Find the area of the triangle formed by the line $y-x=0, x+y=0$ and $x-k=0$.

The equation of the given lines are

$y-x=0 $.....$(1)$

$x+y=0$.....$(2)$

$x-k=0$.....$(3)$

The point of intersection of lines $(1)$ and $(2)$ is given by

$x=0$ and $y=0$

The point of intersection of lines $( 2 )$ and $( 3 )$ is given by

$x=k$ and $y=-k$

The point of intersection of lines $(3)$ and $(1)$ is given by

$x=k$ and $y=k$

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

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

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

Therefore, area of the triangle formed by the three given lines

$=\frac{1}{2}|0(-k-k)+k(k- 0)+k(0+k)|$square units

$=\frac{1}{2}\left|k^{2}+k^{2}\right|$square units

$=\frac{1}{2}\left|2 k^{2}\right|$ square umits

$=k^{2}$ square units