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

(i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.

$\therefore \frac{1}{2}\left|\begin{array}{lll}1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1\end{array}\right|=0$

$\Rightarrow \frac{1}{2}[1(6-y)-2(3-x)+1(3 y-6 x)]=0$

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

$\Rightarrow 2 y-4 x=0$

$\Rightarrow y=2 x$

Hence, the equation of the line joining the given points is y = 2x.

(ii) Let P (x, y) be any point on the line joining points A (3, 1) and

B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of 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 − 3y = 0.