\right)=\left(\frac{{p}}{\mathrm{3}},\frac{{q}}{\mathrm{3}}\right) \\ $$$${say}\:\:\alpha=\frac{{p}}{\mathrm{3}}\:\:\beta=\frac{{q}}{\mathrm{3}} \\ $$$${now}\:{st}\:{line}\:\frac{{x}}{{p}}+\frac{{y}}{{q}}=\mathrm{1}\:{passes}\:{through}\:\left({a},{b}\right)\:{so} \\ $$$$\frac{{a}}{{p}}+\frac{{b}}{{q}}=\mathrm{1} \\ $$$$\frac{\boldsymbol{{a}}}{\mathrm{3}\alpha}+\frac{{b}}{\mathrm{3}\beta}=\mathrm{1} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{the}}\:\boldsymbol{{locus}}\:\boldsymbol{{is}}\:\:\frac{\boldsymbol{{a}}}{\mathrm{3}\boldsymbol{{x}}}+\frac{\boldsymbol{{b}}}{\mathrm{3}\boldsymbol{{y}}}=\mathrm{1} \\ $$$$\frac{\boldsymbol{{a}}}{\boldsymbol{{x}}}+\frac{\boldsymbol{{b}}}{\boldsymbol{{y}}}=\mathrm{3} \\ $$$$ \\ $$$$ \\ $$
Commented by peter frank last updated on 08/Nov/20
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by TANMAY PANACEA last updated on 08/Nov/20
$${most}\:{welcome} \\ $$