Gradient of a line using 2 points Why is that the gradient of the line? Why is it 1/ (OP/OQ)?
The definition of gradient in this case is the same as the derivative: $\frac{\Delta y}{\Delta x}$. you don't even need to put the limit $\Delta x\to 0$, since we are talking about a line.
$$\frac{\Delta y}{\Delta x}=\frac{y_P-y_Q}{x_P-x_Q}=\frac{y_O-y_P}{x_P-x_O}=\frac{-|y_O-y_P|}{|x_P-x_O|}=-\frac{OQ}{OP}=-\cot\theta$$
The gradient of a line is the same as the slope of the line. The slope of a line is defined to be $$\frac{\text{change in }y}{\text{change in }x },$$

or if we want to we can write this as $$\frac{1}{\left(\frac{\text{change in }x }{\text{change in }y}\right)}.$$

Now, in this problem the change in $x$ is given by $OP$ and the change in $y$ is given by $-OQ$, because $OQ$ only gives the length of that line, but as $x$ increases $y$ decreases, so the change is negative.  Putting it together, then, we get that the slope is $$\frac{1}{\left(\frac{\text{change in }x }{\text{change in }y}\right)} = \frac{1}{\left(\frac{OP}{-OQ}\right)} = -\frac{1}{\left(\frac{OP}{OQ}\right)}.$$