Question

A point $P(x,y)$ moves so that the sum of the distance from $P$ to the coordinate axes is equal to 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)

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

We know that,

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

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

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

Now,

According to given question

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

$x+y=\sqrt{x^2+1^2-2x+y^2+1^2-2y}$

On squaring both side and we get,

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

$x^2+y^2+2xy=x^2+y^2-2x-2y+2$

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

$xy=-x-y+1$

$xy+x+y-1=0$

$x(y+1)+y-1+(1-1)=0$ on adding and subtracting 1 (one)

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

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

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

This the locus of point$P(x,y)$.

Option $\left(B\right)$ is the correct answer.