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.