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.
\(\frac{1}{2}\)\(\begin{vmatrix}1&2&1\\3&6&1\\x&y&1\end{vmatrix}\)=0
\(\Rightarrow\)\(\frac{1}{2}\)[1(6-y)-2(3-x)+1(3y-6x)]
=6-y-6+2x+3y-6x=0
\(\Rightarrow\) y=2x
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.
\(\frac{1}{2}\)\(\begin{vmatrix}3&1&1\\9&3&1\\x&y&1\end{vmatrix}\)
\(\Rightarrow\)\(\frac{1}{2}\)[3(3-y)-1(9-x)+1(9y-3x)]
\(\Rightarrow\) 9-3y-9+x+9y-3x=0
\(\Rightarrow\) x-3y=0
Hence, the equation of the line joining the given points is x − 3y = 0
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