Question Number 40847 by LXZ last updated on 28/Jul/18
$${If}\:{the}\:{coordinate}\:{of}\:{the}\:{points}\:{A}\: \\ $$$${and}\:{B}\:{be}\:\left(\mathrm{3},\mathrm{3}\right)\:{and}\:\left(\mathrm{7},\mathrm{6}\right)\:{then}\:{the} \\ $$$${length}\:{of}\:{the}\:{portion}\:{of}\:{the}\:{line}\: \\ $$$${AB}\:{intercepted}\:{between}\:{the}\:{axes}\:{is} \\ $$$$\left({a}\right)\:\:\frac{\mathrm{5}}{\mathrm{4}}\:\:\:\left({b}\right)\:\:\:\frac{\sqrt{\mathrm{10}}}{\mathrm{4}}\:\:\:\:\left({c}\right)\:\:\frac{\sqrt{\mathrm{13}}}{\mathrm{3}}\:\:\:\:\left({d}\right)\:{none} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18
$${eqn}\:{of}\:{st}\:{line}\:{is}\:{y}−\mathrm{3}=\frac{\mathrm{6}−\mathrm{3}}{\mathrm{7}−\mathrm{3}}\left({x}−\mathrm{3}\right) \\ $$$$\mathrm{4}\left({y}−\mathrm{3}\right)=\mathrm{3}\left({x}−\mathrm{3}\right) \\ $$$$\mathrm{4}{y}−\mathrm{12}=\mathrm{3}{x}−\mathrm{9} \\ $$$$−\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{12}−\mathrm{9} \\ $$$$−\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{3} \\ $$$$\frac{{x}}{−\mathrm{1}}+\frac{{y}}{\frac{\mathrm{3}}{\mathrm{4}}}=\mathrm{1} \\ $$$${so}\:{the}\:{line}\:{cuts}\:{x}\:{axis}\:{at}\left(−\mathrm{1},\mathrm{0}\right)\:\:{yaxis}\:{at}\left(\mathrm{0},\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$${so}\:{length}\:{of}\:{intercept}\:{between}\:{axes} \\ $$$$\sqrt{\left(\mathrm{0}+\mathrm{1}\right)^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{0}\right)^{\mathrm{2}} }\:\: \\ $$$$=\sqrt{\mathrm{1}+\frac{\mathrm{9}}{\mathrm{16}}}\:\:=\frac{\mathrm{5}}{\mathrm{4}} \\ $$