Find the equation of line joining (3,2) and (-1,-3) using determinants

1.	Find the equation of line joining (3,2) and (-1,-3) using determinants.
Answer» Let arbitrary point (x ,y) lies on the line joining points (3,2) and (-1,-3). \(\therefore\) \(\begin{vmatrix}x & y & 1\\[0.3em] 3 & 2 &1 \\[0.3em] -1& -3 & 1 \end{vmatrix} = 0\) \(\Rightarrow\) \(x\begin{vmatrix}2 &1 \\[0.3em] -3 & 1 \end{vmatrix} - y\begin{vmatrix}3&1 \\[0.3em]-1&1\end{vmatrix}+1\begin{vmatrix}3 &2 \\[0.3em] -1 & -3 \end{vmatrix}=0\) \(\Rightarrow\) x(2 + 3) - y(3 + 1) + 1(-9 + 2) = 0 \(\Rightarrow\) 5x - 4y - 7 = 0 Which is equation of required line passes through points (3,2) & (-1,-3).

Find the equation of line joining (3,2) and (-1,-3) using determinants.

Answer»

Let arbitrary point (x ,y) lies on the line joining points (3,2) and (-1,-3).

\(\therefore\) \(\begin{vmatrix}x & y & 1\\[0.3em] 3 & 2 &1 \\[0.3em] -1& -3 & 1 \end{vmatrix} = 0\)

\(\Rightarrow\) \(x\begin{vmatrix}2 &1 \\[0.3em] -3 & 1 \end{vmatrix} - y\begin{vmatrix}3&1 \\[0.3em]-1&1\end{vmatrix}+1\begin{vmatrix}3 &2 \\[0.3em] -1 & -3 \end{vmatrix}=0\)

\(\Rightarrow\) x(2 + 3) - y(3 + 1) + 1(-9 + 2) = 0

\(\Rightarrow\) 5x - 4y - 7 = 0

Which is equation of required line passes through points (3,2) & (-1,-3).

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