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(A) The equations 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 $(0, 0)$.The poi

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$k^2$ square units

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(A) The equations 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 $(0, 0)$.The point of intersection of lines $(2)$ and $(3)$ is found by substituting $x=k$ into $x+y=0$,which gives $y=-k$. So,the point is $(k, -k)$.The point of intersection of lines $(3)$ and $(1)$ is found by substituting $x=k$ into $y-x=0$,which gives $y=k$. So,the point is $(k, k)$.Thus,the vertices of the triangle are $(0, 0)$,$(k, -k)$,and $(k, k)$.The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by:$Area = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$Substituting the vertices $(0, 0)$,$(k, -k)$,and $(k, k)$:$Area = \frac{1}{2} |0(-k-k) + k(k-0) + k(0-(-k))|$$Area = \frac{1}{2} |0 + k^2 + k^2|$$Area = \frac{1}{2} |2k^2| = k^2$ square units.

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

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