Find the equation of the line joining (1, 2) and (3, 6) using determinant.
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Find the equation of the line joining (1, 2) and (3, 6) using determinant.
Answer
Let \(p\left( {x,y} \right)\) be any point on the line joining the points (1, 2) and (3, 6).
Then, Area of triangle that could be formed by these points is zero.
\(\therefore\) Area of triangle = \(\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0\)
\(\Rightarrow\) \(\frac{1}{2}\left| {\begin{array}{*{20}{c}} x&y&1 \\ 1&2&1 \\ 3&6&1 \end{array}} \right| = 0\)
\(\Rightarrow \frac{1}{2}\left[ {x\left( {2 - 6} \right) - y\left( {1 - 3} \right) + 1\left( {6 - 6} \right)} \right] = 0\)
\(\Rightarrow - 4x + 2y = 0\)
\(\Rightarrow - 2x + y = 0\)
\( \Rightarrow y = 2x\) which is required line.
Then, Area of triangle that could be formed by these points is zero.
\(\therefore\) Area of triangle = \(\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0\)
\(\Rightarrow\) \(\frac{1}{2}\left| {\begin{array}{*{20}{c}} x&y&1 \\ 1&2&1 \\ 3&6&1 \end{array}} \right| = 0\)
\(\Rightarrow \frac{1}{2}\left[ {x\left( {2 - 6} \right) - y\left( {1 - 3} \right) + 1\left( {6 - 6} \right)} \right] = 0\)
\(\Rightarrow - 4x + 2y = 0\)
\(\Rightarrow - 2x + y = 0\)
\( \Rightarrow y = 2x\) which is required line.
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