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# Find equation of line joining (3, 1) and (9, 3) using determinants.

$\begin{array}{1 1}x+3y = 0 \\ x=3y \\ x=2y \\ x+2y=0 \end{array}$

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A)
• The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
• $\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
• $|\bigtriangleup|=\frac{1}{2}\begin{bmatrix}x_1\begin{vmatrix}y_2 & 1\\y_3 & 1\end{vmatrix}-y_1\begin{vmatrix}x_2 & 1\\x_3 & 1\end{vmatrix}+1\begin{vmatrix}x_2 & y_2\\x_3 & y_3\end{vmatrix}\end{bmatrix}$
Let the points (x,y),(3,1),(9,3) be $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$respectively.

The area of the triangle is given by

$\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$

Now substituting the respective values,and since this is a line,the points are taken to be collinear.

Hence the area is 0.

$0=\frac{1}{2}\begin{vmatrix}x & y & 1\\3 & 1 & 1\\9 & 3 & 1\end{vmatrix}$

Now expanding along the row $R_1$ we get

$0=\frac{1}{2}\begin{bmatrix}x\begin{vmatrix}1 & 1\\3 & 1\end{vmatrix}-y\begin{vmatrix}3 & 1\\9 & 1\end{vmatrix}+1\begin{vmatrix}3 & 1\\9& 3\end{vmatrix}\end{bmatrix}$

0=x(1-3)-y(3-9)+1(9-9)

$\Rightarrow x(-2)-y(-6)+0.$

$\Rightarrow -2x+6y=0.\Rightarrow x=3y.$