Question

Question 8
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.
And distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
Therefore, $\sqrt{(2-10{)}^2+(-3-y{)}^2}=10$
$\sqrt{(-8{)}^2+(3+y{)}^2}=10$
$64+(y+3{)}^2=100$
$(y+3{)}^2=36$
$y+3=\pm 6$
$y+3=+6ory+3=-6$
Therefore, $y=3or-9$