The distance between (2, -3) and (10, y) is 10, then y = ?

(2, -3)   $\overline{<----10---->}$  (10, 4)

PQ = $\sqrt{{(10-2)}^2+{(y-3)}^2 }$= $\sqrt{{\left(8\right)}^2+{(y+3)}^2}$

Since PQ=10

$\sqrt{{\left(8\right)}^2+{(y+3)}^2}=10$

Simplify the above equation and find the value of y.

Squaring both sides

64+(y+3)²=100

(y+3)²=36

$y+3=\pm 6$

y+3=+6   or   y+3=-6

therefore, y= 3  or -9