Question

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

• A. $x+y=1$
• B. $x+y=2$
• C. $x+y=2 x y$
• D. $2 x+2 y=1$

Answer 1

Answer: (A)

The equation of line $A B$ is $y-1=m(x-1)$.
Therefore, the equation of line $A C$ is
$y-1=-\frac{1}{m}(x-1)$

$2 h=1-\frac{1}{m} $
$2 k=1+\frac{1}{m}$
Eliminating $m$, we have locus $x+y=1$.