Question

Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units?

Answer 1

It is given that the distance between (2,-3) and (10,y) is 10.
Therefore, $\sqrt{(2-10)^2+(-3-y)^2}=10$
$\sqrt{(-8)^2+(3+y)^2}=10$
$64+(y+3)^2=100$
$(y+3)^2=36$
$y+3=\pm 6$
$y+3=6$ or $y+3=-6$
Therefore, $y=3$ or $-9$