$\bullet$ If we have $2$ points $(x_1,y_1)$ and $(x_2,y_2) $ on $X-Y$ plane,
Then equation of line will be :- $(y - y_1) = \left ( \frac{y_2 - y_1}{x_2 - x_1} \right )(x-x_1)$
$\bullet$ Definition of $y=|x|$ says $y = \left\{\begin{matrix} x & x\geqslant 0 \\ -x & x < 0 \end{matrix}\right.$
Now,
$(i)$ Let, $x_1 = -3, x_2 = -1 , y_1 = -2 , y_2 = 2 $
So, Equation of Line will be :- $y - (-2) = \left ( \frac{2-(-2)}{-1 -(-3)} \right )(x-(-3))$
$\Rightarrow y+2 = \frac{4}{2}(x+3)$
$\Rightarrow y= 2x+4$
So, option (i) is correct.
$(ii)$ $y = \left\{\begin{matrix} (x-1), &\; (x-1) \geqslant 0 \; (or) x \geqslant 1 \\ -(x-1)= -x + 1, & \; (x-1) < 0 \; (or) x < 1 & \end{matrix}\right.$
For, $x \geqslant 1,$
Let, $x_1 = 1, x_2 = 2 , y_1 = 0 , y_2 = 1 $
So, Equation of Line will be :- $y - 0 = \left ( \frac{1-0}{2 -1} \right )(x-1)$
$\Rightarrow y =(x-1)$
For, $x < 1,$
Let, $x_1 = -1, x_2 = 1 , y_1 = 2 , y_2 = 0 $
So, Equation of Line will be :- $y - 2 = \left ( \frac{0-2}{1-(-1)} \right )(x-(-1))$
$\Rightarrow y =-x+1$
Both equations of lines are drawn in the given graph.
So, option (ii) is correct.
$(iii)$ $y = \left\{\begin{matrix} x-1, & when \; x \geqslant 0 \; and \; (x-1) \geqslant 0 \; i.e. \; x \geqslant 1 \\ -(x-1) & when \; x \geqslant 0 \; and \; (x-1) < 0 \; i.e. \; x < 1 \\ x+1 & when \; x < 0 \; and \; (x+1) \geqslant 0 \;\;\; i.e. \; \; x \geqslant -1 \\ -(x+1) & when \; x < 0 \; and \; (x+1) < 0 \; \; i.e. \; x < -1 \end{matrix}\right.$
$\Rightarrow y = \left\{\begin{matrix} x-1, & \; x \geqslant 1 \\ -x+1, & 0 \leqslant x <1 \\ x+1, & -1 \leqslant x < 0 \\ -x-1, & x < -1 \end{matrix}\right.$
Suppose, $x= -1/2$ , $y= 1/2$ but it is not showing correctly in the graph.
So, option (iii) is incorrect.
(iv) is correct as we can see in the given graph. It is constant function $y=1$ between $x=2$ and $x=3$ (both are inclusive)
Hence, Answer is (B)