Let $(x_0,y_0)\in \mathbb{Z}^2$ be a point on the straight line $8x-3y=11$ which is equidistant from the coordinate axes. Then, the point $(x_0,y_0)$ will lie only in:

A point equidistant from the coordinate axes satisfies:

$|x|=|y|$

So, either

$y=x$

or

$y=-x$

Given line is:

$8x-3y=11$

Case 1:

$y=x$

Put $y=x$ in the line:

$8x-3x=11$

$5x=11$

$x=\frac{11}{5}$

This is not an integer, so this case is rejected.

Case 2:

$y=-x$

Put $y=-x$ in the line:

$8x-3(-x)=11$

$8x+3x=11$

$11x=11$

$x=1$

Then,

$y=-1$

So, the point is:

$(1,-1)$

Here $x>0$ and $y<0$, so the point lies in the IV quadrant.