Given that the coordinates of point A is $(-2,11)$ and point B is $(5,-11)$. Find

1. Substitute the coordinates into the distance formula.

For $A(-2,11)$ and $B(5,-11)$,
$AB=\sqrt{(5-(-2))^2+(-11-11)^2}$

2. Evaluate and simplify the length.

$AB=\sqrt{7^2+(-22)^2}$
$=\sqrt{49+484}$
$=\sqrt{533}$

1. Find the gradient of line $AB$.

Using $A(-2,11)$ and $B(5,-11)$,
$m=\frac{y_2-y_1}{x_2-x_1}=\frac{-11-11}{5-(-2)}=\frac{-22}{7}=-\frac{22}{7}.$

2. Substitute the gradient and one point into $y=mx+c$ to find $c$.

Using $y=mx+c$ and point $A(-2,11)$,
$11=-\frac{22}{7}(-2)+c$
$11=\frac{44}{7}+c$
$c=11-\frac{44}{7}=\frac{77}{7}-\frac{44}{7}=\frac{33}{7}.$

3. Write the equation of the straight line.

Therefore,
$y=-\frac{22}{7}x+\frac{33}{7}.$