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, |

0	0	1
k	k	1
k	−k	1

| = [1(−k2−k2)] = −2k2
Area of triangle =

1

2

|−2k2| =

1

2

×2k2 = k2 sq. units.