Question

The length of a line segment is of 10 units and the coordinates of one end-point are (2, -3). If the abscissa of the other end is 10, find the ordinate of the other end.

Answer 1

Let the ordinate of other point be y.
Then the two points are (2,-3) and (10,y)
$D=\sqrt{(10-2{)}^2+(y+3{)}^2}=10$
$\sqrt{8^2+(y+3{)}^2}=10$
$64+y^2+9+6y=100$
$y^2+6y-27=0$
$y^2+9y-3y-27=0$
$y(y+9)-3(y+9)=0$
$(y-3)(y+9)=0$
$y=3,-9$

So, the other point is $(10,3)$ or $(10,-9)$