Question Number 50625 by rahul 19 last updated on 18/Dec/18
$$\mathrm{2}\:{opposite}\:{vertices}\:{of}\:{a}\:{square}\:{are}\: \\ $$$$\left(\mathrm{5},\mathrm{4}\right)\:{and}\:\left(\mathrm{1},−\mathrm{6}\right).\:{Find}\:{coordinates} \\ $$$${of}\:{remaining}\:{two}\:{vertices}\:? \\ $$
Commented by $@ty@m last updated on 18/Dec/18
$${Plot}\:{these}\:{points}\:{in}\:{graph}\:{paper}. \\ $$
Commented by rahul 19 last updated on 18/Dec/18
$${Sir},\:{can}\:{you}\:{pl}\:{show}\:{your}\:{method}.. \\ $$
Answered by ajfour last updated on 18/Dec/18
$${center}\:\equiv\left(\mathrm{3},−\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×{diagonal}\:=\:\sqrt{\mathrm{29}} \\ $$$${slope}\:{m}=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${slope}\:{of}\:{CD}\:=\:−\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${x}_{{C}} \:=\:\mathrm{3}+\sqrt{\mathrm{29}}\left(\frac{−\mathrm{5}}{\:\sqrt{\mathrm{29}}}\right)\:=\:−\mathrm{2} \\ $$$${y}_{{C}} \:=\:−\mathrm{1}+\sqrt{\mathrm{29}}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{29}}}\right)\:=\:\mathrm{1} \\ $$$${x}_{{D}} \:=\:\mathrm{3}−\sqrt{\mathrm{29}}\left(\frac{−\mathrm{5}}{\mathrm{29}}\right)\:=\:\mathrm{8} \\ $$$${y}_{{D}} \:=\:−\mathrm{1}−\sqrt{\mathrm{29}}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{29}}}\right)\:=\:−\mathrm{3} \\ $$$${C}\:\equiv\left(−\mathrm{2},\:\mathrm{1}\right) \\ $$$${D}\:\equiv\left(\mathrm{8},\:−\mathrm{3}\right)\:. \\ $$
Commented by rahul 19 last updated on 18/Dec/18
thanks sir!��