Question Number 133937 by rexford last updated on 25/Feb/21
Answered by EDWIN88 last updated on 25/Feb/21
$$\boldsymbol{{AB}}\:=\:\hat {\boldsymbol{\mathrm{i}}}+\mathrm{6}\hat {\boldsymbol{\mathrm{j}}}\:,\:\mathrm{let}\:\mathrm{vector}\:\boldsymbol{\mathrm{u}}\:=\:\boldsymbol{\mathrm{AC}}\:\mathrm{where}\:\mathrm{C}\left(\mathrm{x},\mathrm{y}\right) \\ $$$$\boldsymbol{\mathrm{u}}=\left(\mathrm{x}−\mathrm{1},\mathrm{y}+\mathrm{1}\right)\:=\left(\mathrm{x}−\mathrm{1}\right)\hat {\boldsymbol{\mathrm{i}}}+\left(\mathrm{y}+\mathrm{1}\right)\hat {\boldsymbol{\mathrm{j}}} \\ $$$$\Rightarrow\:\boldsymbol{\mathrm{u}}.\boldsymbol{\mathrm{AB}}\:=\mathrm{0}\:\Rightarrow\mathrm{x}−\mathrm{1}+\mathrm{6y}+\mathrm{6}\:=\:\mathrm{0} \\ $$$$\:\mathrm{x}+\mathrm{6y}\:=\:−\mathrm{5}\:,\:\mathrm{x}=−\mathrm{6y}−\mathrm{5} \\ $$$$\boldsymbol{\mathrm{u}}\:=\:\left(−\mathrm{6y}−\mathrm{6}\right)\hat {{i}}+\left(\mathrm{y}+\mathrm{1}\right)\hat {{j}}\: \\ $$$$ \\ $$