Question Number 66746 by John Kaloki Musau last updated on 20/Aug/19
$${A}\:{point}\:{T}\:\:{divides}\:{a}\:{line}\:{AB}\:{internally}\:{in}\:{the}\:{ratio}\:\mathrm{5}:\mathrm{2}.\:{Given}\:{that}\:{A}\:{is}\:\left(-\mathrm{4},\mathrm{10}\right)\:{and}\:{B}\:{is}\:\left(\mathrm{10},\mathrm{3}\right),\:{find}\:{the}\:{coordinates}\:{of}\:{T}. \\ $$
Answered by mr W last updated on 19/Aug/19
$${x}_{{T}} ={x}_{{A}} +\frac{\mathrm{5}}{\mathrm{7}}\left({x}_{{B}} −{x}_{{A}} \right)=−\mathrm{4}+\frac{\mathrm{5}}{\mathrm{7}}\left(\mathrm{10}+\mathrm{4}\right)=\mathrm{6} \\ $$$${y}_{{T}} ={y}_{{A}} +\frac{\mathrm{5}}{\mathrm{7}}\left({y}_{{B}} −{y}_{{A}} \right)=\mathrm{10}+\frac{\mathrm{5}}{\mathrm{7}}\left(\mathrm{3}−\mathrm{10}\right)=\mathrm{5} \\ $$$$\Rightarrow{T}\left(\mathrm{6},\mathrm{5}\right) \\ $$
Commented by John Kaloki Musau last updated on 19/Aug/19
correct
Answered by Rio Michael last updated on 19/Aug/19
$$\:{T}\left({x},{y}\right)\:=\left(\:\frac{\mu{x}_{\mathrm{1}} \:+\:\lambda{x}_{\mathrm{2}} }{\mu\:+\:\lambda},\:\frac{\mu{y}_{\mathrm{1}} \:+\:\lambda{y}_{\mathrm{2}} }{\mu\:+\:\lambda}\right)……..{internal}\:{division} \\ $$$${T}\left({x},{y}\right)\:=\:\left(\frac{\mathrm{5}\left(−\mathrm{4}\right)\:+\:\mathrm{2}\left(\mathrm{10}\right)}{\mathrm{5}+\mathrm{2}}\:,\:\frac{\mathrm{5}\left(\mathrm{10}\right)\:+\:\mathrm{2}\left(\mathrm{3}\right)}{\mathrm{5}\:+\:\mathrm{2}}\right) \\ $$$${T}\left({x},{y}\right)\:=\:\left(\mathrm{0},\:\frac{\mathrm{56}}{\mathrm{7}}\right) \\ $$$$ \\ $$
Commented by mr W last updated on 20/Aug/19
$${please}\:{check}\:{sir}: \\ $$$${it}\:{should}\:{be} \\ $$$${T}\left({x},{y}\right)\:=\:\left(\frac{\mathrm{2}\left(−\mathrm{4}\right)\:+\:\mathrm{5}\left(\mathrm{10}\right)}{\mathrm{5}+\mathrm{2}}\:,\:\frac{\mathrm{2}\left(\mathrm{10}\right)\:+\:\mathrm{5}\left(\mathrm{3}\right)}{\mathrm{5}\:+\:\mathrm{2}}\right) \\ $$$${T}\left({x},{y}\right)\:=\:\left(\mathrm{6},\:\mathrm{5}\right) \\ $$
Commented by mr W last updated on 20/Aug/19
$$\:{T}\left({x},{y}\right)\:=\left(\:\frac{\mu{x}_{\mathrm{2}} \:+\:\lambda{x}_{\mathrm{1}} }{\mu\:+\:\lambda},\:\frac{\mu{y}_{\mathrm{2}} \:+\:\lambda{y}_{\mathrm{1}} }{\mu\:+\:\lambda}\right)……..{internal}\:{division} \\ $$
Commented by Rio Michael last updated on 21/Aug/19
$${your}\:{right}\:{sir}\:{thanks} \\ $$