Question

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

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

Answer 1

The correct option is A $x+y=1$

Let point $B$ be $(b,0)$

Slope of $AB$ is $\frac{1}{1-b}$

$AC$ is perpendicular to $AB$ $\therefore$ its slope $=b-1$

Equation of $AC=y-1=(b-1)(x-1)⟹y=(b-1)x+2-b$

$C\equiv (0,2-b)$

Midpoint of $BC$ is $(\frac{b}{2},1-\frac{b}{2})$
Satisfy the Point above in the given option

thus locus is $x+y=1$