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

• Correct answer
  A

  $x=y-1$
• Correct answer
  B

  $x=-(y-1)$
• C

  $x=1+y$
• D

  $x=-(1+y)$

Solution

Given, ar $(\Delta A B C)=h^{2}$ $\Rightarrow \quad \frac{1}{2}\left|\begin{array}{ccc}1 & 1 & 1 \\ k & k & 1 \\ 2-k & k & 1\end{array}\right|=\pm h^{2}$ $\Rightarrow 1(k-k)-1(k-2+k)+1\left(k^{2}-2 k+k^{2}\right)=\pm 2 h^{2}$ $\Rightarrow \quad-(2 k-2)+\left(2 k^{2}-2 k\right)=\pm 2 h^{2}$ $\Rightarrow \quad 2-2 k+2 k^{2}-2 k=\pm 2 h^{2}$ $\Rightarrow \quad 2 k^{2}-4 k+2=\pm 2 h^{2}$ $\Rightarrow$ $k^{2}-2 k+1=\pm h^{2}$ Hence, locus of a point is $\Rightarrow$ $(k-1)^{2}=h^{2}$ $y-1=\pm x$ $\Rightarrow \quad x=y-1$ or $x=-(y-1)$

Hence, the correct answer is options: 'A', 'B'.