$\begin {array} {1 1} (A)\;k\: sq\: units & \quad (B)\;2k\: sq\: units \\ (C)\;k^2\: sq\: units & \quad (D)\;2k^2\: sq\: units \end {array}$

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- Area of a triangle whose vertices are $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is $ \large\frac{1}{2}$$ | x_1 (y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Equations of the given lines are :

$ \qquad y-x=0$---------(1)

$ \qquad x+y=0$----------(2)

$ \qquad x-k=0$----------(3)

Let us now find the points of intersection of the above lines

Solving equation (1) and (2) we get,

$\qquad x+y=0$

$ \qquad y-x=0$

We get $x=0, y=0$

Hence the point of intersection of line (1) and (2) is (0,0)

Next let us solve equation (2) and (3)

$ \qquad x+y=0$

$ \qquad x = k$

$\: \: \Rightarrow y = -k$

Hence the point of intersection of equation (2) and (3) is (k, -k)

Next let us solve equation (3) and (1)

$ \qquad x-k=0$

$ \qquad y-x=0$

$ \qquad y-k=0$

$ \Rightarrow y=k$ also $ x = k$

$ \therefore $ The point of intersection is (k, k)

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

Now substituting the values in

area = $\large\frac{1}{2}$$ | x_1 (y_2-y_1)+x_2(y_3-y_1)+x_3(y_1-y_2)|$ sq. units

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

$ = \large\frac{1}{2}|k^2+k^2|$ sq units

$ = \large\frac{1}{2}$$ \times 2k^2$ sq. units or $k^2$ sq. units

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