Question

A point $P(x,y)$ moves so that the sum of the distances from $P$ to the coordinates axes is equal to the distance from $P$ to the point $A(1,1)$. The equation of the locus of $P$ in the first quadrant is

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

Answer 1

Answer: (B)

$(x+1)(y+1)=2$

Distance between $P$ and $X-$axis$y$

Distance between $P$ and $Y-$axis$x$

Distance between $P$ and $A(1,1)$axis$=\sqrt{{(x-1)}^2+{(y-1)}^2}$

According to the question, $x+y=\sqrt{{(x-1)}^2+{(y-1)}^2}$

${(x+y)}^2=x^2+1-2x+y^2-2y+1$

$\Rightarrow x^2+y^2+2xy=x^2+1-2x+y^2-2y+1$

$\Rightarrow 2xy=-2x-2y+2$

$\therefore x+y+xy=1$

$\therefore x+y+xy+1=2$

$\therefore x(1+y)+1(y+1)=2$

or $(x+1)(y+1)=2$ is the locus of point $P$