Question

For points $P=(x_1{y}_1)$ and $Q=(x_2,y_2)$ of the coordinate plane, a new distance $d(P,Q)$ is defined by $d(P,Q)=|x_1-x_2|+|y_1-y_2|.$
Let $O=(0,0)$ and $A=(3,2)$. Prove that the set of points in the first quadrant which are equidistant (with respect to the new distance) from $O$ and $A$ consists of the union of a line segment of finite length and an infinite ray. Sketch this set in a labelled diagram.

Answer 1

NOTE $d:(P,Q)=|x_1-x_2|+|y_1-y_2|$
It is new method of representing distance between two
points $P$ and $Q$ and in future very important in coordinate geometry.
Now, let $P(x,y)$ be any point in the first quadrant. We have
$d(P,0)=|x-0|+|y-0|=|x|+|y|=x+y$
$[\because x,y>0]$
$d(P,A)=|X-3|+|Y-2|$ [given]
$d(P,0)=D(P,A)$ [given]
$\Rightarrow x+y=|x-3|+|y-2|...(i)$
Case I When $0<x<3,0<y<2$
In this case, Eq. (i) becomes
$x+y=3-x+2-y\Rightarrow 2x+2y=5$
or $x+y=5/2$
Case II When $0<x<3,y\ge 2$
Now, Eq. (i) becomes
$x+y=3-x+y-2$
$\Rightarrow 2x=1\Rightarrow x=1/2$
Case III When $x\ge 3,0<y<2$

Now, Eq. (i) becomes
$x+y=x-3+2-y$
$\Rightarrow 2y=-1$ or $y=-1/2$
Hence, no solution.
Case IV When $x\ge 3,y\ge 2$
In this case, case I changes to
$x+y=x-3+y-2\Rightarrow 0=-5$
which is not possible.
Hence, the solution set is
$\{(x,y)|x=12,y\ge 2\}\cup \{(x,y)\}|$
$x+y=5/2,0<x<3,0<y>2\}$
The graph is given in adjoining figure.