The given equation of lines are y – x = 0 ... (i) x + y = 0 ... (ii) x – k = 0 ... (iii) Solving (i) and (ii), we get x = y = 0. ∴ (i) and (ii) meets at (0, 0) ... (iv) Solving (i) and (iii), we get x = y = k ... (v) ∴ (i) and (iii) meets at (k, k) Similarly, (ii) and (iii) meets at (k, –k) ....(vi) Thus, (iv), (v) and (vi) shows the vertices of the triangle formed by (i), (ii) & (iii) i.e., (0, 0), (k, k) and (k, – k). Now, $\begin{vmatrix} 0 & 0 & 1 \\ k & k & 1 \\ k & -k & 1 \end{vmatrix}$ = $[1(-k^{2}-k^{2})]$ = $-2k^{2}$ Area of triangle = $\frac{1}{2}\left|-2k^{2}\right|$ = $\frac{1}{2} \times 2k^{2}$ = $k^{2}$ sq. units.