The area of the triangle formed by the inters | vedclass

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(A) The line parallel to the $X$-axis passing through $P(h, k)$ is $y=k$.The intersection of $y=k$ and $y=x$ is $B(k, k)$.The intersection of $y=k$ an

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$x=y-1$

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(A) The line parallel to the $X$-axis passing through $P(h, k)$ is $y=k$.The intersection of $y=k$ and $y=x$ is $B(k, k)$.The intersection of $y=k$ and $x+y=2$ is $C(2-k, k)$.The intersection of $y=x$ and $x+y=2$ is $A(1, 1)$.The area of $\Delta ABC$ is given by $\frac{1}{2} |x_A(y_B-y_C) + x_B(y_C-y_A) + x_C(y_A-y_B)| = h^2$.$\frac{1}{2} |1(k-k) + k(k-1) + (2-k)(1-k)| = h^2$$\frac{1}{2} |0 + k^2 - k + 2 - 2k - k + k^2| = h^2$$\frac{1}{2} |2k^2 - 4k + 2| = h^2$$|k^2 - 2k + 1| = h^2$$(k-1)^2 = h^2$Taking square root on both sides, $k-1 = \pm h$.Replacing $(h, k)$ with $(x, y)$, we get $y-1 = \pm x$.Thus, $x = y-1$ or $x = -(y-1)$.

The area of the triangle formed by the intersection of a line parallel to $X$-axis and passing through $P(h, k)$ with the lines $y=x$ and $x+y=2$ is $h^{2}$. The locus of the point $P$ is

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