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Question
Show that the points 2 \[\hat{i}, - \hat{i}-4 \] \[\hat{j}\] and \[-\hat{i}+4\hat{j}\] form an isosceles triangle.
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Solution
Given:- The points A, B, C with position vectors \[\vec{a} ,\vec{b} , \vec{c}\] respectively.
Also,
\[\vec{a} = 2 \hat{i}\]
\[\vec{b} = - \hat{i} - 4 \hat{j}\]
\[\vec{c} = - \hat{i} + 4 \hat{j}\]
Then,
\[\overrightarrow{AB} = \vec{b} - \vec{a} \]
\[ \Rightarrow \overrightarrow{AB} = \left( - \hat{i} - 4 \hat{j} \right) - 2 \hat{i} \]
\[ \Rightarrow \overrightarrow{AB} = - 3 \hat{i} - 4 \hat{j} \]
\[\text{ Now, }\left| \overrightarrow{AB} \right| = \sqrt{\left( - 3 \right)^2 + \left( - 4 \right)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\]
\[ \overrightarrow {BC} = \vec{c} - \vec{b} \]
\[ \Rightarrow \overrightarrow {BC} = \left( - \hat{i} + 4 \hat{j} \right) - \left( - \hat{i} - 4 \hat{j} \right)\]
\[ \Rightarrow \overrightarrow {BC} = - \hat{i} + 4 \hat{j} + \hat{i} + 4 \hat{j} \]
\[ \Rightarrow \overrightarrow{BC} = 8 \hat{j}\]
and
\[\overrightarrow{AC} = \vec{c} - \vec{a} \]
\[ \Rightarrow \overrightarrow{AC} = \left( - \hat{i} + 4 \hat{j} \right) - 2 \hat{i} \]
\[ \Rightarrow \overrightarrow{AC} = - 3 \hat{i} + 4 \hat{j} \]
\[\text{Now, }\left| \overrightarrow {AC} \right| = \sqrt{\left( - 3 \right)^2 + \left( 4 \right)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\]
Since, the magnitude of AB and AC is equal.
Hence, the points 2 \[\hat{i}, - \hat{i}-4 \] \[\hat{j}\] and \[\hat{i}+4\] form an isosceles triangle.
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