Question

For points $P=(x_1,y_1)\text{and}Q=(x_2,y_2)$ of the co-ordinate 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 $A$ consinsts of the union of a line segment of finite length and an infinite ray. Sketch this set in a labelled diagram.

Answer 1

$\text{Let}P=(h,k)\text{be a general point in the first quadrant such that}d(P,A)=d(P,O)\Rightarrow |h-3|+|k-2|=|h|+|k|=h+k..\left(1\right)\left[\text{h and k are + ve, pt}\right(h,k)\text{being in I quadrant}.]$


$\text{If}h<3,k<2\text{then}(h,k)\text{lies in region I}.\text{If}h>3,k<2,(h,k)\text{lies in region II}.\text{If}h>3,k>2,(h,k)\text{lies in region III}.\text{If}h<3,k>2,(h,k)\text{lies in region IV}.\text{In region I, eq. (1)}\Rightarrow 3-h+2-k=h+k\Rightarrow h+k=\frac{5}{2}\text{In region II, eq. (1) becomes}\Rightarrow h-3+2-k=h+k\Rightarrow k=-\frac{1}{2}\text{;not possible}.\text{In region III, eq. (1) becomes}\Rightarrow h-3+k-2=h+k\Rightarrow -5=0\text{;not possible}.\text{In region IV, eq. (1) becomes}\Rightarrow 3-h+k-2=h+k\Rightarrow h=1/2$
$\Rightarrow$ Hence required set consists of line segment $x+y=5/2$ of finite length as shown in the first region and the ray $x=1/2$ in the fourth region.