Solving Lesson 81 Page 108 Math 10 – Kite >


For death consciousness ABCD. USA is the point of change in the plane satisfying \(\left( {\overrightarrow {MA} + \overrightarrow {MB} } \right).\left( {\overrightarrow {MC} + \overrightarrow {MD} } \right) = 0\). Prove that point USA always lie on a fixed circle.

Solution method – See details

SUse the midpoint property of the line segment to find the set of points USA

Detailed explanation

By assumption, \(\left( {\overrightarrow {MA} + \overrightarrow {MB} } \right).\left( {\overrightarrow {MC} + \overrightarrow {MD} } \right) = 0\)\ ( \Leftrightarrow \left[\begin{array}{l}\overrightarrow{MA}+\overrightarrow{MB}=0\\\overrightarrow{MC}+\overrightarrow{MD}=0\end{array}\right\)[\begin{array}{l}\overrightarrow{MA} +\overrightarrow{MB} =0\\\overrightarrow{MC} +\overrightarrow{MD} =0\end{array}\right\)

Call P, Q is the midpoint of ., respectively AB and CD \( \Rightarrow \left\{ \begin{array}{l}\overrightarrow {MA} + \overrightarrow {MB} = 2\overrightarrow {MP} \\\overrightarrow {MC} + \overrightarrow {MD} = 2\ overrightarrow {MQ} \end{array} \right.\)

\( \Rightarrow \left( {\overrightarrow {MA} + \overrightarrow {MB} } \right).\left( {\overrightarrow {MC} + \overrightarrow {MD} } \right) = 0 \Leftrightarrow 2\overrightarrow {MP} .2\overrightarrow {MQ} = 0 \Leftrightarrow \overrightarrow {MP} .\overrightarrow {MQ} = 0\)

+ If USA does not coincide with P or Q then \(\overrightarrow {MP} .\overrightarrow {MQ} = 0 \Leftrightarrow MP \bot MQ\)

\( \Rightarrow \) Set of points USA is the diameter circle PQ

+ If USA coincides with P or Q it’s obvious USA belonging to the circle of diameter PQ

So USA always belongs to the circle of diameter PQ permanent

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