Solution

Let \(P(x,y)\) be any point on AB. Then, area of triangle ABP is zero (Why?). So \(\frac{1}{2}|\begin{smallmatrix}0&0&1\\ 1&3&1\\ x&y&1\end{smallmatrix}|=0\) This gives \(\frac{1}{2}(y-3x)=0\) or \(y=3x\), which is the equation of required line AB. Also, since the area of the triangle ABD is 3 sq. units, we have \(\frac{1}{2}|\begin{smallmatrix}1&3&1\\ 0&0&1\\ k&0&1\end{smallmatrix}|=\pm3\) This gives, \(\frac{-3k}{2}=\pm3\) , i.e., \(k=\mp2\)