Find the area of the triangle formed by the lines y−x=0, x+y=0 and x−k=0.

Simplification of given data.

The equation of the given lines are

$\begin{array}{l} y-x=0 \ldots(1) \\ x+y=0 \ldots (2)\\ x-k=0 \ldots(3) \end{array}$

The point of intersection of lines (1) and (2):

From (1), $y=x$

Putting it in (2), $x+x=0 \Rightarrow x=0$

Putting it in (1), $y=0$

So, point of intersection is $(0,0)$

The point of intersection of lines (2) and (3):

From (3), $\mathrm{x}=\mathrm{k}$

Putting it in (1), $y-k=0 \Rightarrow y=k$

So, point of intersection is $(k, k)$

Thus, the vertices of the triangle formed by the three given lines are $(0,0),(k,-k)$, and $(k, k)$.

Area of triangle

We know that area of triangle whose vertices are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is

$=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

Area of triangle $\triangle O A B$ whose vertices are $O(O, O), A(k, k)$ and $B(k,-k)$

$\begin{array}{l} =\frac{1}{2}|O(k-(-k))+k(-k-0)+k(O-k)| \\ =\frac{1}{2}|O+k(-k)+k(-k)| \\ =\frac{1}{2}\left|-k^{2}-k^{2}\right| \\ =\frac{1}{2}\left|-2 k^{2}\right| \\ =\frac{2 k^{2}}{2} \\ =k^{2} \text { square units. } \end{array}$

Final answer:

Therefore, required area of triangle is $k^{2}$ square units.