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# Find equation of line joining (1, 2) and (3, 6) using determinants

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

Toolbox:
• 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),(1,2),(3,6) be $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$respectively.

The area of the triangle is given by

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

Since this is a line,points are taken to be collinear,Hence the area is 0.Substituting the other values

$0=\frac{1}{2}\begin{vmatrix}x & y & 1\\1 & 2 & 1\\3 & 6 & 1\end{vmatrix}$

Now expanding along the row $R_1$ we get

$0=\frac{1}{2}\begin{bmatrix}x\begin{vmatrix}2 & 1\\6 & 1\end{vmatrix}-y\begin{vmatrix}1 & 1\\3 & 1\end{vmatrix}+1\begin{vmatrix}1 & 2\\3& 6\end{vmatrix}\end{bmatrix}$

0=x(2-6)-y(1-3)+1(6-6)

$\Rightarrow -4x-y(-2)+0=0.$

$\Rightarrow -4x+2y=0.\Rightarrow y=2x.$

edited Feb 24, 2013