Question Number 121422 by faysal last updated on 08/Nov/20
$${a}\:{rectangle}'{s}\:{a}\:{side}'{s}\:{terminal} \\ $$$${points}\:{coordinate}\:{are}\:\left(\mathrm{2},−\mathrm{1}\right),\:\left(\mathrm{6},\mathrm{5}\right)\:{and}\:{the} \\ $$$${length}\:{of}\:{the}\:{diagonal}\:{is}\:\mathrm{8}\:{unit}. \\ $$$${what}\:{are}\:{coordinate}\:{of}\:{terminal}\:{point} \\ $$$${of}\:{paralel}\:{side}\:{that}\:{side} \\ $$
Commented by mr W last updated on 08/Nov/20
$${length}\:{of}\:{diagonal}\:{is} \\ $$$$\sqrt{\left(\mathrm{6}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{5}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{13}}\neq\mathrm{8} \\ $$$${with}\:{given}\:{conditions}\:{you}\:{can}\:{not} \\ $$$${determine}\:{the}\:{other}\:{diagonal}. \\ $$
Answered by $@y@m last updated on 08/Nov/20
$${The}\:{length}\:{of}\:{the}\:{line}\:{joining} \\ $$$$\left(\mathrm{2},−\mathrm{1}\right)\:{and}\:\left(\mathrm{6},\mathrm{5}\right)\:{is} \\ $$$$\left.\sqrt{\left\{\right.}\left(\mathrm{2}−\mathrm{6}\right)^{\mathrm{2}} +\left(−\mathrm{1}−\mathrm{5}\right)^{\mathrm{2}} \right\}=\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} }=\sqrt{\mathrm{52}} \\ $$$$\neq\mathrm{8} \\ $$$${Something}\:{wrong}\:{with}\:{question}. \\ $$$${Please}\:{check}. \\ $$