STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

Then, the points $A, B$ and $P$ are collinear. Therefore, the area of the triangle $ABP$ will be zero.

$\therefore \frac{1}{2}\left|\begin{array}{lll}3 & 1 & 1 \\ 9 & 3 & 1 \\ x & y & 1\end{array}\right|=0$

$\Rightarrow \frac{1}{2}[3(3-y)-1(9-x)+1(9 y-3 x)]=0$

$\Rightarrow 9-3 y-9+x+9 y-3 x=0$

$\Rightarrow 6 y-2 x=0$

$\Rightarrow x-3 y=0$

Hence, the equation of the line joining the given points is $x-3 y=0$