It is given that

\[y\text{ }\text{ }x\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 1 \right)\]

\[x\text{ }+\text{ }y\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 2 \right)\]

\[x\text{ }\text{ }k\text{ }=\text{ }0\text{ }\ldots \text{ }.\text{ }\left( 3 \right)\]

Here the mark of crossing point of

Lines (1) and (2) is

\[x\text{ }=\text{ }0\text{ }and\text{ }y\text{ }=\text{ }0\]

Lines (2) and (3) is

\[x\text{ }=\text{ }k\text{ }and\text{ }y\text{ }=\text{ }\text{ }k\]

Lines (3) and (1) is

\[x\text{ }=\text{ }k\text{ }and\text{ }y\text{ }=\text{ }k\]

So the vertices of the triangle framed by the three given lines are \[\left( 0,\text{ }0 \right),\text{ }\left( k,\text{ }-\text{ }k \right)\text{ }and\text{ }\left( k,\text{ }k \right)\]

Here the space of triangle whose vertices are \[\left( x1,\text{ }y1 \right),\text{ }\left( x2,\text{ }y2 \right)\text{ }and\text{ }\left( x3,\text{ }y3 \right)\] is

\[{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\left| x1\text{ }\left( y2\text{ }\text{ }y3 \right)\text{ }+\text{ }x2\text{ }\left( y3\text{ }\text{ }y1 \right)\text{ }+\text{ }x3\text{ }\left( y1\text{ }\text{ }y2 \right) \right|\]

So the space of triangle framed by the three given lines

\[=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\left| 0\text{ }\left( -\text{ }k\text{ }\text{ }k \right)\text{ }+\text{ }\left( k\text{ }\text{ }0 \right)\text{ }+\text{ }k\text{ }\left( 0\text{ }+\text{ }k \right) \right|\] square units

By additional computation

\[=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\left| k2\text{ }+\text{ }k2 \right|\] square units

So we get

\[=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\left| 2k2 \right|\]

\[=\text{ }k2\] square units