Given $A \equiv (1,1)$ and $AB$ is any line through it cutting the $x$-axis in $B$. If $AC$ is perpe

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Given $A \equiv (1,1)$ and $AB$ is any line through it cutting the $x$-axis in $B$. If $AC$ is perpendicular to $AB$ and meets the $y$-axis in $C$, then the equation of locus of mid-point $P$ of $BC$ is:

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Sol. [Key Concept: Midpoint locus of $BC$ with $AB \perp AC$ and $B,C$ on the axes is $x + y = 1$.]$$y - 1 = m(x - 1)$$$$\begin{matrix} & y - 1 = - \frac{1}{m}(x - 1) \ & 2h = 1 - \frac{1}{m} \end{matrix}$$$$2k = 1 + .\frac{1}{m} $$locus is $x + y = 1$

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