Answer
$\frac{\triangle y}{\triangle x}$ is not constant over the interval $[x_1, x_2]$ unless $x_2 = x_1$.
Work Step by Step
Since $\frac{\triangle y}{\triangle x} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{(x_2)^2 -(x_1)^2}{x_2 - x_1} = \frac{(x_2 - x_1)(x_2 + x_1)}{(x_2 - x_1)}$, then
$\frac{\triangle y}{\triangle x} = (x_2 +x_1) = a \, value \, depending \, on \, both \, of \, \, x_1 \, and \, \, x_2$
(and $x_2 - x_1 \neq 0$)
Thus, generally $\frac{\triangle y}{\triangle x}$ is not constant over the interval $[x_1, x_2]$ unless $x_2 = x_1$.
In this case, the interval reduces to a singleton, and $\frac{\triangle y}{\triangle x}=0$.
(Note that you can not divide by $0$).