A line segment is of length 10 units and one end is $\left(2,-3\right)$ . If the abscissa of the other end is 10, then its ordinate is

$A-B=10$

Length of$\overline{AB}=10$

One end point is $\left(2,-3\right)$

abscissa of other end of the point is $10$

Let ordinate be $x$

$\sqrt{{\left(10-2\right)}^2+{\left(x+3\right)}^2}=10$

$64+x^2+9+6x=100$

$x^2+6x=100-73=27$

$x^2+6x-27=0$

$x^2+9x-3x-27=0$

$x\left(x+9\right)-3\left(x+9\right)=0$

$\left(x-3\right)\left(x+9\right)=0$

$x=3\left(\text{or}\right)-9$