Find the equation of the line joining A(2,1) and B(6,3) using determinants

The correct option is 2y-x=0.

To find the equation of a line joining two given points using determinants, we can use the condition of collinearity. Let $P(x,y)$ be any general point on the line joining the points $A(2,1)$ and $B(6,3)$. Since the points $P(x,y)$, $A(2,1)$, and $B(6,3)$ are collinear, the area of the triangle formed by them must be zero.

The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ in determinant form is given by:
$\Delta =\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|$

Setting the area to zero, we get the equation of the line:
$\left|\begin{array}{ccc}x& y& 1\\ 2& 1& 1\\ 6& 3& 1\end{array}\right|=0$

Now, we expand the determinant along the first row:
$x·\left|\begin{array}{cc}1& 1\\ 3& 1\end{array}\right|-y·\left|\begin{array}{cc}2& 1\\ 6& 1\end{array}\right|+1·\left|\begin{array}{cc}2& 1\\ 6& 3\end{array}\right|=0$

Evaluating each 2×2 determinant:
$x(1·1-1·3)-y(2·1-1·6)+1(2·3-1·6)=0$

Simplifying inside the parentheses:
$x(1-3)-y(2-6)+1(6-6)=0$

This simplifies further to:
$-2x-y(-4)+0=0$
$-2x+4y=0$

Dividing the entire equation by 2:
$-x+2y=0$

Rearranging the terms, we get:
$2y-x=0$

Thus, the required equation of the line is indeed 2y-x=0.