Find the equation of the line joining A(5,1), B(4,0) using determinants

The correct option is x-y=4.

To find the equation of a line joining two points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ using determinants, we can assume a general point $P(x,y)$ on the line. Since these three points are collinear (lie on the same straight line), the area of the triangle formed by them is zero.

The condition for three points $(x,y)$, $(x_1,y_1)$, and $(x_2,y_2)$ to be collinear is given by the determinant equation:
$$\left|\begin{array}{ccc}x& y& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|=0$$

Given the points $A(5,1)$ and $B(4,0)$, we substitute $x_1=5$, $y_1=1$, $x_2=4$, and $y_2=0$ into the determinant:
$$\left|\begin{array}{ccc}x& y& 1\\ 5& 1& 1\\ 4& 0& 1\end{array}\right|=0$$

Now, we expand the determinant along the first row:
$$x·\left|\begin{array}{cc}1& 1\\ 0& 1\end{array}\right|-y·\left|\begin{array}{cc}5& 1\\ 4& 1\end{array}\right|+1·\left|\begin{array}{cc}5& 1\\ 4& 0\end{array}\right|=0$$

Evaluating the 2×2 determinants:
$\left|\begin{array}{cc}1& 1\\ 0& 1\end{array}\right|=(1·1)-(1·0)=1-0=1$
$\left|\begin{array}{cc}5& 1\\ 4& 1\end{array}\right|=(5·1)-(1·4)=5-4=1$
$\left|\begin{array}{cc}5& 1\\ 4& 0\end{array}\right|=(5·0)-(1·4)=0-4=-4$

Substitute these values back into the expanded equation:
$$x\left(1\right)-y\left(1\right)+1(-4)=0$$

Simplifying the terms, we get:
$$x-y-4=0$$

Adding 4 to both sides gives:
$$x-y=4$$

Thus, the equation of the line joining the given points is x-y=4.