Answer
$x= 1$ or $-5$
Work Step by Step
To find x, we will calculate distance between given points, i. e. (-2, 3) and (x, 1), using distance formula and equate it to $\sqrt 13$. Thus-
$ \sqrt { (x_{2} - x_{1}) ^{2} + (y_{2} - y_{1}) ^{2}}$ = $\sqrt 13$
$ \sqrt { (x-(-2)) ^{2} + (1- 3) ^{2}}$ = $\sqrt 13$
$ \sqrt { (x+2) ^{2} + (-2) ^{2}}$ = $\sqrt 13$
$ \sqrt { ( x^{2} + 4x + 4) + 4}$ = $\sqrt 13$
$ \sqrt { ( x^{2} + 4x + 8}$ = $\sqrt 13$
Squaring on both sides, we get-
$ x^{2}+ 4x + 8$ = $13$
OR
$ x^{2}+ 4x-5$ = 0 ( adding -13 on both sides)
$(x-1) (x+5)$ = 0 (On factorizing)
Equating each factor to zero, we get-
$x= 1$ or $-5$
Therefore x may be 1 or -5
